Problem source

I choose a number from the integers $1, 2, \ldots, (2n-1)$ and
the outcome is the random variable $N$. Calculate $ \E(N)$ and $\E(N^2)$.
I then  repeat a certain experiment $N$ times, the outcome of the  $i$th experiment being the random variable $X_i$ ($1\le i \le N$). For each $i$, the random variable $X_i$ has mean $\mu$ and variance $\sigma^2$, and $X_i$ is independent of $X_j$ for $i\ne j$ and also independent of $N$. The random variable $Y$ is defined by $Y= \sum\limits_{i=1}^NX_i$. Show that $\E(Y)=n\mu$ and that $\mathrm{Cov}(Y,N) = \frac13n(n-1)\mu$. Find  $\var(Y) $ in terms of $n$, $\sigma^2$ and $\mu$.

Solution source

\begin{align*}
&& \E[N] &= \sum_{i=1}^{2n-1} \frac{i}{2n-1} \\
&&&= \frac{2n(2n-1)}{2(2n-1)} = n\\
&& \E[N^2] &= \sum_{i=1}^{2n-1} \frac{i^2}{2n-1} \\
&&&= \frac{(2n-1)(2n)(4n-1)}{6(2n-1)} \\
&&&= \frac{n(4n-1)}{3} \\
&& \var[N] &= \frac{n(4n-1)}{3} - n^2 \\
&&&= \frac{n^2-n}{3}
\end{align*}

\begin{align*}
&& \E[Y] &= \E \left [ \E \left [ \sum_{i=1}^N X_i | N = k\right] \right]\\
&&&= \E \left[ N\mu \right] = n\mu \\
\\
&& \mathrm{Cov}(Y,N) &= \mathbb{E}[XY] - \E[X]\E[Y] \\
&&&= \E \left [ \E \left [N \sum_{i=1}^N X_i | N = k\right] \right] - n^2 \mu \\
&&&= \E[N^2\mu] - n^2 \mu \\
&&&= \left ( \frac{n^2(4n-1)}{3} - n^2  \right) \mu \\
&&&= \frac{n^2-n}{3}\mu \\
\\
&& \E[Y^2] &= \E \left [ \E \left [ \left ( \sum_{i=1}^N X_i \right) ^2\right ] \right] \\
&&&= \E \left [ \E \left [  \sum_{i=1}^N X_i ^2 + 2\sum_{i,j} X_iX_j\right ] \right] \\
&&&= \E \left [   \sum_{i=1}^N \left ( \E[X_i ^2] + 2\sum_{i,j} \E[X_i]\E[X_j]\right ) \right] \\
&&&= \E \left [ N(\sigma^2 + \mu^2) + (N^2-N)\mu^2\right] \\
&&&= n(\sigma^2+\mu^2) +  \left ( \frac{n^2-n}{3}-n \right)\mu^2 \\
&&&= n\sigma^2 + \frac{n^2-n}{3} \mu^2 \\
\Rightarrow && \var[Y] &= n\sigma^2 + \frac{n^2-n}{3} \mu^2 - n^2\mu^2 \\
&&&= n\sigma^2 - \frac{2n^2+n}{3} \mu^2

\end{align*}