Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 25 insertions, 25 deletions

diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md
index 14ba4cb..1193a93 100644
--- a/source/know/concept/binomial-distribution/index.md
+++ b/source/know/concept/binomial-distribution/index.md
@@ -9,11 +9,11 @@ layout: "concept"
 ---
 
 The **binomial distribution** is a discrete probability distribution
-describing a **Bernoulli process**: a set of independent $N$ trials where
+describing a **Bernoulli process**: a set of independent $$N$$ trials where
 each has only two possible outcomes, "success" and "failure",
-the former with probability $p$ and the latter with $q = 1 - p$.
+the former with probability $$p$$ and the latter with $$q = 1 - p$$.
 The binomial distribution then gives the probability
-that $n$ out of the $N$ trials succeed:
+that $$n$$ out of the $$N$$ trials succeed:
 
 $$\begin{aligned}
     \boxed{
@@ -22,8 +22,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The first factor is known as the **binomial coefficient**, which describes the
-number of microstates (i.e. permutations) that have $n$ successes out of $N$ trials.
-These happen to be the coefficients in the polynomial $(a + b)^N$,
+number of microstates (i.e. permutations) that have $$n$$ successes out of $$N$$ trials.
+These happen to be the coefficients in the polynomial $$(a + b)^N$$,
 and can be read off of Pascal's triangle.
 It is defined as follows:
 
@@ -33,10 +33,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The remaining factor $p^n (1 - p)^{N - n}$ is then just the
+The remaining factor $$p^n (1 - p)^{N - n}$$ is then just the
 probability of attaining each microstate.
 
-The expected or mean number of successes $\mu$ after $N$ trials is as follows:
+The expected or mean number of successes $$\mu$$ after $$N$$ trials is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -49,7 +49,7 @@ $$\begin{aligned}
 <label for="proof-mean">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-mean">Proof.</label>
-The trick is to treat $p$ and $q$ as independent until the last moment:
+The trick is to treat $$p$$ and $$q$$ as independent until the last moment:
 
 $$\begin{aligned}
     \mu
@@ -61,12 +61,12 @@ $$\begin{aligned}
     = N p (p + q)^{N - 1}
 \end{aligned}$$
 
-Inserting $q = 1 - p$ then gives the desired result.
+Inserting $$q = 1 - p$$ then gives the desired result.
 </div>
 </div>
 
-Meanwhile, we find the following variance $\sigma^2$,
-with $\sigma$ being the standard deviation:
+Meanwhile, we find the following variance $$\sigma^2$$,
+with $$\sigma$$ being the standard deviation:
 
 $$\begin{aligned}
     \boxed{
@@ -79,7 +79,7 @@ $$\begin{aligned}
 <label for="proof-var">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-var">Proof.</label>
-We use the same trick to calculate $\overline{n^2}$
+We use the same trick to calculate $$\overline{n^2}$$
 (the mean squared number of successes):
 
 $$\begin{aligned}
@@ -96,7 +96,7 @@ $$\begin{aligned}
     &= N p + N^2 p^2 - N p^2
 \end{aligned}$$
 
-Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$:
+Using this and the earlier expression $$\mu = N p$$, we find the variance $$\sigma^2$$:
 
 $$\begin{aligned}
     \sigma^2
@@ -105,11 +105,11 @@ $$\begin{aligned}
     = N p (1 - p)
 \end{aligned}$$
 
-By inserting $q = 1 - p$, we arrive at the desired expression.
+By inserting $$q = 1 - p$$, we arrive at the desired expression.
 </div>
 </div>
 
-As $N \to \infty$, the binomial distribution
+As $$N \to \infty$$, the binomial distribution
 turns into the continuous normal distribution,
 a fact that is sometimes called the **de Moivre-Laplace theorem**:
 
@@ -124,8 +124,8 @@ $$\begin{aligned}
 <label for="proof-normal">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-normal">Proof.</label>
-We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$
-around the mean $\mu = Np$:
+We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$
+around the mean $$\mu = Np$$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
@@ -134,7 +134,7 @@ $$\begin{aligned}
     D_m(n) = \dvn{m}{\ln\!\big(P_N(n)\big)}{n}
 \end{aligned}$$
 
-We use Stirling's approximation to calculate the factorials in $D_m$:
+We use Stirling's approximation to calculate the factorials in $$D_m$$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
@@ -143,8 +143,8 @@ $$\begin{aligned}
     &\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big)
 \end{aligned}$$
 
-For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation
-to get a non-zero result. We take advantage of $N - N p = N q$:
+For $$D_0(\mu)$$, we need to use a stronger version of Stirling's approximation
+to get a non-zero result. We take advantage of $$N - N p = N q$$:
 
 $$\begin{aligned}
     D_0(\mu)
@@ -161,7 +161,7 @@ $$\begin{aligned}
     = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
 \end{aligned}$$
 
-Next, we expect that $D_1(\mu) = 0$, because $\mu$ is the maximum.
+Next, we expect that $$D_1(\mu) = 0$$, because $$\mu$$ is the maximum.
 This is indeed the case:
 
 $$\begin{aligned}
@@ -176,7 +176,7 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-For the same reason, we expect that $D_2(\mu)$ is negative.
+For the same reason, we expect that $$D_2(\mu)$$ is negative.
 We find the following expression:
 
 $$\begin{aligned}
@@ -189,7 +189,7 @@ $$\begin{aligned}
     = - \frac{1}{\sigma^2}
 \end{aligned}$$
 
-The higher-order derivatives tend to zero for $N \to \infty$, so we discard them:
+The higher-order derivatives tend to zero for $$N \to \infty$$, so we discard them:
 
 $$\begin{aligned}
     D_3(n)
@@ -201,7 +201,7 @@ $$\begin{aligned}
     \cdots
 \end{aligned}$$
 
-Putting everything together, for large $N$,
+Putting everything together, for large $$N$$,
 the Taylor series approximately becomes:
 
 $$\begin{aligned}
@@ -210,7 +210,7 @@ $$\begin{aligned}
     = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2}
 \end{aligned}$$
 
-Taking $\exp$ of this expression then yields a normalized Gaussian distribution.
+Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution.
 </div>
 </div>