# Distribution that minimises ties

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?

• 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. – Jake Westfall Oct 21 '18 at 15:35
• 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 – Henry Oct 21 '18 at 15:48
• @Henry How so? That guarantees ties -- $A_1$ always ties with $A_2$, etc. – Ron Davis Oct 21 '18 at 15:51
• 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? – Henry Oct 21 '18 at 17:16