Update latex rendering in md, emph * or _ to only *

Author: localhost433

notes/courses/MATH-UA-333/04-05-drv.md

@@ -37,6 +37,7 @@ No independence required.
 
 **Counting-by-indicators trick**  
 If $X=\sum_{i=1}^n \mathbf{1}_{A_i}$ counts how many events $A_i$ occur, then
+
 $$
 \E[X] = \sum_{i=1}^n \Pr(A_i).
 $$
@@ -62,7 +63,8 @@ $$
 
 ## Inclusion–Exclusion via indicators
 
-Using $\mathbf{1}_{A^c}=1-\mathbf{1}_A$ and expanding $\mathbf{1}_{\cup_i E_i}=1-\prod_{i=1}^n(1-\mathbf{1}_{E_i})$, then taking expectations yields
+Using $\mathbf{1}\_{A^c}=1-\mathbf{1}\_A$ and expanding $\mathbf{1}\_{\cup\_i E\_i}=1-\prod\_{i=1}^n(1-\mathbf{1}\_{E\_i})$, then taking expectations yields
+
 $$
 \Pr \Big(\bigcup_{i=1}^n E_i\Big)
 =\sum_{k=1}^n (-1)^{k-1} \sum_{1\le i_1<\cdots<i_k\le n} \Pr(E_{i_1}\cap\cdots\cap E_{i_k}).
@@ -104,14 +106,22 @@ so $\Var{X}=\E[X^2]-(\E X)^2=np(1-p)$.
 2) **Factorial moment**  
 $\E[X(X-1)]=n(n-1)p^2$, hence $\E[X^2]=n(n-1)p^2+np$ and $\Var{X}=np(1-p)$.
 
-> Examples  
-> - Three fair flips: $Y\sim\mathrm{Bin}(3,\tfrac12)$.  
+> Examples
+>
+> - Three fair flips: $Y\sim\mathrm{Bin}(3,\tfrac12)$.
 > - Ten dice, number of sixes: $X\sim\mathrm{Bin}(10,\tfrac16)$.  
 > - If you win \$5 per six, $Z=5X$ (not binomial). Then
+>
+> $$
+> \Pr(Z=10)=\Pr(X=2)=\binom{10}{2} \left(\tfrac16\right)^2 \left(\tfrac56\right)^8
+> $$
+> 
+> $$
+> \E[Z]=5\,\E[X]=\tfrac{25}{3}
+> $$
+> 
 > $$
-> \Pr(Z=10)=\Pr(X=2)=\binom{10}{2} \left(\tfrac16\right)^2 \left(\tfrac56\right)^8,\quad
-> \E[Z]=5\,\E[X]=\tfrac{25}{3},\quad
-> \Var{Z}=25\,\Var{X}=25\cdot 10\cdot\tfrac16\cdot\tfrac56.
+> \Var{Z}=25\,\Var{X}=25\cdot 10\cdot\tfrac16\cdot\tfrac56
 > $$
 
 ---

notes/courses/MATH-UA-333/09-sum-indep-rv.md

@@ -79,7 +79,7 @@ Let $X,Y$ be independent $\mathrm{Unif}(0,1)$, $Z=X+Y$.
 Then
 
 $$
-f_Z(t) = \int_{-\infty}^{\infty} \mathbf{1}_{[0,1]}(x)\mathbf{1}_{[0,1]}(t-x) dx.
+f\_Z(t) = \int\_{-\infty}^{\infty} \mathbf{1}\_{[0,1]}(x)\mathbf{1}\_{[0,1]}(t-x) dx.
 $$
 
 The integrand is $1$ exactly when $0 \le x\le1$ and $0 \le t-x\le1$.

notes/courses/MATH-UA-333/10-cond-dist-exp.md

@@ -91,11 +91,7 @@ $$
 Thus
 $$
 f_{X\mid Y}(x\mid b)=
-\begin{cases}
-\displaystyle
-\frac{1}{2\sqrt{1-b^2}}, & -\sqrt{1-b^2}\le x\le\sqrt{1-b^2},\\[6pt]
-0, & \text{otherwise.}
-\end{cases}
+\frac{1}{2\sqrt{1-b^2}}, \quad \text{for }-\sqrt{1-b^2}\le x\le\sqrt{1-b^2} \quad \text{0 otherwise}
 $$
 
 So given $Y=b$, $X$ is uniform on the horizontal chord.

