How do i show that the a [X1 has a chi square distribution with n degrees of freedom] + [X2 has a chi square distribution with m degrees of freedom] is a [X1+X2 has a chi square distribution with n+m degrees of freedom]?
How can i use moment generating functions to do this?
the mgf for a chi square with n freedoms is
(1-2t)^(-n/2)
the mgf for a chi square with m freedoms is
(1-2t)^(-m/2)
The MGF of the sum of independent random variables is the product of the MGFs of the random variables. So the product is
(1-2t)^(-(n+m)/2)
which is a chi squared distribution with n+m degrees of freedom
so in conclusion [X1 has a chi square distribution with n degrees of freedom] + [X2 has a chi square distribution with m degrees of freedom] is a [X1+X2 has a chi square distribution with n+m degrees of freedom]
Is this right?
I know the thread is cloes and why, but I like the following proof which uses a completely different method (and I know that some will argue that its not a proof, but that is another discussion):
Since A RV which is the sum of the squares of n independent standard normals has a chi-squared distribution with n degrees of freedom, the sum of two such independednt RV with n and m degrees of freedom respectively has a chi-squared distribution with n+m degrees of freedom since it is the sum of the squares of n+m independent standard normals..
Since there is no way of telling how our chi-square RV were produced the result holds in general.
CB