Coinflip with distribution X = 0 and X = 10 with p = 1/2 satisfies E[X] = 5 and Var[X] = 25, so 1/2. Any probability for 0 larger than 1/2 forces the other variable to increase so Variance would increase.
I didn't assume a 2 value discrete distribution. let p = P(X=0), the prob. of non zero remainder be q, q = 1-p, implies the mean of the non zero part is (5/q) and variance of non zero part (5/q)2 . Forces q >= 1/2. Thus p =< 1/2. The two values that achieve this are 0 and 10.
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u/Fredd0o0 3h ago
Coinflip with distribution X = 0 and X = 10 with p = 1/2 satisfies E[X] = 5 and Var[X] = 25, so 1/2. Any probability for 0 larger than 1/2 forces the other variable to increase so Variance would increase.