Jekyll2020-09-27T11:45:12+00:00https://fromatogra.com/feed.xmlfromatograSoftware and mathematicsMarc Flerackershttp://fromatogra.comIntroduction to statistics: Part 22019-06-23T03:27:01+00:002019-06-23T03:27:01+00:00https://fromatogra.com/math/2019/06/23/qualitative-statistics<h2 id="nonparametric-and-qualitative-data">Nonparametric and qualitative data</h2> <p>In a lot of cases, we don’t only have numbers, but categories as well. This can happen when we place measurable values into buckets, like [low, medium, high] or when we have distinct possibilities which are not necessarily ordered, like [red, green, blue, yellow]. These are qualitative parameters, while in part 1 we only worked with quantitative parameters.</p> <h3 id="spearmans-rank-correlation-coefficient">Spearman’s rank correlation coefficient</h3> <p>When the data we use is nonparametric, but still has a certain order, we can use Spearman’s rank correlation coefficient instead of the Pearson correlation coefficient.</p> <h4 id="calculating-the-rank">Calculating the rank</h4> <p>The Spearman’s rank correlation coefficient is obtained by calculating the rank of every value, and then using these values to calculate the Pearson correlation coefficient.</p> <p>There are several methods to calculate the rank of a list of values. The simplest method is to sort the values in ascending or descending order and to assign the values numbers from 1 until n, the amount of values. However if a value appears more than once, all these duplicates are assigned the rank of the first duplicate, while the numbering keeps going up silently, so when a new unique value is encountered, the numbering jumps.</p> <table> <thead> <tr> <th>Value</th> <th>Rank</th> </tr> </thead> <tbody> <tr> <td>2</td> <td>1</td> </tr> <tr> <td>3</td> <td>2</td> </tr> <tr> <td>4</td> <td>3</td> </tr> <tr> <td>4</td> <td>3</td> </tr> <tr> <td>4</td> <td>3</td> </tr> <tr> <td>5</td> <td>6</td> </tr> </tbody> </table> <p>Every value 4 receives rank 3. Rank 4 and 5 are skipped here, while value 5 receives the next rank, rank 6.</p> <p>Alternatively the duplicates can be assigned the average of their lowest and highest rank.</p> <table> <thead> <tr> <th>Value</th> <th>Rank</th> </tr> </thead> <tbody> <tr> <td>2</td> <td>1</td> </tr> <tr> <td>3</td> <td>2</td> </tr> <tr> <td>4</td> <td>4</td> </tr> <tr> <td>4</td> <td>4</td> </tr> <tr> <td>4</td> <td>4</td> </tr> <tr> <td>5</td> <td>6</td> </tr> </tbody> </table> <p>The values 4 would have received rank 3, 4 and 5. The average of 3 and 5 is 4, so all values receive rank 4.</p> <h4 id="calculating-the-correlation-coefficient">Calculating the correlation coefficient</h4> <p>Now that we can calculate the rank of each element in a list of values, we can calculate the Spearman’s rank correlation coefficient as follows</p> $\frac{cov(x',y')}{std(x')std(y')}$ <p>Where $$x'$$ and $$y'$$ are the list of values obtained by replacing each value in $$x$$ and $$y'$$ by its rank.</p> <h4 id="significance-of-the-correlation-coefficient">Significance of the correlation coefficient</h4> <p>Just having the coefficient isn’t enough to make a conclusion. To know how significant the result is, which depends on how much samples we have, we need to calculate the probability. To do this, we first calculate the tscore as</p> $t=\frac{r*\sqrt{n-2}}{\sqrt{1-r^2}}$ <p>Then we can calculate the probability using TDIST(t, n-2, 2) in Excel or pt(-abs(x), n-2) * 2 in R. The lower the probability, the more unusual our result is. Usually a cutoff value of 0.05 is used. If our probability is lower than 0.05, we say that there is a correlation between the parameters, as there is a very small chance that our samples come from a distribution where the parameters are unrelated.</p> <h3 id="cramers-v">Cramer’s V</h3> <p>This method can be used when both our data series are using categories rather than numbers.</p> <p>To calculate Cramer’s V, we first calculate the frequency of each pair. If series x has categories from 1 to k, and series y has categories from 1 to r, then $$n_{ij}$$ holds the count of all data pairs where x is i and y is j. $$n_i$$ is the amount of data pairs where x is i, and $$n_j$$ is the amount of data pairs where y is j. $$n$$ itself is the amount of samples.</p> <p>Then we calculate $$\chi^{2}$$ or Chi squared from these frequencies</p> $\chi^2=\sum_{i,j}{\frac{(n_{ij}-\frac{n_{i}n_{j}}{n})^2}{\frac{n_{i}n_{j}}{n}}}$ <p>And we can calculate V from $$\chi^{2}$$ as follows</p> $V=\sqrt{\frac{\chi^2/n}{\min(k-1,r-1)}}$ <p>Where 0 means that the parameters are independent, and 1 that the parameters are dependent. So the value of V tells us how much the two parameters are related.</p> <p>What exactly are we calculating though? Let’s take a step back, and use an example. Let’s say we have a list of records with people’s gender and hair color. We can create a table listing the frequency of each combination</p> <table> <thead> <tr> <th> </th> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>Male</td> <td>56</td> <td>143</td> <td>34</td> <td>46</td> </tr> <tr> <td>Female</td> <td>52</td> <td>143</td> <td>37</td> <td>81</td> </tr> </tbody> </table> <p>So 56 of the people in our set have black hair and are male.</p> <p>Our table now contains the $$n_{ij}$$ values, from which we can calculate $$n_i$$ (108, 286, 71, 127), $$n_j$$ (279, 313) and $$n$$ (592)</p> <table> <thead> <tr> <th> </th> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> <th>Sum</th> </tr> </thead> <tbody> <tr> <td>Male</td> <td>56</td> <td>143</td> <td>34</td> <td>46</td> <td>279</td> </tr> <tr> <td>Female</td> <td>52</td> <td>143</td> <td>37</td> <td>81</td> <td>313</td> </tr> <tr> <td>Sum</td> <td>108</td> <td>286</td> <td>71</td> <td>127</td> <td>592</td> </tr> </tbody> </table> <p>Now we create a table which contains $$\frac{n_{i}n_{j}}{n}$$ for each i and j</p> <table> <thead> <tr> <th> </th> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>Male</td> <td>50.89864865</td> <td>134.7871622</td> <td>33.46114865</td> <td>59.85304054</td> </tr> <tr> <td>Female</td> <td>57.10135135</td> <td>151.2128378</td> <td>37.53885135</td> <td>67.14695946</td> </tr> </tbody> </table> <p>This table contains what we call expected values. It is the table you would get if the parameters gender and hair color were completely unrelated. For example males with black hair. we know there are 108 people with black hair, and 279 people of 592 are male. This means that in a perfect case of unrelated parameters 108*279/592 or 50.89864865 would be males with black hair.</p> <p>If we replace $$\frac{n_{i}n_{j}}{n}$$ with $$E_{ij}$$ or the expected value at ij, we can rewrite the formula for $$\chi^2$$ as</p> $\chi^2=\sum_{i,j}{\frac{(n_{ij}-E_{ij})^2}{E_{ij}}}$ <p>So we are looking at the scaled squared distance between the observed and expected values, and summing these scaled distances.