Sum of squares of normal distribution

Question

I have a problem in which I need to apply weak law of large numbers to sum of squares of normal distribution with mean $1$ and variance $1$.
$X_1, X_2,\dots \sim \mathcal{N}(1,1).$ They are iid.

$V(X_1^2+X_2^2+\dots\X_n^2)/n$ what will be the limit value of this when $n$ tends to infinite.

My views:
I know sum of squares of SNV is chi square.
Variance of this chi square will be $2n$.
But I don’t know if it follows central chi square in case when mean is not $0$.

In this particular case, will my variance be $2n$?

Answer 1

My views: I know sum of squares of SNV is chi square. Variance of this chi square will be 2n. But I don’t know if it follows central chi square in case when mean is not 0.

sum of square of SNV is a chi-squared but your Gaussian are not centered thus the sum of your iid reduced gaussian is a Noncentral chi-squared distribution with variance $2(k+2\lambda)$ where $\lambda$ is the noncentrality parameter.

As @Kavi Rama Murthy already asked, we have to know what is there in the numerator:

• $\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{\infty}X_k^2\right]$

• $\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{n}X_k^2\right]$

• Something else?

EDIT

I have a finite sum up to n

Thus the result is simply

$$\frac{1}{n}\mathbb{V}\left[\sum_{k=1}^{n}X_k^2\right]=\frac{1}{n}\cdot n\mathbb{V}[X_1^2]=2(1+2)=6$$