Let $Y_n = X_n + X_{n+1}$ and $T_n = \frac{1}{n} \sum_{i=1}^n Y_i$. I want to find $Var[T_n]$

Question

Let $Y_n = X_n + X_{n+1}$ with $X_n$ ~ Bernoulli$(p)$ independent and identically distributed. Let $T_n = \frac{1}{n} \sum_{i=1}^n Y_i$. I want to find $Var[T_n]$.

This is what I have done so far:

I know that $Y_n$ ~ Bin(2,p). Then I can write this:

$$ Var[T_n] = Var[\frac{1}{n}\sum_{i=1}^n Y_i]= \frac{1}{n^2}Var[\sum_{i=1}^n Y_i] =$$

$$=\frac{1}{n^2}\sum_{i=1}^nVar[ Y_i] = \frac{1}{n^2}*n* 2p(1-p) $$

However, the result should actually contain a $\frac{2n-1}{n^2}$. Can somebody point out my mistake?

Answer 1

Hints:

• The $Y_i$s are not independent so you cannot just sum their variances

• Consider $Var[T_n] = Var\left[\frac{1}{n}\sum\limits_{i=1}^n Y_i\right] = Var\left[\frac{1}{n}\left(X_1+X_{n+1}+\sum\limits_{i=2}^n 2X_i\right)\right]$ where the $X_i$s are independent