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?


1 Answer 1


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 $B$.

  • $\begingroup$ 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? $\endgroup$ Sep 28, 2018 at 2:06
  • $\begingroup$ I learned a lot about MGFs and a bit about $\chi^2$ distributions while trying to figure out the answer to your question. :) $\endgroup$
    – irchans
    Sep 28, 2018 at 8:24
  • $\begingroup$ @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). $\endgroup$
    – pre-kidney
    Sep 28, 2018 at 9:01

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