# Random variable conv. in prob. to c. How to find c?

1. Mar 19, 2014

### SithsNGiggles

1. The problem statement, all variables and given/known data

Let $Y_1,...Y_n$ be independent standard normal random variables.

What is the distribution of $\displaystyle\sum_{i=1}^n{Y_i}^2$ ?

Let $W_n=\displaystyle\frac{1}{n}\sum_{i=1}^n {Y_i}^2$. Does $W_n\xrightarrow{p}c$ for some constant $c$? If so, what is the value of $c$?

2. Relevant equations

3. The attempt at a solution

In response to the first question, I say that $W_n$ has a Chi Square distribution with $n$ degrees of freedom.

For the second, I use a theorem stated in my book that says if $\text{Var}(\hat{\theta}_n)\to 0$, then $\hat{\theta}_n\xrightarrow{p}\theta$.

I have that $\text{Var}(W_n)=\dfrac{1}{n^2}\text{Var}\left(\displaystyle\sum_{i=1}^n {Y_i}^2\right)=\dfrac{2n}{n^2}=\dfrac{2}{n}$, which approaches $0$ as $n\to\infty$. Thus $W_n$ indeed converges in probability to $c$.

The problem I'm having now is with finding the value of $c$. There don't appear to be any examples in my text that describe the process to finding the limiting probability (if that's the right term). Any help is appreciated

2. Mar 19, 2014

### Ray Vickson

What is $E(W_n)$?

3. Mar 19, 2014

### SithsNGiggles

Judging by $W_n$'s distribution, that would be $n$. Does that mean $c=n$, and if so, is that always the case (that the mean is the constant I'm supposed to find)?

4. Mar 19, 2014

### haruspex

No, E(Wn) is not n.

5. Mar 19, 2014

### SithsNGiggles

Oh right, I forgot that $W_n=\frac{1}{n}\sum\cdots$. Thanks! I meant to say $E(W_n)=1$.

6. Mar 20, 2014

### haruspex

Right - so all done here?

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