- Thread starter omega
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(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?

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.

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?