</p> <p>So the $$\chi^2$$ value tells us how much the two sample distributions are apart. In our case $$\chi^2$$ is 7.994244189.</p> <p>This is actually what CHITEST(observed, expected) uses in Excel. It calculates $$\chi^2$$ like we did and calculates the degrees of freedom df</p> $df=max(1, r-1)*max(1, c-1)$ <p>Then it calculates the probability. We can do this using CHIDIST($$\chi^2$$, df) in Excel or pchisq($$\chi^2$$, df, lower.tail=FALSE) in R. In our case p is 0.04613081084.</p> <p>If the probability is beneath a threshold, 0.05 for example, it means that the probability that the samples are from the same population is lower than 5%, so our observed values are probably not from a population where the categories are unrelated, thus there might be a relation between the two parameters. Since 0.04613081084 is slightly less than 0.05, we might say that there is more than 0 correlation between gender and hair color. How much though, is hard to tell.</p> <p>Cramer’s V however, gives an easier to interpret value between 0 and 1, without the need to choose a threshold. In our case V is 0.1162058125, which means there is a slight correlation between gender and hair color.</p> <p>In R, cramersV(x) gives V from the table x.</p> <h3 id="the-uncertainty-coefficient">The Uncertainty Coefficient</h3> <p>Cramer’s V gives us the same value irrespective of the order in we pass our parameters. So we can’t see the direction of a possible relation. However it might be that it is easier to guess whether a person is female when we know the hair color is blond than knowing the hair color when we know the person is female, V won’t provide this information.</p> <p>The uncertainty coefficient will tell us how certain it is that we can guess x given y.</p> <p>To calculate this, we need a few other concepts first. The uncertainty coefficient is based on information theory and the concept of entropy.</p> <p>The entropy of a list of data can be calculated as</p> $H(X)=-\sum_{i=1}^{n}{P(x_{i})\log_{e}P(x_{i})}$ <p>With $$P(x_{i})$$ the probability of $$x_{i}$$ occurring.</p> <p>The higher the entropy, the more uncertainty or surprise there is. This happens when some probabilities are small. This is because the smaller the probability, the larger the logarithm of the probability is. Since we use $$log_e$$, our probability of 0 to 1 maps onto -∞ to 0. So the - sign makes sure the entropy is positive.</p> <p>When we have events involving the probability that one parameter has a certain value and another parameter has a certain value, we can calculate the conditional entropy.</p> <p>The entropy of the data pairs with parameters X and Y can be written as</p> $H(X,Y)=-\sum_{i=1,j=1}^{n}{P(x_{i},y_{j})\log_{e}{P(x_{i},y_{j})}}$ <p>With $$P(x_{i},y_{j})$$ the probability of $$x_{i}$$ and $$y_{i}$$ occurring simultaneously.</p> <p>Now we can calculate conditional entropy. If we subtract the entropy of Y from H(X, Y), because we know Y, we get</p> $H(X|Y)=H(X,Y)-H(Y)$ <p>or</p> $H(X|Y)=-\sum_{i=1,j=1}^{n}{P(x_{i},y_{j})\log_{e}{P(x_{i},y_{j})}}+\sum_{i=1}^{n}{P(y_{i})\log_{e}{P(y_{i})}}$ <p>Which gives us</p> $H(X|Y)=\sum_{i,j}P(x_{i},y_{j})\log_{e}{\frac{P(y_{j})}{P(x_{i},y_{j})}}$ <p>This tells us how much uncertainty remains in X when we know Y.</p> <p>Given these definitions, we can now calculate the uncertainty coefficient as</p> $U(X|Y)=\frac{H(X)-H(X|Y)}{H(X)}$ <p>This gives us a number between 0 and 1 which is a measure of how good we can predict X from Y.</p> <p>To use this on our example, we first calculate the frequency lists like we did with Cramer’s V. For X we get</p> <table> <thead> <tr> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>108</td> <td>286</td> <td>71</td> <td>127</td> </tr> </tbody> </table> <p>And for Y</p> <table> <thead> <tr> <th>Male</th> <th>Female</th> </tr> </thead> <tbody> <tr> <td>279</td> <td>313</td> </tr> </tbody> </table> <p>We can calculate probabilities from these by dividing by the amount of samples, 592.</p> <table> <thead> <tr> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>108</td> <td>286</td> <td>71</td> <td>127</td> </tr> <tr> <td>0.1824324324</td> <td>0.4831081081</td> <td>0.1199324324</td> <td>0.214527027</td> </tr> </tbody> </table> <table> <thead> <tr> <th>Male</th> <th>Female</th> </tr> </thead> <tbody> <tr> <td>279</td> <td>313</td> </tr> <tr> <td>0.4712837838</td> <td>0.5287162162</td> </tr> </tbody> </table> <p>Now we can calculate the entropy H(X) = 1.2464359225967288, as well as H(Y) = 0.6914970305474704, which give the entropy of X and Y respectively.</p> <p>Next we calculate the conditional entropy of H(X|Y). For this we need our pair data</p> <table> <thead> <tr> <th> </th> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>Male</td> <td>56</td> <td>143</td> <td>34</td> <td>46</td> </tr> <tr> <td>Female</td> <td>52</td> <td>143</td> <td>37</td> <td>81</td> </tr> </tbody> </table> <p>We calculate probabilities for these by dividing by our sample count, 592.</p> <table> <thead> <tr> <th> </th> <th>Black</th> <th>Brown</th> <th>Red</th> <th>Blond</th> </tr> </thead> <tbody> <tr> <td>Male</td> <td>0.09459459459</td> <td>0.2415540541</td> <td>0.05743243243</td> <td>0.0777027027</td> </tr> <tr> <td>Female</td> <td>0.08783783784</td> <td>0.2415540541</td> <td>0.0625</td> <td>0.1368243243</td> </tr> </tbody> </table> <p>And calculate H(X|Y) which gives 1.239600755.</p> <p>Finally we calculate the uncertainty coefficient U(X|Y) which is 0.00548376956. This measures how good we can guess the hair color given that we know the gender. Not null, but not very high, so the correlation is there, but very low.</p> <p>We can also calculate U(Y|X) to see if the reverse is really different. U(Y|X) is 0.009884593959. This measures how good we can guess the gender given that we know the hair color. Also not very high, but double of U(X|Y), which means that it is slightly easier to guess the gender than the hair color.</p> <p>In R, given a table x, the uncertainty coefficient can be found using nom.uncertainty(x).</p> <p>There is also a symmetric form of the uncertainty coefficient</p> $U(X,Y)=2\frac{H(X)+H(Y)-H(X,Y)}{H(X)+H(Y)}$ <p>Which, like Cramer’s V is not dependent on the order of the parameters, so</p> $U(X,Y)=U(Y,X)$ <p>While this is not necessarily true for U(X|Y) and U(Y|X).</p> <p>In our case H(X,Y) is 0.003527040169, which again points to a very weak correlation.</p> <h3 id="correlation-ratio">Correlation Ratio</h3> <p>When we have one qualitative parameter and one quantitative parameter, we can use the correlation ratio to look for correlation.</p> <p>We start by sorting all the quantitative values into buckets, one for each possible value of the qualitative parameter. So if our qualitative parameter is gender, and our quantitative parameter is the salary, we make two lists, one for each gender, with all the salaries for that particular gender.