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Here's a result I'm trying to get as part of a larger problem I'm solving: The random variables $A_1,B_1...J_1$ and $A_2,B_2,...J_2$ can take integer values between 0 and 100 such that $A_1+...+J_1=100$, $A_2+...+J_2=100$ and random variables with the same letter (e.g. $C_1,C_2$) must be identically distributed.

What is the distribution that minimises the expected weighted sum of ties, where the weights are given as "A : 1", ... "J: 10"? Intuitively, it should be the distribution with the highest variance. Is this right?

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  • $\begingroup$ I don't think you'd want the distribution with the highest variance. I haven't checked, but I'd guess that the highest-variance distribution would be something like having $A_i$=91 and $B_i,...,J_i$ = 1, which has many ties. $\endgroup$ – Jake Westfall Oct 21 '18 at 15:35
  • $\begingroup$ while $A_i=5, B_i=6, C_i=7, D_i=8, E_i=9,F_i=11,G_i=12, H_i=13, I_i=14, J_i=15$ with probability $1$ presumably minimises $\sum P(X_i=Y_i)$ - though your edit may have made "ties" more ambiguous $\endgroup$ – Henry Oct 21 '18 at 15:48
  • $\begingroup$ @Henry How so? That guarantees ties -- $A_1$ always ties with $A_2$, etc. $\endgroup$ – Ron Davis Oct 21 '18 at 15:51
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    $\begingroup$ In your pre-edit version, a "tie" involved $X_i=Y_i$. Are you now saying a "tie" is $X_1=X_2$, or even $X_1=Y_2$? You say $X_1$ and $X_2$ are identically distributed; are they independent? $\endgroup$ – Henry Oct 21 '18 at 17:16

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