Given independence, is the median of a product equal to the product of the medians? 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.
 A: 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.
A: 
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:


*

*x>0, y>0

*x>0, y<0

*x<0, y>0

*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
