minor fix to the solution of problem 5.51

Author: RamonAraujo

src/chapters/5/sections/mixed_practice/problems/51.tex

@@ -1,23 +1,23 @@
 (a) We know that, \(X^2 \le X\) with probability 1.
 So \(\mathbb{E}[X^2] \le \mathbb{E}{X}\)
-\begin{flalign}
+\begin{flalign*}
     V(X) & = \mathbb{E}[X^2] - \mathbb{E}[X]^2 \\
     & \le \mu - \mu^2 \\
-    & \le \frac{1}{4} \text{taking the minimum of the above quadratic}
-\end{flalign}
+    & \le \frac{1}{4} \text{ (taking the maximum of the above quadratic)}
+\end{flalign*}
 (b) I have to show that \(V(X) = 1/4\) leads to a unique distribution.
 From (a), \(V(X) \le \mu - \mu^2 \le 1/4\) implies that, \(\mu = 1/2\)
 Now \[\mathbb{E}[(X - 1/2)^2] = 1/4\]
-But \[0 \ge (X - 1/2)^2 \le 1/4\] with probability 1. 
+But \[0 \le (X - 1/2)^2 \le 1/4\] with probability 1.
 To get \(\mathbb{E}[(X - 1/2)^2] = 1/4\), we need \((X - 1/2)^2 = 1/4\) with probability 1.
 So, 
-\begin{equation}
+\begin{equation*}
     X = 
     \begin{cases}
         0 & \text{with prob p} \\
         1 & \text{with prob 1 - p}
     \end{cases}
-\end{equation}
+\end{equation*}
 Using \(\mu = 1/2\) gives us, \(p = 1/2\).
 
-
+The distribution of $X$ is Bernoulli. More specifically, $X \sim \mathrm{Bern}(1/2)$.