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Question: Assume $X$ and $Y$ are independent random variables. Is $Median(XY) = Median(X) \cdot Median(Y)$? If so, how would one prove this? If not, what conditions would be sufficient for this relationship to hold?

Additional question: Does the relationship hold for $\alpha$-trimmed means?

Update: Based on a conversation with @Glen_b in the comments on his answer, as well as the contribution of @nikie, it appears that sufficient conditions for the relationship to hold are: 1) independence, and 2) at least one of the distributions of $X$ and $Y$ has a median of zero.

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    $\begingroup$ Short answer: no. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 5:20
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    $\begingroup$ There is no basis that for any of these claims to be valid. And they are not. $\endgroup$
    – crogg01
    Jan 29, 2014 at 5:29
  • $\begingroup$ @HansRoggeman I didn't think I made any claims... but just to be clear, I've edited the question to make it clear it is a question. $\endgroup$
    – Colin T Bowers
    Jan 29, 2014 at 6:10
  • $\begingroup$ I like the edited version, it's a better question. Actually, I think it's quite a good question for the site for a number of reasons - the sort of question I wish I'd thought to ask. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 6:27
  • $\begingroup$ @Glen_b Thanks for the validation. A google search of the question title reveals nothing obviously relevant, so for that reason alone I didn't think it was a bad question. $\endgroup$
    – Colin T Bowers
    Jan 29, 2014 at 8:14

2 Answers 2

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Counterexample:

Consider $X_i\sim\text{Unif}(0,1)$, $i=1,2$. Their common median is $\frac{1}{2}$.

Let $Y=X_1\, X_2$. The median of $Y$ is about $0.1867$, which is smaller than $(\frac{1}{2})^2\,\text{:}$

The log of a uniform is the negative of a standard exponential. The sum of two exponential random variables is gamma-distributed with shape 2, which (for scale 1) has median 1.67834... Hence the median of the log of the product of two uniforms is -1.67834. Exponentiation is monotonic, so the median of the product of two uniforms is $\exp(-1.67834...)\approx 0.1867$

More directly, it's relatively easy to derive the density of the product ($f(y) = \log(1/y),\quad 0<y<1$), which means the median is found by solving $m - m\log m =\frac{1}{2}$ for $m$ (which has two solutions, but only one in $(0,1)$ ).

Additional question: Does a similar relationship exist for α-trimmed means?

Yes, in the sense that it's also not true in general for trimmed means.

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    $\begingroup$ I didn't actually know any specific ones when I typed my comment (in spite of understanding it wasn't generally true) but it didn't take long to figure one out. I wanted to try to think of a continuous example for which all the calculations were simple. Didn't quite do that (since the median of a $\text{Gamma}(2)$ random variable involves the incomplete gamma function), but if I think of a simpler continuous one I will probably come back and include it. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 6:17
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    $\begingroup$ It's also a good counter-example because it demonstrates that even symmetry of the underlying densities is not sufficient for the result. Thanks again. $\endgroup$
    – Colin T Bowers
    Jan 29, 2014 at 6:23
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    $\begingroup$ In fact, symmetry about anything but zero is going to tend to lead to counterexamples. Not in every case, but pretty often. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 6:25
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    $\begingroup$ Consider the belief expressed in your first sentence ("the product of any two...") in the light of the counterexample in the answer. The product of two standard uniforms is not symmetric. But your final sentence is correct, so I suspect you meant to express a slightly different belief. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 6:35
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    $\begingroup$ It's easy to find examples where your conjecture holds, of course - an easy way is to take random variables with symmetric distributions (not necessarily the same for both) and exponentiate them. $\endgroup$
    – Glen_b
    Jan 29, 2014 at 6:40
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I suspect, but have not proven, that sufficient conditions for the relationship to hold are: 1) independence, 2) X and Y both have symmetric distributions, and 3) At least one of the distributions of X and Y is centred on zero.

I don't think you need condition 2).

Let's say X has median zero. Then we have 4 cases:

  1. x>0, y>0

  2. x>0, y<0

  3. x<0, y>0

  4. x<0, y<0

x*y > 0 will be true in cases 1 and 4.

If X has median 0, then p(x>0) = 0.5

If X and Y are independent, then p(x>0, y>0) = p(x>0) * p(y>0) (for all 4 combinations)

so p(x*y>0) = p(x>0)*p(y>0) + p(x<0)*p(y<0) = 0.5 (p(y>0)+p(y<0)) = 0.5

=> the median of x*y is also 0

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  • $\begingroup$ I just realized that I forgot the cases x=0 / y=0. I think they don't change anything except making the proof more complicated (9 cases instead of 4) $\endgroup$
    – nikie
    Jan 30, 2014 at 7:15
  • $\begingroup$ +1 This is interesting, thank you. I've updated the update. $\endgroup$
    – Colin T Bowers
    Jan 31, 2014 at 0:43

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