# Thread: help with chi square freedom sums

1. ## help with chi square freedom sums

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?

2. Originally Posted by Sneaky
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?
Are X1 and X2 independent. If so, then use the well known theorem for the moment generating function of the sum.

3. yes they are independent

4. Originally Posted by Sneaky
yes they are independent
Then apply the appropriate theorem.

5. 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?

6. Originally Posted by Sneaky
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?
Yes.

8. Originally Posted by Sneaky
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?

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