# Does the difference between two symmetric r.v.'s also have a symmetric distribution?

If I have two different symmetric (with respect to the median) distributions $X$ and $Y$, is the difference $X-Y$ also a symmetric (with respect to the median) distribution?

• The distribution of $X-Y$ is not a "difference between two distributions", it's the distribution of the difference between symmetrically-distributed random variables; The difference in distributions would be $F_X(t)-F_Y(t)$; which is not a distribution; similarly a difference of pdfs would not be a pdf... please amend your title description Dec 15, 2017 at 22:42
• @Glen_b: I edited the OP's title to say so, but in future please go ahead and edit it yourself. Colloquially I think everyone understood what the OP meant.
– smci
Dec 16, 2017 at 9:23
• @smci Actually, I chose to ask the OP to do it rather than doing it myself for a reason (if you check my profile you'll see I have over 3100 posts edited -- I understand the general rules about editing). Thanks for helping out, though. I also think a little more care with expressing what is meant would solve a substantial fraction of the novice questions on site; and I think clarity is especially important in a title. Dec 17, 2017 at 7:34

Let $$X \sim f(x)$$ and $$Y \sim g(y)$$ be PDFs symmetric about medians $$a$$ and $$b$$ respectively. As long as $$X$$ and $$Y$$ are independent, the probability distribution of the difference $$Z = X - Y$$ is the convolution of $$X$$ and $$-Y$$, i.e.

$$p(z) = \int_{-\infty}^\infty f(z + y) g(-y) dy,$$

where $$h(y) = g(-y)$$ is simply the PDF over $$-Y$$ with median $$-b.$$

Intuitively, we would expect the result to be symmetric about $$a - b$$ so let's try that.

$$\begin{split} p(a - b - z) &= \int_{-\infty}^\infty f(a - b - z + y) g(-y) dy \\ &= \int_{-\infty}^\infty f(a + b + z - y) g(y - 2 b) dy \\ &= \int_{-\infty}^\infty f(a - b + z + v) g(-v) dv \\ &= p(a - b + z). \end{split}$$

In the second line I used both the symmetry of $$f(x)$$ about $$a$$ and of $$g(-y)$$ about $$-b.$$ In the third line, I used the substitution $$v = 2 b - y$$ in the integral. This proves that $$p(z)$$ is symmetric about $$a - b$$ if $$f(x)$$ is symmetric about $$a$$ and $$g(y)$$ is symmetric about $$b.$$

If $$X$$ and $$Y$$ were not independent, and $$f$$ and $$g$$ were simply marginal distributions, then we would need to know the joint distribution, $$X,Y \sim h(x,y).$$ Then, in the integral, we would have to replace $$f(z + y) g(-y)$$ with $$h(z + y, -y).$$ However, just because the marginal distributions are symmetric, that does not imply that the joint distribution is symmetric about each of its arguments. So you could not apply similar reasoning.

• This example appears to prove that p(a - b - z) = p(z), but if p is symmetric around ( a - b), then I would have thought that you would need to prove that p(a - b - z) = p(a - b + z)? Apr 28, 2022 at 7:36
• @Obromios Yes, you're right! I must have had a brain fart when first writing this, by thinking the condition for symmetry about $a$ was $f(a - x) = f(x)$, because I applied that to both $f$ and $g$ when doing the derivation, and, when doing this it just so happened by weird luck that this fake condition for symmetry held for $p$! But I just edited it to make the correction. It turns out when applying the real condition for symmetry to $f$ and $g$ you still get the condition for symmetry in $p$. Thanks for pointing that out! Apr 29, 2022 at 0:30

This is going to depend on the relationship between $x$ and $y$, here is a counter example where $x$ and $y$ are symmetric, but $x-y$ is not:

$$x=[-4, -2, 0, 2, 4]$$ $$y=[-1, -3, 0, 1, 3]$$ $$x-y = [-3, 1, 0, 1, 1]$$

So here the median of $x-y$ is not the same as the difference in the medians and $x-y$ is not symmetric.

Edit

This may be clearer in @whuber's notation:

Consider the discrete uniform distribution where $x$ and $y$ are related such that you can only select one of the following pairs:

$$(x,y)=(-4,-1); (-2,-3); (0,0); (2,1); (4,3)$$

If you insist on thinking in a full joint distribution then consider the case where $x$ can take on any of the values $(-4, -2, 0, 2, 4)$ and $y$ can take the values $(-3, -1, 0, 1, 3)$ and the combination can take on any of the 25 pairs. But the probability of the given pairs above is 16% each and all the other possible pairs have probability of 1% each. The marginal distribution of $x$ will be discrete uniform which each value having 20% probability and therefore symmetric about the median of 0, the same is true for $y$. Take a large sample from the joint distribution and look at just $x$ or just $y$ and you will see a uniform marginal distribution (symmetric), but take the difference $x-y$ and the result will not be symmetric.

• I don't understand this example at all. If $X$ can be equal to 4 and $Y$ can be equal to e.g. 1, then $X-Y$ should be able to be 3, but you don't list this possibility. Maybe I misunderstand your example; what are these three vectors? Dec 15, 2017 at 19:33
• $x$ and $y$ are not independent in his example. Think of $x$, $y$, and $x-y$ as being functions of some random variable $i$ which indexes into each vector. Then if $i=0$, $x=-4$, $y=-1$, and $x-y=-3$ Dec 15, 2017 at 21:05
• If you are considering $x$ and $y$ not to be independent, then you are really viewing $(x,y)$ as a bivariate random variable. As such what you demonstrate is that symmetric marginals do not imply the joint distribution is symmetric. That's a fine observation, but the notation in this answer is confusing. It might be clearer to describe the data in a bivariate notation as $(x,y)=(-4,-1),(-2,-3),(0,0),(2,1),(4,3)$.
– whuber
Dec 15, 2017 at 21:20
• @amoeba, It depends on the relationship between $X$ and $Y$, if they are independent or weakly dependent, then yes there could be a case like you say, but my example is strong dependence between the 2 variables. If X were height in inches and y were height in centimeters then $X=10$ is a possible value, and $Y=1$ is a possible value, but not at the same time for the same object. Dec 15, 2017 at 22:23
• The comments and the edit have clarified what you meant. Thanks. Dec 16, 2017 at 0:04

You'll need to assume independence between X and Y for this to hold in general. The result follows directly since the distribution of $X-Y$ is a convolution of symmetric functions, which is also symmetric.