</p> <p>Then we calculate the average for each list, which is the average salary by gender, $$\overline{y_i}$$. We also calculate the average of all salaries, $$\overline{y}$$.</p> <p>Now we can calculate the correlation ratio as</p> $\frac{\sum_i{n_i(\overline{y_i}-\overline{y})^2}}{\sum_i{(y-\overline{y})^2}}$ <p>We can see that the numerator gives us the deviation of each bucket’s average to the total average, scaled by the amount of elements in the bucket. While the denominator gives us the deviation of all values to the total average. What does this mean? If one or more buckets have a disproportionate amount of larger numbers, the average of those buckets will be greater, and thus the deviation to the total average will be greater. This happens when there is a correlation, as in the case of a correlation, the bucket a value is in will influence its magnitude.</p>Marc Flerackershttp://fromatogra.comA short introduction to statistical methods for qualitative data.Fake voxel rendering2018-08-14T02:10:01+00:002018-08-14T02:10:01+00:00https://fromatogra.com/math/2018/08/14/fake-voxels<h2 id="rendering-an-individual-object">Rendering an individual object</h2> <p>A fake voxel object is rendered as a stack of sprites. The bottom sprite is drawn at the position of the object, every next sprite is drawn a few pixels above the previous one. Most logically, the distance between sprites is the same as the size of a voxel. So if you scale up sprites by 4, you offset each next sprite by 4 pixels. Each sprite is drawn rotated, the angle is the sum of the camera angle and the angle of the object, the origin is the center of the sprite.</p> <div class="language-lua highlighter-rouge"><div class="highlight"><pre class="highlight"><code><span class="kd">local</span> <span class="n">offset</span> <span class="o">=</span> <span class="n">self</span><span class="p">.</span><span class="n">y</span> <span class="kd">local</span> <span class="n">angle</span> <span class="o">=</span> <span class="n">self</span><span class="p">.</span><span class="n">angle</span> <span class="o">+</span> <span class="n">camera</span><span class="p">.</span><span class="n">angle</span> <span class="k">for</span> <span class="n">_</span><span class="p">,</span> <span class="n">layer</span> <span class="k">in</span> <span class="nb">ipairs</span><span class="p">(</span><span class="n">self</span><span class="p">.</span><span class="n">layers</span><span class="p">)</span> <span class="k">do</span> <span class="n">love</span><span class="p">.</span><span class="n">graphics</span><span class="p">.</span><span class="n">draw</span><span class="p">(</span><span class="n">self</span><span class="p">.</span><span class="n">image</span><span class="p">,</span> <span class="n">layer</span><span class="p">,</span> <span class="n">self</span><span class="p">.</span><span class="n">x</span><span class="p">,</span> <span class="n">offset</span><span class="p">,</span> <span class="n">angle</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="n">self</span><span class="p">.</span><span class="n">ox</span><span class="p">,</span> <span class="n">self</span><span class="p">.</span><span class="n">oy</span><span class="p">)</span> <span class="n">offset</span> <span class="o">=</span> <span class="n">offset</span><span class="o">-</span><span class="mi">4</span> <span class="k">end</span> </code></pre></div></div> <h2 id="transforming-objects--translating-the-camera">Transforming objects : translating the camera</h2> <p>If we want to roam the world by translating the camera, we need to transform each object (technically we could put the translation on the transformation stack and let the vertex shader do it, but since we’ll be doing more than just translating later on, we do it on the CPU). Our camera has a position $$(cx, cy)$$, and we want to make sure the center of our world ends up in the middle of the screen $$(sx, sy)$$. So our transformation is</p> $\begin{pmatrix} 1 &amp; 0 &amp; sx - cx \\ 0 &amp; 1 &amp; sy - cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>This translates the point in the world at $$(cx, cy)$$ to the middle of the screen $$(sx, sy)$$. Applying this matrix to the position of the object is nothing but a vector addition, since we are not scaling or rotating yet.</p> <h2 id="transforming-objects--rotating-the-camera">Transforming objects : rotating the camera</h2> <p>First note that when we talk about the camera angle $$\alpha$$ here, it is actually the reverse angle, because just like the translation, when the camera rotates clockwise, the world which we see actually rotates counterclockwise.</p> <p>When we rotate the camera, we want it to rotate not around $$(0, 0)$$, but the place where the camera is. So we actually need to translate, rotate and translate back (like you do when rotating a sprite around a different origin than its left top). So our rotation around $$(cx, cy)$$ is</p> $\begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; cx - cos(\alpha) * cx + sin(\alpha) * cy \\ sin(\alpha) &amp; cos(\alpha) &amp; cy - sin(\alpha) * cx - cos(\alpha) * cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>When we combine this with our camera translation from before, we multiply these matrices (which ends up to be just an addition of the translation vectors since our translation is independent of the rotation) and get</p> $\begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; sx - cx + cx - cos(\alpha) * cx + sin(\alpha) * cy \\ sin(\alpha) &amp; cos(\alpha) &amp; sy - cy + cy - sin(\alpha) * cx - cos(\alpha) * cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>Or after simplifying</p> $\begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; sx - cos(\alpha) * cx + sin(\alpha) * cy \\ sin(\alpha) &amp; cos(\alpha) &amp; sy - sin(\alpha) * cx - cos(\alpha) * cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>The two left rows will rotate our position, and the right row will offset it by the screen and the camera position. Since our camera is expressed in coordinates of the world, which is rotated now, the camera vector has been rotated as well. If you don’t see it yet, we can rewrite</p> $sx - cos(\alpha) * cx + sin(\alpha) * cy\\sy - sin(\alpha) * cx - cos(\alpha) * cy$ <p>as</p> $sx - (cos(\alpha) * cx - sin(\alpha) * cy)\\sy - (sin(\alpha) * cx + cos(\alpha) * cy)$ <p>which shows it as being an offset composed of the screen offset subtracted by the camera offset which is first rotated by the camera angle.</p> <p>In this final transformation, we can notice the following</p> <ol> <li>It contains cosine and sine, we don’t want to calculate these often, so we better cache them.