# difference of two random variables: symmetric distribution if only location parameter is different?

In Box and Tiao (1973) page# 156, authors write that if the distributions of two random variables are identical except location, then the distribution of the differences would certainly be symmetric. In other words, if two random variables are identical except in their mean, then the difference of the two random variables would be symmetric.

But authors have not provided any proof for this claim. Maybe because it is supposed to be obvious. But I am not able to understand why this statement is true? It will be helpful if someone can show a proof for this claim or an intuitive explanation

Suppose $Y$ and $Y^\prime$ are the random variables in question, where there is a number $\mu$ for which $Z=Y^\prime-\mu$ and $Y$ are identically distributed. (That's what "identical except location" means.) The question concerns the random variable $$X=Y^\prime-Y = \mu + Z - Y.$$

The underlying idea is that when $Z$ and $Y$ can be swapped without changing their joint distribution (they are "exchangeable"), then

$$2\mu - X = \mu + Y - Z$$

obviously has the same distribution as $X,$ because it arises through exchanging $Y$ and $Z.$ (It is likely the authors implicitly assumed--or even previously stated--that $Y$ and $Y^\prime$ are independent, which easily implies the exchangeability of $Z$ and $Y.$)

The equivalence in distribution of $X$ and $2\mu-X$ is the very definition of symmetric.

If a random variable $T$ has a distribution symmetric about zero, then $T\stackrel{D}{=} -T$, meaning that $T$ and $-T$ has the same distribution. If $T$ has a distribution symmetric about some value $\mu$, then $T-\mu$ is symmetric about zero, so that $T-\mu \stackrel{D}{=} -(T-\mu) = -T+\mu$.

Let $X_1, Y_1$ be two independent random variables, identical in distribution. Then define $X=X_1-\theta, Y=Y_1$. Then $X$ and $Y$ will be "identical except location" (in distribution) (I take this to be the authors meaning). Then $D_1 = X_1-Y_1$ is symmetric about zero: $D_1 = X_1-Y_1\stackrel{D}{=} Y_1-X_1=-(X_1-Y_1)=-D_1$.

Let $D=X-Y$. To show that $D$ has a symmetric distribution (about some center $\mu$), so: $X-Y= X_1-\theta -Y_1=(X_1-Y_1)-\theta=D_1-\theta \stackrel{D}{=} -D_1-\theta = -(D+\theta)-\theta$ and by adding $\theta$ on both sides we finally get $D+\theta \stackrel{D}{=} -D-\theta$ concluding the proof.

Note that we dont even need the independence of $X_1$ and $Y_1$. It is enough to assume that $X_1, Y_1$ are excahgeable, that is: $(X_1,Y_1)\stackrel{D}{=} (Y_1, X_1)$, and everything goes through.