Probability that one weighted mean of iid random variables is greater than the other I read somewhere that if $X_1,\dots, X_n,Y_1,\dots,Y_m$ are all i.i.d. and admit probability densities w.r.t the Lebesgue measure and we choose weights $\omega_1,\dots,\omega_n,\rho_1,\dots,\rho_m$ such that $\sum_{i=1}^n\omega_i=\sum_{i=1}^m\rho_i=1$, we have:
$$\mathbb{P}\left(\sum_{i=1}^n\omega_iX_i\leq \sum_{i=1}^m\rho_iY_i\right)=1/2.$$
I have tested this numerically and it seems to hold, but I cannot seem to prove it, does someone have an idea on how to do this?
 A: I don't think the claim is true in general.
Example 1.
Let $n=2$, $m=2$, weights $\omega=(0.9, 0.1)$ and $\rho=(0.5, 0.5)$. Let $X_i,Y_j \sim \text{Exp}(1)$. Then simulations suggest that the probability is about $0.539$.
Or another example, because the weighted sums of exponentials are a bit cumbersome to handle analytically:
Example 2. Let $F$ be the mixture distribution of 90% uniform over $[0, 0.001]$ and 10% uniform over $[1,10$]. Let $n=1$, $m=20$, $\omega_1=1$, $\rho_j=1/20$ and $X_i,Y_j \sim F$.
Now the left-hand side $X_1$ has 90% chance of being less than $0.001$. But on the right-hand side, there is $1-0.9^{20} \approx 87.8\%$ chance that at least one of the $Y_j$ is greater than 1, so then the RHS is greater than $0.05$. Clearly there is greater than 75% chance that LHS<RHS. Indeed, if you try it out, you get LHS<RHS about 81.8% of the time.
Note: If the common distribution of $X_i,Y_j$ is symmetric, then it is not difficult to show (by convolution) that also LHS and RHS have symmetric distributions; also their difference has a symmetric distribution, with mean $0$, so (by continuity) it is negative with probability $1/2$. So in that case the claim holds.
