What is the variance of $X_1+X_2+\cdots+X_{N+1}$?

Question

Let $X_1,X_2,\cdots$ be a sequence of independent normally distributed random variables with mean 1 and variance 1. Let $N$ be a Poisson random variable with mean $2$, independent of $X_1,X_2,\cdots$. Then, the variance of $X_1+X_2+\cdots+X_{N+1}$ is ?

My attempt:

We know that $\text{Var}(aX_1+bX_2)=a^2\text{Var}(X_1)+b^2\text{Var}(X_2)$ when $X_1$ and $X_2$ are independent. And for a poisson random variable $X_n$, mean=variance=$\lambda$. So, in our case $\text{Var}(X_1+X_2+\cdots+X_{N+1})=n+2$. I believe this is wrong, as the answer is a constant (independent of $N$).

Answer 1

Use the total law of variance:

\begin{align} Var\left[\sum_{i=1}^{N+1} X_i\right] &= E\left[ Var\left[ \sum_{i=1}^{N+1}X_i|N\right] \right] + Var\left[ E\left[ \sum_{i=1}^{N+1}X_i|N\right] \right]\\ &= E \left[ N+1\right] + Var(N+1) \end{align}

I will leave the rest to you.

Answer 2

By defining $$Y_N=X_1+X_2+\cdots +X_{N+1}$$we can say that $$Y_N|N=n\sim \mathcal{N}(n+1,n+1)$$ hence $$E\{Y_N\}{=E_N\Big\{E_{Y_N}\{Y_n|N\}\Big\}\\=E_N\{N+1\}=3}$$also$$E\{Y^2_N\}{=E_N\Big\{E_{Y^2_N}\{Y^2_n|N\}\Big\}\\=E_N\{N+1+(N+1)^2\} \\=E_N\{N^2+3N+2\} \\=6+6+2 =14}$$hence $$\sigma^2_{Y_n}=5$$