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

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Homework Help Overview

The discussion revolves around the convergence of a sequence of random variables, specifically examining the behavior of the average of squared independent standard normal random variables. Participants are exploring the distribution of the sum of squares and the limiting behavior of the average as the number of variables increases.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the distribution of the sum of squared standard normal variables and the implications for the average of these variables. There is an exploration of the variance of the average and its convergence in probability. Questions arise regarding the value of the limiting constant and the relationship between the mean of the distribution and the constant sought.

Discussion Status

The discussion is active, with participants providing insights into the variance and expected value of the average of the squared variables. Some participants express uncertainty about the value of the constant to which the average converges, while others clarify misconceptions regarding the expected value.

Contextual Notes

There is a focus on the convergence properties of random variables and the need for clarity on the expected value of the average. Participants are navigating through assumptions about the relationship between the mean and the limiting constant.

SithsNGiggles
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Homework Statement



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##?

Homework Equations



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
 
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SithsNGiggles said:

Homework Statement



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##?

Homework Equations



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

What is ##E(W_n)##?
 
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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)?
 
SithsNGiggles said:
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)?
No, E(Wn) is not n.
 
haruspex said:
No, E(Wn) is not n.

Oh right, I forgot that ##W_n=\frac{1}{n}\sum\cdots##. Thanks! I meant to say ##E(W_n)=1##.
 
SithsNGiggles said:
Oh right, I forgot that ##W_n=\frac{1}{n}\sum\cdots##. Thanks! I meant to say ##E(W_n)=1##.
Right - so all done here?
 

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