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?

up vote 0 down vote accepted

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

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.