</li> <li>The last row is a vector which is only dependent on the camera rotation and position, as long as the camera doesn’t change don’t change, we can cache this too.</li> </ol> <p>Given this, we can write our transform function, which transforms a world position to a screen position (in other words, which tells us where to draw an object) as follows</p> <div class="language-lua highlighter-rouge"><div class="highlight"><pre class="highlight"><code><span class="k">function</span> <span class="nf">Camera</span><span class="p">:</span><span class="n">transform</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span><span class="p">)</span> <span class="k">return</span> <span class="n">x</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">cos</span> <span class="o">-</span> <span class="n">y</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">sin</span> <span class="o">+</span> <span class="n">self</span><span class="p">.</span><span class="n">offsetX</span><span class="p">,</span> <span class="n">x</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">sin</span> <span class="o">+</span> <span class="n">y</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">cos</span> <span class="o">+</span> <span class="n">self</span><span class="p">.</span><span class="n">offsetY</span> <span class="k">end</span> </code></pre></div></div> <p>Where cos, sin and the offset vector are only calculated when the viewpoint changes</p> <div class="language-lua highlighter-rouge"><div class="highlight"><pre class="highlight"><code><span class="k">function</span> <span class="nf">Camera</span><span class="p">:</span><span class="n">recalc</span><span class="p">(</span><span class="n">angleChanged</span><span class="p">)</span> <span class="k">if</span> <span class="n">angleChanged</span> <span class="k">then</span> <span class="n">self</span><span class="p">.</span><span class="n">cos</span> <span class="o">=</span> <span class="nb">math.cos</span><span class="p">(</span><span class="n">self</span><span class="p">.</span><span class="n">angle</span><span class="p">)</span> <span class="n">self</span><span class="p">.</span><span class="n">sin</span> <span class="o">=</span> <span class="nb">math.sin</span><span class="p">(</span><span class="n">self</span><span class="p">.</span><span class="n">angle</span><span class="p">)</span> <span class="k">end</span> <span class="n">self</span><span class="p">.</span><span class="n">offsetX</span> <span class="o">=</span> <span class="n">self</span><span class="p">.</span><span class="n">screenX</span> <span class="o">-</span> <span class="n">self</span><span class="p">.</span><span class="n">cos</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">x</span> <span class="o">+</span> <span class="n">self</span><span class="p">.</span><span class="n">sin</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">y</span> <span class="n">self</span><span class="p">.</span><span class="n">offsetY</span> <span class="o">=</span> <span class="n">self</span><span class="p">.</span><span class="n">screenY</span> <span class="o">-</span> <span class="n">self</span><span class="p">.</span><span class="n">sin</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">x</span> <span class="o">-</span> <span class="n">self</span><span class="p">.</span><span class="n">cos</span> <span class="o">*</span> <span class="n">self</span><span class="p">.</span><span class="n">y</span> <span class="k">end</span> </code></pre></div></div> <h2 id="moving-around">Moving around</h2> <p>Rotating the camera is easy, we just need to change the angle. To pan the camera however, is a bit more involved as we want it to always pan left-right and up-down according to the screen, not how the camera is rotated, as it will confuse the player.</p> <p>This means rather than adding $$(0, -speed * dt)$$ to the camera position when pressing up (remember, when the camera goes up, the world goes down), we need to counter rotate the vector $$(speed * dt, 0)$$ first, so it is not influenced by the camera angle.</p> <p>So we need the following transformation matrix</p> $\begin{pmatrix} cos(-\alpha) &amp; -sin(-\alpha) &amp; 0 \\ sin(-\alpha) &amp; cos(-\alpha) &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>Unfortunately, we need the cosine and sine of $$-\alpha$$, while we only have the cosine and sine of $$\alpha$$. Lucky for us there’s the following identity</p> $cos(-\alpha)=cos(\alpha)\\ sin(-\alpha)=-sin(\alpha)$ <p>so we can use</p> $\begin{pmatrix} cos(\alpha) &amp; sin(\alpha) &amp; 0 \\ -sin(\alpha) &amp; cos(\alpha) &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>for which we have everything cached already. Transforming our camera displacement $$(dx, dy)$$ by this matrix gives us</p> $cx' = cx + dx*cos(\alpha) + dy*sin(\alpha)\\ cy' = cy - dx*sin(\alpha) + dy*cos(\alpha)$ <p>For the up key, which uses $$(0, -speed * dt)$$ as $$(dx, dy)$$ that gives</p> $cx' = cx - speed * dt *dy*sin(\alpha)\\ cy' = cy - speed * dt *dy*cos(\alpha)$ <p>For the other direction, down $$(0, speed * dt)$$, or strafing left $$(-speed * dt, 0)$$ and right $$(speed * dt, 0)$$ we can do the same.</p> <h2 id="intermezzo--dropping-the-cosine-and-sine-calculations">Intermezzo : dropping the cosine and sine calculations</h2> <p>While they are by no means taxing calculations since they are only done while rotating the camera, we can in fact remove them. With one caveat however, we won’t be able to rotate arbitrary angles.</p> <p>Why is it important to be able to rotate by an arbitrary angle? Mainly because in most cases, our time between updates will not be exactly fixed, like 1/60th of a second. The frame rate might be a lot higher on a PC, and lower on mobile. This means that if you would turn each update at a fixed angle, the rotation speed will be linked to the speed of the device. Since we don’t want that, we usually multiply the angular speed by the time it took since the last update.</p> <p>But let’s say we are OK with fixed angles. We might for example accumulate time until we have enough to rotate a bit, or might do multiple fixed rotations to cover the time on a slow device.</p> <p>We have the current angle $$\alpha$$ and want to add the fixed angle $$\beta$$. This means we need to calculate $$cos(\alpha+\beta)$$ and $$sin(\alpha+\beta)$$, but we don’t want to actually call the cosine and sine functions. What we do have is $$cos(\alpha)$$ and $$sin(\alpha)$$ since we were using it for the last update, and another handy identity</p> $cos(\alpha+\beta)=cos(\alpha) * cos(\beta) − sin(\alpha) * sin(\beta) \\ sin(\alpha+\beta)=sin(\alpha) * cos(\beta) + cos(\alpha) * sin(\beta)$ <p>So if we just calculate $$cos(\beta)$$ and $$sin(\beta)$$ at load time, we can use it to rotate the current $$cos(\alpha)$$ and $$sin(\alpha)$$ in any direction (remember the $$-\alpha$$ identity from above) by $$\beta$$ without using cosine or sine during run-time.