Let $X_1,X_2$, and $X_3$ be independent random variables such that $X_1+X_2$ and $X_1+X_3$ have the same distribution. Does it follow that $X_2$ and $X_3$ have the same distribution?

Can this be answered without referring to characteristic functions?

Due to independence, the PGF of $U=X_1+X_2$ is given by $P_U(t)=P_{X_1}(t)P_{X_2}(t)$ and the same for $V=X_1+X_3$ is given by $P_V(t)=P_{X_1}(t)P_{X_3}(t)$

So we have $P_U(t)=P_V(t)\qquad$ (assuming convergence of the PGF's for some $t$)

$\Rightarrow P_{X_1}(t)P_{X_2}(t)=P_{X_1}(t)P_{X_3}(t)$

Now can I conclude from here that $P_{X_2}(t)=P_{X_3}(t)$? Moreover, I am restricting the random variables to be non-negative integer valued if I proceed like this.


2 Answers 2


OK, for this answer I will assume the independent random variables have moment generating functions (mgf) existing in some open interval containing zero. To dispel one possible doubt, mgf's can never be zero, since they are given by expectation of an exponential function $\DeclareMathOperator{\E}{\mathbb{E}} M_X(t) = \E e^{t X}$. Exponential functions are everywhere positive so that expectation can never be zero (or negative).

So, the assumption is that $X_1 +X_2$ have the same distribution as $X_1+X_3$. Write the mgf's as $M_i(t)=M_{X_i}(t)$. It follows that $$ M_1(t) M_2(t) = M_1(t) M_3(t) $$ for all $t$ (where all three are defined, which will be some open interval containing zero). Since these functions can never be zero, we can divide on both sides by $M_1(t)$, obtaining $M_2(t)=M_3(t)$. We can conclude that $X_2$ and $X_3$ have the same distribution. For a nice review of properties of mgf's, see Existence of the moment generating function and variance

Now, what to do if mgf's do not exist? Replace the mgf's in the proof with characteristic functions. That becomes somewhat more involved, since that involves complex numbers, and my proof above that mgf's can never take the value zero do not work in that case. But we can save the argument in that case in the following way: a characteristic function is always (uniformly) continuous, on the entire space, see https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) . Since it always take the value 1 at the argument zero, we can find an open interval containing zero where none of the three characteristic functions have zeros, so the cancellation works as above (this argument also works for moment generating functions).

  • $\begingroup$ Yes I wanted to see it explicitly mentioned whether they have the same distribution or not. One more thing is that if I use pgf, then the RVs are assumed to be discrete. What about when they are continuous? I suppose using mgf nullifies this issue. $\endgroup$ Commented Sep 17, 2017 at 17:02
  • $\begingroup$ For pgf versus mgf see stats.stackexchange.com/questions/297711/… $\endgroup$ Commented Sep 17, 2017 at 17:03
  • $\begingroup$ I am probably missing something, but tollowing Polya's theorem (see en.wikipedia.org/wiki/…) it seems to be possible to build two characteristic functions with the same values in a neighborhood of 0. Therefore if $M_1$ vanishes where $M_2$ and $M_3$ are not identical any more, the equation holds... $\endgroup$
    – RUser4512
    Commented Mar 14, 2018 at 17:35

In general, no.

The proof given by @kjetil b halvorsen is fine, as long as the moment generating function exists.

However, there are counter examples when it does not. First, we need to recall a theorem by Polya. (More details can be found in the article Remarks on characterisic functions by G. Polya)

If φ is a real-valued, even, continuous function which satisfies the conditions

  • $\varphi(0) = 1$
  • $\varphi$ is convex for $ t<0$,
  • $\varphi(\infty) = 0 $,

then φ(t) is the characteristic function of an absolutely continuous symmetric distribution.

Now from this theorem, the two functions below are characteristic functions of some random variables, call them $X_1$ and $X_2$.

enter image description here

Now call $X$ the random variable whose characteristic function is $ x \rightarrow (1-|x|)^+$ (again, using Polya's theorem).

Now, $\phi_{X_1}\phi_{X}=\phi_{X_2}\phi_{X}$ everywhere, so that $X_1+X\stackrel{d}=X_2+X$ but $X_1$ and $X_2$ have different distributions.

  • $\begingroup$ Very helpful, thanks. I was looking for a counterexample when I posted the question. $\endgroup$ Commented Mar 19, 2018 at 15:09

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