# Is it necessary to assume a moment generating function exists?

Consider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?

Addition: Does this change if we know A and C both come from (different) chi-squared distributions? I am tasked with finding the distribution of B. If I can just do MGF(A) / MGF (C) = MGF (B) then it's simple... but can I even write this statement without assuming MGF (B) exists?

Generally -- You can't compute the MGF of B

In general, you can't compute the MGF of $$B$$ if you only know the MGFs of $$A$$ and $$C$$. For example, consider two possible joint distributions of $$A$$ and $$C$$:

Case 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is 1.

Case 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is $$\frac{\exp(-t)+\exp(t)}2=\cosh t$$.

Notice that in both Case 1 and Case 2 the MGFs for $$A$$ and $$C$$ were $$(1+exp(t))/2$$, but the MGF for $$B$$ changed from Case 1 to Case 2.

 Generally -- You can prove existence of an MGF for B

Although you can't compute the MGF of $$B$$, you can prove that $$M_B(t)$$ exists for $$(*)\quad t\in D=\frac12 (Dom(M_A)\cap (-Dom(M_C)).$$ Suppose $$t\in D$$. If the MGF's of $$A$$ and $$C$$ exist, then for all $$t\in Dom(M_A)$$, $$M_A(t)=||\exp(ta)||_1<\infty$$ and for all $$t\in (-Dom(M_C))$$, $$M_C(-t)=||\exp(-tc)||_1<\infty$$ where $$||g||_p=\left(\int\int |g(a,c)|^p\; f(a,c)\; da dc\right)^{1/p}$$ is the $$L_p$$-norm of $$g$$ over the joint probability space and $$f(a,c)$$ is the joint pdf of $$A$$ and $$C$$. That implies $$||\exp(ta/2)||_2 < \infty$$ and $$||\exp(-tc/2)||_2 < \infty$$. By the Hölder's inequality or the Schwarz inequality, $$||\exp(ta)\exp(-tc)||_1<\infty$$. But, $$||\exp(ta)\exp(-tc)||_1= ||\exp(t(a-c)||_1= E[\exp(tB)]=M_B(t).$$ This proves that $$M_B(t)$$ exists for $$t\in D$$.

 If A and C are independent, you can compute the MGF of B

If $$A$$ and $$C$$ are independent and $$B = A-C$$, then it must be the case that $$(**) \quad M_B(t) = M_A(t)\cdot M_C(-t)$$ whenever $$t\in Dom(M_A)\cap(-Dom(M_C))$$ (see e.g. Wikipedia). Here is a rough proof.

If $$t\in Dom(M_A)\cap(-Dom(M_C))$$, then $$M_A(t)\cdot M_C(-t) = \int_{a=-\infty}^\infty \exp(t a) dF_A(a) \cdot \int_{c=-\infty}^\infty \exp(-t c) dF_C(c)$$ $$= \int_{a=-\infty}^\infty \int_{c=-\infty}^\infty \exp(t (a-c)) dF_A(a) dF_C(c)$$ $$= \int_{b=-\infty}^\infty \exp(t b) dF_B(b) = M_B(t)$$ where $$F_A, F_B$$, and $$F_C$$ are the cumulative distribution functions of $$A, B$$, and $$C$$ respectively.

 If A and C have $$\chi^2$$ distributions

In general, if $$A$$ and $$C$$ have $$\chi^2$$ distributions, you can only state that $$B$$ has an MGF and the domain of $$B$$'s MGF is \begin{align} Dom(B) &= \frac12 (Dom(M_A)\cap (-Dom(M_C))\\ &= \frac12 ((-\infty,2)\cap (-(-\infty,2))\\ &= \frac12 ((-\infty,2)\cap (-2,\infty))\\ &= \frac12 (-2,2)=(-1,1)\\ \end{align} by (*) and the fact that the domain of the MGF of a $$\chi^2$$ distribution is $$(-\infty,2)$$.

If you know that $$A$$ and $$C$$ have $$\chi^2$$ distributions, and you know that they are independent, then you can look up the MGFs for $$\chi^2$$ distributions and apply formula (**) to compute the formula for the MGF of $$B$$.

• Thank you. The question asks to prove that B is chi-squared distributed. It sounds like we know that a MGF exists, but that we can't prove it's specifically chi-squared? – purpleostrich Sep 28 at 2:06
• I learned a lot about MGFs and a bit about $\chi^2$ distributions while trying to figure out the answer to your question. :) – irchans Sep 28 at 8:24
• @purpleostrich, I suspect that if this was posed as an exercise to you then there is a missing assumption, for instance that $B$ and $C$ are independent, or as explored by irchans the assumption could be that $A$ and $C$ are independent (although I find this latter assumption somewhat less likely than the former). – pre-kidney Sep 28 at 9:01