</p> <h2 id="z-ordering">Z-ordering</h2> <p>Now that we can move and rotate our camera around, we notice that objects are not always drawn in the right order. Sometimes a further object obscures a closer one. This is because we draw our objects in the order they were placed in the list. We actually need to draw them from back to front. So we will need to sort our list before drawing.</p> <p>For this we need to split the transform and draw stages. First we transform all objects and update their screen coordinates. Then we sort the list and finally we draw them in the order of the newly sorted list.</p> <p>This example uses lua’s built-in table.sort which uses quicksort internally. While quicksort is not a bad algorithm, it has the disadvantage that it has a worst case of $$O(n)$$ and that it is not a stable sort. Not being stable means that two objects at the exact same y may switch places after a second sort, which may cause flicker.</p> <p>Another thing to note is that, while we rotate, not all objects should be re-sorted. Depending on the distance between objects, their order stays the same for many angles. A better sort algorithm would be for example <a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">timsort</a> which is a stable sort algorithm based on mergesort with a worst case of $$O(n\log{}n)$$. It can cope much better with partially sorted lists.</p> <p>LuaJit has been planning to move to <a href="https://en.wikipedia.org/wiki/Smoothsort">smoothsort</a>, a sort algorithm with a worst case of $$O(n\log{}n)$$ that is based on heapsort. This makes it good for partially sorted data, but like heapsort it is not a stable sort.</p> <h2 id="picking--inverse-transform">Picking : inverse transform</h2> <p>If we want to build an editor (of course we want to) for our stages, we need to be able to place objects. We can do this easily when the camera is not rotated, we just need to subtract the center of the screen, since this is where the camera is drawn and add the camera position, since the camera is placed at $$(0, 0)$$ after subtracting the screen center. What we did just now is building a transformation from screen to world coordinates.</p> $\begin{pmatrix} 1 &amp; 0 &amp; - sx + cx \\ 0 &amp; 1 &amp; - sy + cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>This was easy, but once we add a rotation, things get more complex. Let’s look at the general solution for an inverse matrix first.</p> <h3 id="general-inverse">General inverse</h3> <p>For any general $$3\times3$$ matrix</p> $\begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; i \end{pmatrix}$ <p>for which the determinant is not 0, we can calculate the inverse by calculating</p> $\frac{1}{det}*\begin{pmatrix} (ei-fg) &amp; -(bi-ch) &amp; (bf-ce) \\ -(di-fg) &amp; (ai-cg) &amp; -(af-cd) \\ (dh-eg) &amp; -(ah-bg) &amp; (ae-bd) \end{pmatrix}$ <p>The determinant of a matrix can be calculated by adding the product of the left-top to right-bottom diagonals and subtracting the products of the left-bottom to top-right diagonals. Another way is taking the sum of the top row elements multiplied by the determinants of the $$2\times2$$ matrices obtained by removing the row and columns from which said elements are part. Either way gives us as determinant</p> $det=a(ei−fh)−b(di−fg)+c(dh−eg)$ <p>That’s quite a lot of calculations for getting an inverse. Luckily our transformation matrices are more specific, as the last row is always</p> $\begin{vmatrix} 0 &amp; 0 &amp; 1 \end{vmatrix}$ <p>If we take this into consideration, by replacing g, h, i with 0, 0, 1 respectively we get the following much simplified solution</p> $\begin{pmatrix} a &amp; b &amp; c \\ d &amp; e &amp; f \\ 0 &amp; 0 &amp; 1 \end{pmatrix}^{-1}=\frac{1}{det}*\begin{pmatrix} e &amp; -b &amp; bf−ce \\ -d &amp; a &amp; cd−af \\ 0 &amp; 0 &amp; ae−bd \end{pmatrix}$ <p>The determinant can be simplified as well</p> $a∗e−b∗d$ <p>In the case of our world to screen and screen to world transformation the determinant will 1, as we don’t scale or shear. The determinant is an indicator of how the transformation influences the area between 3 or more points. Just translating and rotating doesn’t change the area, so the determinant is 1.</p> <p>We can now fill in the values of our world to screen matrix to calculate the inverse.</p> <p>However instead of doing this right now, we’re going to approach the problem with some logic instead and develop the inverse transformation ourselves. While solutions like we just saw are handy for general cases , they don’t help us to develop a deeper understanding of what is going on.</p> <h3 id="getting-to-the-inverse-step-by-step">Getting to the inverse step by step</h3> <p>Our world to screen transform is as follows</p> $\begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; sx - cos(\alpha) * cx + sin(\alpha) * cy \\ sin(\alpha) &amp; cos(\alpha) &amp; sy - sin(\alpha) * cx - cos(\alpha) * cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>It takes a point in world coordinates and subtracts the camera position from it, thus bringing the camera to $$(0, 0)$$. Then it rotates everything around $$(0, 0)$$ and finally the screen center or camera screen position $$(sx, sy)$$ is added to bring the camera to the middle of the screen. This sequence can be written as the multiplication of three matrices</p> $\begin{pmatrix} 1 &amp; 0 &amp; sx \\ 0 &amp; 1 &amp; sy \\ 0 &amp; 0 &amp; 1 \end{pmatrix} * \begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; 0 \\ sin(\alpha) &amp; cos(\alpha) &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix} * \begin{pmatrix} 1 &amp; 0 &amp; -cx \\ 0 &amp; 1 &amp; -cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>To go from screen to world, we need to do everything in reverse. Both the order, and every individual transformation.</p> $\begin{pmatrix} 1 &amp; 0 &amp; cx \\ 0 &amp; 1 &amp; cy \\ 0 &amp; 0 &amp; 1 \end{pmatrix} * \begin{pmatrix} cos(\alpha) &amp; sin(\alpha) &amp; 0 \\ -sin(\alpha) &amp; cos(\alpha) &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix} * \begin{pmatrix} 1 &amp; 0 &amp; -sx \\ 0 &amp; 1 &amp; -sy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>The inverse of a translation is translating in the opposite direction, so we just flip signs. The inverse of a rotation is rotating in the opposite direction. This means we just need to flip the signs of the sines. If we combine these matrices again, we get the following inverse transformation matrix.