# help with chi square freedom sums

#### omega

##### New Member
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?

#### Dason

The MGF of the sum of independent random variables is the product of the MGFs of the random variables. You should be able to show this pretty easily using the definition of independence and the definition of the MGF.

#### omega

##### New Member
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?

#### Dason

Looks fine to me. But it'll only pass as a proof for a class if you've proved or at least have been shown that fact about MGFs that I mentioned before. Otherwise you'll have to derive that result yourself.

#### omega

##### New Member
Looks fine to me. But it'll only pass as a proof for a class if you've proved or at least have been shown that fact about MGFs that I mentioned before. Otherwise you'll have to derive that result yourself.
would the proof of that be:
m[X1+X2](t) = E(e^(t(X1+X2)))
=E(e^(tX1)e^(tX2))
=E(e^(tX1))E(e^(tX2)) (since they are independent)
=m[X1](s)m[X2](s)

is this fine?