notes/courses/MATH-UA-333/12-mgf.md

@@ -10,11 +10,14 @@ $$
 \mathbb{E}(X^n).
 $$
 
-For a (deterministic) sequence $\{a_n\}_{n\ge 0}$, the **(ordinary) generating function** is
+For a (deterministic) sequence $(a_n)_{n\ge 0}$, the **(ordinary) generating function** is
+
 $$
 F(z) = \sum_{n=0}^{\infty} a_n z^n.
 $$
+
 Within the radius of convergence,
+
 $$
 a_n = \frac{1}{n!} \frac{d^n}{dz^n} F(z)\bigg|_{z=0}.
 $$
@@ -100,7 +103,7 @@ $$
 > $$
 > M_X(t) =
 > \begin{cases}
-> \dfrac{\lambda}{\lambda - t}, & t < \lambda,\\[4pt]
+> \dfrac{\lambda}{\lambda - t}, & t < \lambda,\\
 > \infty, & t \ge \lambda.
 > \end{cases}
 > $$
@@ -117,25 +120,29 @@ $$
 > M_X(t) = \sum_{k=0}^{n} \binom{n}{k} (p e^t)^k (1-p)^{n-k}
 > = (1 - p + p e^t)^n.
 > $$
+>
 > Differentiate:
 > $$
 > M_X'(t) = n(1 - p + p e^t)^{n-1} \cdot p e^t,
 > $$
+>
 > $$
 > M_X''(t)
-> = n(n-1)(1 - p + p e^t)^{n-2} (p e^t)^2
-> + n(1 - p + p e^t)^{n-1} p e^t.
+> = n(n-1)(1 - p + p e^t)^{n-2} (p e^t)^2 + n(1 - p + p e^t)^{n-1} p e^t.
 > $$
+>
 > At $t=0$,
 > $$
 > \mathbb{E}(X) = M_X'(0)
 > = n(1 - p + p)^{n-1} p
 > = np,
 > $$
+>
 > $$
 > \mathbb{E}(X^2) = M_X''(0)
 > = n(n-1)p^2 + np.
 > $$
+>
 > Hence
 > $$
 > \operatorname{Var}(X)

notes/courses/MATH-UA-333/13-inequalities-exp.md

@@ -32,20 +32,21 @@ These monotonicity properties are the starting point for Markov’s inequality.
 Let $X$ be a random variable with $X \ge 0$ almost surely and let $a>0$.
 
 We have
+
 $$
-X \ge X \mathbf{1}_{\{X\ge a\}} \ge a \mathbf{1}_{\{X\ge a\}}.
+X \ge X \mathbf{1}\_{\{X \ge a\}} \ge a \mathbf{1}\_{\{X \ge a\}}.
 $$
 
 Taking expectations,
+
 $$
-\mathbb{E}X \;\ge\; \mathbb{E}\big(X \mathbf{1}_{\{X\ge a\}}\big)
-\;\ge\; a\,\mathbb{E}\big(\mathbf{1}_{\{X\ge a\}}\big)
-= a\,\mathbb{P}(X\ge a).
+\mathbb{E}X \ge \mathbb{E} \big(X \mathbf{1}\_{\{X\ge a\}} \big) \ge a \mathbb{E}\big(\mathbf{1}\_{\{X \ge a\}} \big)
+= a \mathbb{P}(X \ge a)
 $$
 
 Therefore
 $$
-\mathbb{P}(X\ge a) \le \frac{\mathbb{E}X}{a}.
+\mathbb{P}(X \ge a) \le \frac{\mathbb{E}X}{a}.
 $$
 
 If $X$ is not necessarily nonnegative, we can apply Markov to $|X|$:

notes/courses/MATH-UA-333/14-lln.md

@@ -92,19 +92,22 @@ $$
 First compute the mean and variance of $\overline{X}_n$:
 
 - Linearity of expectation:
-  $$
-  \mathbb{E}\overline{X}_n
-  = \mathbb{E}\left(\frac{1}{n}\sum_{k=1}^n X_k\right)
-  = \frac{1}{n}\sum_{k=1}^n \mathbb{E}X_k
-  = \mu.
-  $$
+
+$$
+\mathbb{E}\overline{X}\_n
+= \mathbb{E} \left(\frac{1}{n}\sum\_{k=1}^n X\_k\right)
+= \frac{1}{n}\sum\_{k=1}^n \mathbb{E}X\_k
+= \mu
+$$
+
 - Independence and identical distribution give
-  $$
-  \operatorname{Var}(\overline{X}_n)
-  = \operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^n X_k\right)
-  = \frac{1}{n^2} \sum_{k=1}^n \operatorname{Var}(X_k)
-  = \frac{\sigma^2}{n}.
-  $$
+
+$$
+\operatorname{Var}(\overline{X}_n)
+= \operatorname{Var}\left(\frac{1}{n}\sum\_{k=1}^n X\_k\right)
+= \frac{1}{n^2} \sum\_{k=1}^n \operatorname{Var}(X\_k)
+= \frac{\sigma^2}{n}.
+$$
 
 Now for any $\varepsilon>0$, Chebyshev’s inequality gives
 $$

notes/courses/MATH-UA-333/16-markov-chains.md

@@ -19,7 +19,7 @@ where $n$ is usually interpreted as discrete time.
 
