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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.

Colin T Bowers
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    Short answer: no. – Glen_b Jan 29 '14 at 05:20
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    There is no basis that for any of these claims to be valid. And they are not. – Hans Roggeman Jan 29 '14 at 05:29
  • @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. – Colin T Bowers Jan 29 '14 at 06:10
  • 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. – Glen_b Jan 29 '14 at 06:27
  • @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. – Colin T Bowers Jan 29 '14 at 08:14

2 Answers2

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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.

Glen_b
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  • Interesting. Thank you. I probably should have started off by looking for counter-examples. – Colin T Bowers Jan 29 '14 at 06:12
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    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. – Glen_b Jan 29 '14 at 06:17
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    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. – Colin T Bowers Jan 29 '14 at 06:23
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    In fact, symmetry about anything but zero is going to tend to lead to counterexamples. Not in every case, but pretty often. – Glen_b Jan 29 '14 at 06:25
  • I believe the product of any two random variables both from symmetric distributions is also symmetrically distributed. Given this and the equivalence of means and medians for symmetric distributions, I suspect independence, symmetry of the underlying densities, and at least one density centred on zero is a sufficient set of conditions for the above result to hold. Fairly stringent :-) – Colin T Bowers Jan 29 '14 at 06:34
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    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. – Glen_b Jan 29 '14 at 06:35
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    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. – Glen_b Jan 29 '14 at 06:40
  • Gah! The bit about one density centred on zero was also supposed to be in the first sentence - along with independence actually. Not my best day. – Colin T Bowers Jan 29 '14 at 08:11
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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

nikie
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