Standard normal limiting function chi squared

[davidkong0987]

Hi, I have a question

If X1,X2,...,Xn are independent random variables having chi-square distribution witn v=1 and Yn=X1+X2+...+Xn, then the limiting distribution of

(Yn/n) - 1
Z= --------------- as n->infinity is the standard normal distribution.
sqrt(2/n)

I know that Yn has chi-square distribution with v=n, but how to proceed.

Homework Equations

The Attempt at a Solution

 

[davidkong0987]

I read this post on physics forum but no one had answered it. I finally found an answer and thought I should share it. The methodology might seem obvious - using the method of MGF (moment generating functions).

The MGF of Y is that of a chi-squared distribution with n degrees of freedom:
(1-2t)^(-n/2).

Applying the transformation Y --> Z will give you an MGF of Z of
e^(-√(n/2) t) (1-√(2/n) t)^(-n/2).

Now, this is the million dollar question: you will have to use the series ln⁡(1+x)=x-1/2 x^2+1/3 x^3+⋯.

This will help you simplify to an MGF of e^(1/2t^2) which is the mgf of a standard normal curve.

 

[Ray Vickson]

Hi, I have a question

If X1,X2,...,Xn are independent random variables having chi-square distribution witn v=1 and Yn=X1+X2+...+Xn, then the limiting distribution of

(Yn/n) - 1
Z= --------------- as n->infinity is the standard normal distribution.
sqrt(2/n)

I know that Yn has chi-square distribution with v=n, but how to proceed.

Homework Equations

The Attempt at a Solution

There is nothing new to prove: this is just an application of the Central Limit Theorem. There is nothing special about chi-squared here: the result holds for any distribution having finite mean and variance, and the proof (using MGF) is almost the same.

RGV