Step 1: Find $E[X]$ and $(E[X])^2$
We use the formula $E[X]=E[E[X \mid N]]$.
- Inner Expectation ( $E[X \mid N]$ ): Given $N=n, X$ is the number of heads in $n$ fair coin tosses. This is a Binomial distribution $X \mid N \sim \operatorname{Binomial}(n, p=0.5)$. The expectation of a binomial is $n p$.
$$
E[X \mid N]=N \times 0.5=\frac{N}{2}
$$
- Outer Expectation ( $E[X]$ ): Now we find the expectation of $N / 2$, where $N$ is the die roll.
$$
E[X]=E\left[\frac{N}{2}\right]=\frac{1}{2} E[N]
$$
We know the expected value of a 6 -sided die is $E[N]=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}$.
$$
E[X]=\frac{1}{2} \times \frac{7}{2}=\frac{7}{4}
$$
Calculate $(E[X])^2$ :
$$
(E[X])^2=\left(\frac{7}{4}\right)^2=\frac{49}{16}
$$
Step 2: Find $E\left[X^2\right]$
We use the formula $E\left[X^2\right]=E\left[E\left[X^2 \mid N\right]\right]$.
- Inner Expectation $\left(E\left[X^2 \mid N\right]\right)$ : We need the second moment of the binomial distribution $X \mid N$. We know $\operatorname{Var}(X)=E\left[X^2\right]-(E[X])^2$, so $E\left[X^2\right]=\operatorname{Var}(X)+(E[X])^2$.
- $\operatorname{Var}(X \mid N)=n p(1-p)=N \times 0.5 \times 0.5=\frac{N}{4}$.
- $E[X \mid N]=\frac{N}{2}$.
Substitute these:
$$
E\left[X^2 \mid N\right]=\frac{N}{4}+\left(\frac{N}{2}\right)^2=\frac{N}{4}+\frac{N^2}{4}=\frac{N^2+N}{4}
$$
- Outer Expectation $\left(E\left[X^2\right]\right.$): Now we find the expectation of this expression.
$$
E\left[X^2\right]=E\left[\frac{N^2+N}{4}\right]=\frac{1}{4} E\left[N^2+N\right]
$$
By linearity of expectation: $E\left[X^2\right]=\frac{1}{4}\left(E\left[N^2\right]+E[N]\right)$.
- We know $E[N]=\frac{7}{2}$.
- We need $E\left[N^2\right]$ for the die roll: $E\left[N^2\right]=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}$.
Substitute these back in:
$$
\begin{gathered}
E\left[X^2\right]=\frac{1}{4}\left(\frac{91}{6}+\frac{7}{2}\right) \\\\
E\left[X^2\right]=\frac{1}{4}\left(\frac{91}{6}+\frac{21}{6}\right)=\frac{1}{4}\left(\frac{112}{6}\right)=\frac{28}{6}=\frac{14}{3} \\\\
\operatorname{Var}(X)=E\left[X^2\right]-(E[X])^2 \\\\
\operatorname{Var}(X)=\frac{14}{3}-\frac{49}{16} \\\\
\operatorname{Var}(X)=\frac{224-147}{48}=\frac{77}{48}
\end{gathered}
$$
