weird redundant formulation fix

_site/blog/FlowMatching/index.html

@@ -272,28 +272,25 @@ <h3 id="the-score-in-flow-matching">The Score in Flow Matching</h3>
 <h4 id="characteristic-functions">Characteristic Functions</h4>
 
 <p>What we’re aiming for is to determine the score term $\nabla_x \log p_\tau(x_t)$ from the stochastic process that connects the two random variables $x_0$ and $\epsilon$.
-But the random variable $x_t$ during simpling is the sum of two unknown random variables $x_0$ and $\epsilon$, namely via $x_t = \alpha(t) x_0 + \beta(t) \epsilon$.
-This is the sum of two random variables.</p>
+But the random variable $x_t$ during sampling is the sum of two random variables $x_0$ and $\epsilon$, namely via $x_t = \alpha(t) x_0 + \beta(t) \epsilon$.
+The probability distribution of the sum of two random variables is the convolution of the respective probability density functions.</p>
 
-<p>The probability distribution of the sum of two random variables is the convolution of the probability density functions of the two respective random variables.
-So in our case we have the sum of two independent weighted random variables $\alpha(t) x_0$ and $\beta(t)\epsilon$.
-For the time being we will first consider the sum of two independent random variables $x_0$ and $\epsilon$ without their respective scaling.</p>
-
-<p>Considering only $Z = X + Y$ the PDF of $Y$ is then the convolution of the PDFs of $X$ and $Y$:</p>
+<p>To ease into the topic, we will for starters only consider $Z = X + Y$, where the PDF of $Z$ is consequentially the convolution of the PDFs of $X$ and $Y$:</p>
 <div style="overflow-x: auto;">
 $$
 \begin{align*}
 p(z) &amp;= p(x) \star p(y) \\
-&amp;= \int_{x=-\infty}^{\infty} p(x) \ \cdot \ p(z - x) \ dx \\
+&amp;= \int_{x=-\infty}^{\infty} p(x) \ \cdot \ p(z - x) \ dx
 % &amp;= \hat{p}_{x_0}(k) \cdot \hat{p}_\epsilon(k) \\
 \end{align*}
 $$
 </div>
 
-<p>which is a bit difficult to work with.</p>
+<p>That doesn’t seem to be the most equation to be working with.</p>
+
+<p>Also, why is actually the convolution of the two PDF’s?</p>
 
-<p>Also, why is actually the convolution of the two PDF’s?
-In the case of the sum of $X$ and $Y$ we have to consider all possible values of $X$ and $Y$ that sum up to $Z$.
+<p>In the case of the sum of $X$ and $Y$ we have to consider all possible values of $X$ and $Y$ that sum up to $Z$.
 For example, let’s consider the probability of obtaining $Z=5$.
 We can then first choose $X$ and choose $Y$ as the remainder such that $Y = Z-X$.
 In order to get $Z=5$ we need to consider the joint probability of all permissible combinations of $X$ and $Y$, so for example $X=2$ and $Y=3$, or $X=3$ and $Y=2$, or $X=4$ and $Y=1$ or $X=-1000$ and $Y=995$.

blogposts/_posts/2024-12-08-FlowMatching.md

@@ -206,27 +206,24 @@ In order to derive the score term $\nabla_x \log p_\tau(x_\tau)$ we have to do a
 #### Characteristic Functions
 
 What we're aiming for is to determine the score term $\nabla_x \log p_\tau(x_t)$ from the stochastic process that connects the two random variables $x_0$ and $\epsilon$.
-But the random variable $x_t$ during simpling is the sum of two unknown random variables $x_0$ and $\epsilon$, namely via $x_t = \alpha(t) x_0 + \beta(t) \epsilon$.
-This is the sum of two random variables.
+But the random variable $x_t$ during sampling is the sum of two random variables $x_0$ and $\epsilon$, namely via $x_t = \alpha(t) x_0 + \beta(t) \epsilon$.
+The probability distribution of the sum of two random variables is the convolution of the respective probability density functions.
 
-The probability distribution of the sum of two random variables is the convolution of the probability density functions of the two respective random variables.
-So in our case we have the sum of two independent weighted random variables $\alpha(t) x_0$ and $\beta(t)\epsilon$.
-For the time being we will first consider the sum of two independent random variables $x_0$ and $\epsilon$ without their respective scaling.
-
-Considering only $Z = X + Y$ the PDF of $Y$ is then the convolution of the PDFs of $X$ and $Y$:
+To ease into the topic, we will for starters only consider $Z = X + Y$, where the PDF of $Z$ is consequentially the convolution of the PDFs of $X$ and $Y$:
 <div style="overflow-x: auto;">
 $$
 \begin{align*}
 p(z) &= p(x) \star p(y) \\
-&= \int_{x=-\infty}^{\infty} p(x) \ \cdot \ p(z - x) \ dx \\
+&= \int_{x=-\infty}^{\infty} p(x) \ \cdot \ p(z - x) \ dx
 % &= \hat{p}_{x_0}(k) \cdot \hat{p}_\epsilon(k) \\
 \end{align*}
 $$
 </div>
 
-which is a bit difficult to work with.
+That doesn't seem to be the most equation to be working with.
 
 Also, why is actually the convolution of the two PDF's?
+
 In the case of the sum of $X$ and $Y$ we have to consider all possible values of $X$ and $Y$ that sum up to $Z$.
 For example, let's consider the probability of obtaining $Z=5$.
 We can then first choose $X$ and choose $Y$ as the remainder such that $Y = Z-X$.