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

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 $$N$$).

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.

• I am unable to solve. Can you please complete the solution, it will be very helpful. Apr 11, 2020 at 12:02
• hmmm... do you know the mean and variance of $N$? Apr 11, 2020 at 12:05
• not true. hmmm... do you know the distribution of $N$? Apr 11, 2020 at 12:06
• No, i don't know. Apr 11, 2020 at 12:07
• read the first two lines of the question again. (btw, I need to run some errand, slow in response, but I will get back to you after I get home). Apr 11, 2020 at 12:08

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 $$\sigma^2_{Y_n}=5$$

• I am unable to solve. Can you please complete the solution, it will be very helpful. Apr 11, 2020 at 12:02
• Sure. I've added more details. You can also read more about Poisson distribution here:en.wikipedia.org/wiki/Poisson_distribution Apr 11, 2020 at 12:14