</p> $\begin{pmatrix} cos(\alpha) &amp; -sin(\alpha) &amp; cx - cos(\alpha) * sx - sin(\alpha) * sy \\ sin(\alpha) &amp; cos(\alpha) &amp; cy + sin(\alpha) * sx - cos(\alpha) * sy \\ 0 &amp; 0 &amp; 1 \end{pmatrix}$ <p>This is the same result we get when filling in our world to screen matrix elements into the formula to calculate the inverse. The screen position of the camera gets a rotation, and the rotation of the world position of the camera gets cancelled out because</p> $cos^2(\alpha) + sin^2(\alpha) = 1$ <h2 id="clipping--only-drawing-what-we-see">Clipping : only drawing what we see</h2> <p>Until now we were drawing all objects, whether they were within the screen area or not. If our world or stage is a lot bigger than the screen, we should skip drawing objects which are not visible in order to reduce or CPU usage.</p> <p>There are several ways to approach this. The smallest change to our code would be to check whether an object is within the screen rectangle, after calculating screen coordinates and before sorting.</p> <p>But this means we would still be wasting calculations on transforming all objects to the screen.</p> <p>Since we have the inverse transform as well, we can instead transform the four screen corners of the screen to the world. From those we can construct a new axis aligned (in the world’s coordinate system) rectangle with which we can check whether objects are within the screen by using just their world coordinates.</p> <p>This is better, but if we have a large world, we’re still doing a lot of testing. We can avoid this by subdividing our world in smaller sections. Instead of one big list, we give each section a list of objects which are situated in that section. This allows us to only test whether a section is within the screen, and discard many objects at once if it isn’t. Care should be taken that we do not make our world less organic because of the limits we impose on it. If we use a grid to cut our world in sections for example, an object might be crossing a border. This means that if that object is not allowed to be in both sections, it will disappear when the section it is in is not visible. Placing it in both grid squares is an option, but then care should be taken not to place it twice in the drawing list, which might incur additional processing (using a set instead of a list).</p> <h2 id="drawing-sprites-in-batches">Drawing sprites in batches</h2> <p>Finally, drawing each sprite separately when using a GPU isn’t very good for performance. Instead we should place all sprites into a batch and draw them all at once.</p> <p>You can find three renderers in the example code. One which draws directly, and two which batch drawing calls. One batch renderer only supports a single texture, the other one supports as many textures as needed. Note that both are primitive though. Usually you start a new batch when your draw call needs to be flushed. This is when a textures, blending mode, shader or other non uniform variable is changed.</p> <p>For the code, see the <a href="https://github.com/mflerackers/fakevoxels">github</a> repository.</p>Marc Flerackershttp://fromatogra.comA short explanation on how to render fake voxels for 2D games.Introduction to statistics: Part 12018-05-11T13:03:01+00:002018-05-11T13:03:01+00:00https://fromatogra.com/math/2018/05/11/statistics<h2 id="our-data-set">Our data set</h2> <p>For doing statistical calculations we need data. We will create some fake data points here to use as an example. We’re growing plants and giving them a certain amount of sunlight. We measure the growth after a certain interval and place all this data into a table.</p> <table> <thead> <tr> <th>Light (hours)</th> <th>Growth (mm)</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> </tr> <tr> <td>1</td> <td>1</td> </tr> <tr> <td>2</td> <td>5</td> </tr> <tr> <td>3</td> <td>9</td> </tr> <tr> <td>4</td> <td>12</td> </tr> <tr> <td>5</td> <td>15</td> </tr> </tbody> </table> <h2 id="mean">Mean</h2> <p>The mean or average is calculated by dividing the sum of all samples by the amount of samples.</p> $\overline{x}=\frac{\sum\limits_{i=1}^{n} x_i}{n}$ <p>This is the population mean</p> <p>In case we only have samples, rather than a complete population, we need to divide by $$n-1$$ instead of $$n$$.</p> $\overline{x}=\frac{\sum\limits_{i=1}^{n} x_i}{n-1}$ <p>This is called the sample mean.</p> <p>For our plant growth, the mean of the plant growth is (0+1+5+9+12+15)/6 or 42/6 which is 7.</p> <h2 id="median">Median</h2> <p>The median is calculated by sorting the samples and taking the middle sample. In case the amount of samples n is even, the median of the two middle samples is taken.</p> <p>In our case, the median of the plant growth is (5+9)/2 or 7, which is the mean growth of 2 and 3 hours of light.</p> <h2 id="variance">Variance</h2> <p>The variance is a measure of how much variance or difference there is among our samples. A low variance means most samples are equal, a high variance means there are many samples lower or higher than the median.</p> <p>The variance is calculated by summing the square of the difference between each sample and the mean, and dividing it by $$n-1$$.</p> $\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}{n-1}$ <p>Thus it is the sample mean of the squared differences of each sample $$x$$ with the mean $$\overline{x}$$.</p> <p>Conceptually, the variance is the average squared distance of a sample to the mean.</p> <p>To calculate the variance of the plant growth we take all differences -7, -6, -2, 2, 5, 8 and square them to obtain 49, 36, 4, 4, 25, 64. We add all of these together and divide them by 5 to get (49+36+4+4+25+64)/5 or 36.4.</p> <h2 id="standard-deviation">Standard deviation</h2> <p>If we take the square root of the variance, we get the standard deviation. Since we take the square root of the squared average distance to the mean, this is the average distance of a sample to the mean.</p> $\sqrt{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}{n-1}}$ <p>The square root of the variance of our plant growth is $$\sqrt{36.4}$$ or 6.03, about 6. Since our values of growth go from 0 to 15, a standard deviation of 6 means there is quite some difference between our plant growth samples. It means that the average distance from the mean growth of 7mm is 6mm.</p> <h2 id="covariance">Covariance</h2> <p>Until now, we worked with one measurement, $$x$$. If we want to compare an input parameter with our measurement, we use the covariance. The covariance is calculated by taking the sum of the products of the differences of each sample with its mean, divided by n-1. We basically swapped one $$(x_i-\overline{x})$$ with $$(y_i-\overline{y})$$.</p> $\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})}{n-1}$ <p>The covariance when calculated for the amount of light and plant growth is 11.2. To calculate this, we calculated the mean of our hours of light, which is 2.5. Then we took the sum of all products of the differences between light and light mean and growth and growth mean, finally dividing by 5.