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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?

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    $\begingroup$ 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 $\endgroup$ – Glen_b Dec 15 '17 at 22:42
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    $\begingroup$ @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. $\endgroup$ – smci Dec 16 '17 at 9:23
  • $\begingroup$ @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. $\endgroup$ – Glen_b Dec 17 '17 at 7:34
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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.

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

In the second line I used the substitution $v = b - y$ in the integral. In the third line, I used both the symmetry of $f(x)$ about $a$ and of $g(-y)$ about $-b.$ 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.

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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.

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    $\begingroup$ 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? $\endgroup$ – amoeba Dec 15 '17 at 19:33
  • $\begingroup$ $x$ 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 $x-y=-3$ $\endgroup$ – Moormanly Dec 15 '17 at 21:05
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    $\begingroup$ 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 $(x,y)=(-4,-1),(-2,-3),(0,0),(2,1),(4,3)$. $\endgroup$ – whuber Dec 15 '17 at 21:20
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    $\begingroup$ @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. $\endgroup$ – Greg Snow Dec 15 '17 at 22:23
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    $\begingroup$ The comments and the edit have clarified what you meant. Thanks. $\endgroup$ – amoeba Dec 16 '17 at 0:04
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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.

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