 ### Markov property
 
-We say a process $(X_n)_{n\ge 0}$ with values in $\mathcal{S}$ has the **Markov property** if for every $n\ge 0$ and all states
+We say a process $(X\_n)\_{n\ge 0}$ with values in $\mathcal{S}$ has the **Markov property** if for every $n\ge 0$ and all states
 $$
 i_0, i_1, \dots, i_{n-1}, i, j \in \mathcal{S},
 $$
@@ -76,13 +76,15 @@ Properties:
 
 > **Example (Gambler’s dice game).**  
 > You start with $6$ dollars. At each step:
+>
 > - Pay $1$ dollar to play.  
 > - Roll a fair die:
 >   - If you roll a $6$, you win $6$ dollars.  
 >   - Otherwise, you win nothing.  
 > - You must stop once your money hits $0$.  
 >
-> Let $X_n$ be the amount of money you have after the $n$-th play. Then $(X_n)_{n\ge 0}$ is a Markov chain with state space $\{0,1,2,\dots\}$.  
+> Let $X_n$ be the amount of money you have after the $n$-th play. Then $(X_n)_{n\ge 0}$ is a Markov chain with state space $\{0,1,2,\dots\}$.
+>
 > - State $0$ is absorbing: once you have $0$, you stay there,
 >   $$
 >   p_{00} = 1, \quad p_{0j} = 0 \text{ for } j>0.
@@ -109,10 +111,10 @@ We can compute $v^{(n+1)}$ from $v^{(n)}$ using the transition matrix $P$.
 
 For each state $j$,
 $$
-\mathbb{P}(X_{n+1} = j)
-= \sum_{i=1}^m \mathbb{P}(X_{n+1}=j, X_n = i)
-= \sum_{i=1}^m \mathbb{P}(X_{n+1}=j \mid X_n=i)\, \mathbb{P}(X_n=i)
-= \sum_{i=1}^m v^{(n)}_i p_{ij}.
+\mathbb{P}(X\_{n+1} = j)
+= \sum\_{i=1}^m \mathbb{P}(X\_{n+1}=j, X\_n = i)
+= \sum\_{i=1}^m \mathbb{P}(X\_{n+1}=j \mid X\_n=i)\, \mathbb{P}(X\_n=i)
+= \sum\_{i=1}^m v^{(n)}\_i p\_{ij}.
 $$
 
 In matrix form,

notes/js/note.js

@@ -190,3 +190,30 @@ fetch(`/notes/courses/${course}/${noteSlug}.md`)
         console.error(err);
         container.innerHTML = "<p>Could not load note.</p>";
     });
+
+const tokenizer = {
+  em(src) {
+    // Only treat *...* as emphasis, ignore underscores
+    const match = /^\*([^*]+)\*/.exec(src);
+    if (!match) return;
+    return {
+      type: 'em',
+      raw: match[0],
+      text: match[1],
+      tokens: this.lexer.inlineTokens(match[1])
+    };
+  },
+  strong(src) {
+    // Only treat **...** as strong emphasis, ignore __
+    const match = /^\*\*([^*]+)\*\*/.exec(src);
+    if (!match) return;
+    return {
+      type: 'strong',
+      raw: match[0],
+      text: match[1],
+      tokens: this.lexer.inlineTokens(match[1])
+    };
+  }
+};
+
+marked.use({ tokenizer });

posts/js/post.js

@@ -137,4 +137,31 @@ fetch(`./posts/entries/${slug}.md`)
   .catch(e => {
     console.error(e);
     content.innerHTML = "<h2>Post not found</h2>";
-  });
+  });
+
+const tokenizer = {
+  em(src) {
+    // Only treat *...* as emphasis, ignore underscores
+    const match = /^\*([^*]+)\*/.exec(src);
+    if (!match) return;
+    return {
+      type: 'em',
+      raw: match[0],
+      text: match[1],
+      tokens: this.lexer.inlineTokens(match[1])
+    };
+  },
+  strong(src) {
+    // Only treat **...** as strong emphasis, ignore __
+    const match = /^\*\*([^*]+)\*\*/.exec(src);
+    if (!match) return;
+    return {
+      type: 'strong',
+      raw: match[0],
+      text: match[1],
+      tokens: this.lexer.inlineTokens(match[1])
+    };
+  }
+};
+
+marked.use({ tokenizer });