</p> <h2 id="linear-regression-and-the-regression-coefficient">Linear regression and the regression coefficient</h2> <p>When we have two sets of samples, we can calculate a line which follows the trend of the samples. This line can be used to extrapolate or guess values for which we don’t have samples or just to view the trend.</p> <p>A line can be defined as an equation $$y=ax+b$$, where a is the slope and b is the position where the line crosses the x axis, thus where y is 0.</p> <p>One way to calculate the slope a is by the dividing the covariance of x and y by the variance of x.</p> $\frac{cov(x, y)}{var(x)}$ <p>This is called the regression coefficient.</p> <p>If we don’t have the covariance and variance yet we can use the following formula instead</p> $a=\frac{n\sum\limits_{i=1}^{n}{x_iy_i}-\sum\limits_{i=1}^{n}{y_i}\sum\limits_{i=1}^{n}{x_i}}{n\sum\limits_{i=1}^{n}{x_i^2}-\sum\limits_{i=1}^{n}{_ix}\sum\limits_{i=1}^{n}{x_i}}$ <p>Both formulas are mathematically identical, though limits with floating point values can give slightly different results.</p> <p>Once we have a, we can calculate b since</p> $\frac{\sum\limits_{i=1}^{n}{y_i}}{n}=a\frac{\sum\limits_{i=1}^{n}{x_i}}{n}+b$ <p>We can subtract $$a\frac{\sum\limits_{i=1}^{n}{x_i}}{n}$$ from both sides to get</p> $b=\frac{\sum\limits_{i=1}^{n}{y_i}}{n}-a\frac{\sum\limits_{i=1}^{n}{x_i}}{n}$ <p>or if we group the fractions</p> $b=\frac{\sum\limits_{i=1}^{n}{y_i}-a\sum\limits_{i=1}^{n}{x_i}}{n}$ <h2 id="the-pearson-correlation-coefficient-r">The Pearson correlation coefficient R</h2> <p>The correlation coefficient is to covariance as the standard deviation is to variance. It is a better measure to interpret results from. The correlation coefficient is calculated by dividing the covariance by the product of the standard deviations of each measurement.</p> $\frac{cov(x,y)}{std(x)std(y)}=\frac{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})}{n-1}}{\sqrt{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}{n-1}}\sqrt{\frac{\sum\limits_{i=1}^{n}(y_i-\overline{y})^2}{n-1}}}$ <p>However if we place the square root outside, we see that correlation coefficient is equal to the square root of the product of the regression coefficients for both samples</p> $\sqrt{\frac{(\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})}{n-1})^2}{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}{n-1}\frac{\sum\limits_{i=1}^{n}(y_i-\overline{y})^2}{n-1}}}=\sqrt{\frac{cov(x,y)^2}{var(x)var(y)}}$ <p>If the correlation coefficient is close to 0, there is no correlation between both sets of samples. If it is close to 1, we have a direct correlation, if the value of the x sample goes up, the value of the y sample does too. If it is close to -1, we have an inverse correlation, if the sample of x goes up, the sample of y goes down.</p> <p>To calculate the correlation coefficient between the amount of light and how much our plants grew, we still need to calculate the standard deviation of our hours of light. The variance is 3.5 and the square root of that is 1.87. Now we can calculate the correlation coefficient as 11.2/(1.87*6.03) which is 0.99.</p> <p>Since our correlation coefficient is nearly 1, there is a high correlation between hours of light and plant growth.</p> <h2 id="covariance-matrix">Covariance matrix</h2> <p>The covariance matrix is a matrix which contains all covariances between each two parameters. If we would build a covariance matrix between just two parameters x and y, like our light and plant growth parameters, we would get the matrix</p> $\begin{bmatrix}cov(x, x) &amp; cov(x, y) \\ cov(y, x) &amp; cov(y,y)\end{bmatrix}$ <p>Of course the covariance between two identical parameters is the variance, so we can write</p> $\begin{bmatrix}var(x) &amp; cov(x, y) \\ cov(y, x) &amp; var(y)\end{bmatrix}$ <p>And since the covariance of x and y is equal to the covariance of y and x, the matrix is symmetric .</p> <p>The more parameters we have, the larger the matrix will be as it needs to cover more combinations</p> $\begin{bmatrix}var(x) &amp; cov(x, y) &amp; cov(x, z) &amp; ..\\ cov(y, x) &amp; var(y) &amp; cov(y,z) &amp; ..\\cov(z,x) &amp; cov(z,y) &amp; var(z) &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>We will shortly see why we are interested in this matrix, but before that we will recreate it using matrix math. While it is completely OK to create it using the methods we used before, and even more logical as we only need to calculate one of two symmetric triangles, matrix operations might be faster depending on the hardware used.</p> <p>The steps are as follows. We start with an nxm matrix in which each each row defines a sample from 1 to n and each column defines a parameter from 1 to m</p> $\begin{bmatrix}x_1 &amp; y_1 &amp; z_1 &amp; ..\\ x_2 &amp; y_2 &amp; z_2 &amp; ..\\x_3 &amp; y_3 &amp; z_3 &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>We multiply this matrix with a nxn matrix of ones</p> $\begin{bmatrix}1 &amp; 1 &amp; 1 &amp; ..\\ 1 &amp; 1 &amp; 1 &amp; ..\\1 &amp; 1 &amp; 1 &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix} * \begin{bmatrix}x_1 &amp; y_1 &amp; z_1 &amp; ..\\ x_2 &amp; y_2 &amp; z_2 &amp; ..\\x_3 &amp; y_3 &amp; z_3 &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}=\begin{bmatrix}\sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\ \sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\\sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>The result is that each column, from 1 to m will contain n rows, which all contain the sum of all sample values for that parameter. Then we divide by the amount of samples to get</p> $\begin{bmatrix}\sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\ \sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\\sum\limits_{i=1}^{n}{x_i} &amp; \sum\limits_{i=1}^{n}{y_i} &amp; \sum\limits_{i=1}^{n}{z_i} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}*\frac{1}{n}=\begin{bmatrix}\frac{\sum\limits_{i=1}^{n}{x_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{y_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{z_i}}{n} &amp; ..\\ \frac{\sum\limits_{i=1}^{n}{x_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{y_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{z_i}}{n} &amp; ..\\ \frac{\sum\limits_{i=1}^{n}{x_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{y_i}}{n} &amp; \frac{\sum\limits_{i=1}^{n}{z_i}}{n} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}=\begin{bmatrix}\overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ \overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ \overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>This is a matrix which contains in every column for every row the mean of the samples. We subtract this from our original sample matrix to get the sample value minus the mean.</p> $\begin{bmatrix}x_1 &amp; y_1 &amp; z_1 &amp; ..\\ x_2 &amp; y_2 &amp; z_2 &amp; ..\\x_3 &amp; y_3 &amp; z_3 &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}-\begin{bmatrix}\overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ \overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ \overline{x} &amp; \overline{y} &amp; \overline{z} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}=\begin{bmatrix}x_1-\overline{x} &amp; y_1-\overline{y} &amp; z_1-\overline{z} &amp; ..\\ x_2-\overline{x} &amp; y_2-\overline{y} &amp; z_2-\overline{z} &amp; ..\\x_3-\overline{x} &amp; y_3-\overline{y} &amp; z_3-\overline{z} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>To get the sum of squares, we multiply the transpose of this matrix with itself</p> $\begin{bmatrix}x_1-\overline{x} &amp; x_2-\overline{x} &amp; x_3-\overline{x} &amp; ..\\ y_1-\overline{y} &amp; y_2-\overline{y} &amp; y_3-\overline{y} &amp; ..\\z_1-\overline{z} &amp; z_2-\overline{z} &amp; z_3-\overline{z} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}*\begin{bmatrix}x_1-\overline{x} &amp; y_1-\overline{y} &amp; z_1-\overline{z} &amp; ..\\ x_2-\overline{x} &amp; y_2-\overline{y} &amp; z_2-\overline{z} &amp; ..\\x_3-\overline{x} &amp; y_3-\overline{y} &amp; z_3-\overline{z} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>This gives us</p> $\begin{bmatrix}\sum\limits_{i=1}^{n}{(x_i-\overline{x})^2} &amp; \sum\limits_{i=1}^{n}{(x_i-\overline{x})(y_i-\overline{y})} &amp; \sum\limits_{i=1}^{n}{(x_i-\overline{x})(z_i-\overline{z})} &amp; ..\\ \sum\limits_{i=1}^{n}{(y_i-\overline{y})(x_i-\overline{x})} &amp; \sum\limits_{i=1}^{n}{(y_i-\overline{y})^2} &amp; \sum\limits_{i=1}^{n}{(y_i-\overline{y})(z_i-\overline{z})} &amp; ..\\ \sum\limits_{i=1}^{n}{(z_i-\overline{z})(x_i-\overline{x})} &amp; \sum\limits_{i=1}^{n}{(z_i-\overline{z})(y_i-\overline{y})} &amp; \sum\limits_{i=1}^{n}{(z_i-\overline{z})^2} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>Which we can divide by the amount of samples minus 1 to get</p> $\begin{bmatrix}\frac{\sum\limits_{i=1}^{n}{(x_i-\overline{x})^2}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(x_i-\overline{x})(y_i-\overline{y})}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(x_i-\overline{x})(z_i-\overline{z})}}{n-1} &amp; ..\\ \frac{\sum\limits_{i=1}^{n}{(y_i-\overline{y})(x_i-\overline{x})}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(y_i-\overline{y})^2}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(y_i-\overline{y})(z_i-\overline{z})}}{n-1} &amp; ..\\ \frac{\sum\limits_{i=1}^{n}{(z_i-\overline{z})(x_i-\overline{x})}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(z_i-\overline{z})(y_i-\overline{y})}}{n-1} &amp; \frac{\sum\limits_{i=1}^{n}{(z_i-\overline{z})^2}}{n-1} &amp; ..\\ .. &amp; .. &amp; .. &amp; ..\end{bmatrix}$ <p>Which is the covariance matrix.</p> <p>On the diagonal we can see how much every parameter varies among samples, and the off-diagonal elements show how much relationship there is between parameters.</p> <h2 id="principal-component-analysis-or-pca">Principal Component Analysis or PCA</h2> <p>When we have many parameters per sample, we can find which parameters or parameter groups cause the most change by using principal component analysis.</p> <p>To do a principal component analysis we take the covariance matrix. For this matrix we search for the eigenvectors and the corresponding eigenvalues. The eigenvectors with the largest eigenvalues indicate which parameters work together to influence the most change.</p> <p>Eigenvectors are vectors which don’t change direction when transformed by the matrix. Eigenvalues are the scale factors by which the eigenvectors are scaled when transformed by the matrix.</p> <p>[More content coming soon]</p> <h2 id="deriving-the-regression-formulas">Deriving the regression formulas</h2> <p>Let’s show that both formulas for linear regression are identical. We start with the formula for covariance of x and y over the variance of x</p> $\frac{cov(x, y)}{var(x)}$ <p>or</p> $a=\frac{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})}{n-1}}{\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}{n-1}}$ <p>Since the $$n-1$$ cancel eachother out we get</p> $a=\frac{\sum\limits_{i=1}^{n} (x_i-\overline{x})(y_i-\overline{y})}{\sum\limits_{i=1}^{n} (x_i-\overline{x})^2}$ <p>Then we write out the product</p> $a=\frac{\sum\limits_{i=1}^{n} (x_iy_i-x_i\overline{y}-y_i\overline{x}+\overline{x}\overline{y})}{\sum\limits_{i=1}^{n} (x_ix_i-x_i\overline{x}-x_i\overline{x}+\overline{x}\overline{x})}$ <p>Now we can split the terms</p> $a=\frac{\sum\limits_{i=1}^{n}x_iy_i-\sum\limits_{i=1}^{n}x_i\overline{y}-\sum\limits_{i=1}^{n}y_i\overline{x}+\sum\limits_{i=1}^{n}\overline{x}\overline{y}}{\sum\limits_{i=1}^{n}x_ix_i-\sum\limits_{i=1}^{n}x_i\overline{x}-\sum\limits_{i=1}^{n}x_i\overline{x}+\sum\limits_{i=1}^{n}\overline{x}\overline{x}}$ <p>We can pull out factors independent of i</p> $a=\frac{\sum\limits_{i=1}^{n}x_iy_i-\overline{y}\sum\limits_{i=1}^{n}x_i-\overline{x}\sum\limits_{i=1}^{n}y_i+N\overline{x}\overline{y}}{\sum\limits_{i=1}^{n}x_ix_i-\overline{x}\sum\limits_{i=1}^{n}x_i-\overline{x}\sum\limits_{i=1}^{n}x_i+N\overline{x}\overline{x}}$ <p>Since $$\overline{x}=\frac{\sum\limits_{i=1}^{n}x_i}{n}$$ and $$\overline{y}=\frac{\sum\limits_{i=1}^{n}y_i}{n}$$</p> $a=\frac{\sum\limits_{i=1}^{n}x_iy_i-\frac{\sum\limits_{i=1}^{n}y_i}{n}\sum\limits_{i=1}^{n}x_i-\frac{\sum\limits_{i=1}^{n}x_i}{n}\sum\limits_{i=1}^{n}y_i+n\frac{\sum\limits_{i=1}^{n}x_i}{n}\frac{\sum\limits_{i=1}^{n}y_i}{n}}{\sum\limits_{i=1}^{n}x_ix_i-\frac{\sum\limits_{i=1}^{n}x_i}{n}\sum\limits_{i=1}^{n}x_i-\frac{\sum\limits_{i=1}^{n}x_i}{n}\sum\limits_{i=1}^{n}x_i+n\frac{\sum\limits_{i=1}^{n}x_i}{n}\frac{\sum\limits_{i=1}^{n}x_i}{n}}$ <p>Grouping terms gives</p> $a=\frac{\sum\limits_{i=1}^{n}x_iy_i-2\frac{\sum\limits_{i=1}^{n}y_i\sum\limits_{i=1}^{n}x_i}{n}+\frac{\sum\limits_{i=1}^{n}x_i\sum\limits_{i=1}^{n}y_i}{n}}{\sum\limits_{i=1}^{n}x_ix_i-2\frac{\sum\limits_{i=1}^{n}x_i\sum\limits_{i=1}^{n}x_i}{n}+\frac{\sum\limits_{i=1}^{n}x_i\sum\limits_{i=1}^{n}x_i}{n}}$ <p>Further grouping gives</p> $a=\frac{\sum\limits_{i=1}^{n}x_iy_i-\frac{\sum\limits_{i=1}^{n}y_i\sum\limits_{i=1}^{n}x_i}{n}}{\sum\limits_{i=1}^{n}x_ix_i-\frac{\sum\limits_{i=1}^{n}x_i\sum\limits_{i=1}^{n}x_i}{n}}$ <p>Multiplying both numerator and denominator by n gives</p> $a=\frac{n\sum\limits_{i=1}^{n}x_iy_i-\sum\limits_{i=1}^{n}y_i\sum\limits_{i=1}^{n}x_i}{n\sum\limits_{i=1}^{n}x_ix_i-\sum\limits_{i=1}^{n}x_i\sum\limits_{i=1}^{n}x_i}$ <p>Which is the formula we were aiming for.</p>Marc Flerackershttp://fromatogra.comA short introduction to statistical entities and how to use them. From mean to principal component analysis.