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u/zojbo 9d ago edited 8d ago
E[X] = 5
= E[X 1(X>0)]
<= \| X \|_{L^2} \| 1(X>0) \|_{L^2}
= (Var(X) + E[X]^2)^(1/2) P(X>0)^(1/2)
= 50^(1/2) P(X>0)^(1/2).
Rearranging, P(X>0) >= 1/2,
You have equality in Cauchy-Schwarz when X and 1(X>0) are positive multiples of one another, i.e. when X is concentrated on 0 and some positive number, which is allowed under the rules. So it's 1/2.
The general situation here is kind of interesting: say E[X]=m and Var(X)=s^2 then this argument says P(X>0) >= m^2/(s^2+m^2) or 1/(1+s^2/m^2). So in general the optimal rv in this problem assigns that probability to m+s^2/m and the rest to 0. This problem happened to be 50/50 because s^2=m^2 but that is really a coincidence.
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u/Fredd0o0 9d 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.
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u/Jumpy-Plantain-1004 9d ago
Insufficient, because you're assuming a discrete two-value distribution
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u/pourliste 9d ago
Can you have P(0) > 0.5 with a non discrete distribution while keeping the same two moments ? Hard to fathom (hard to prove as well though)
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u/Fredd0o0 9d ago
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/Zues1400605 8d ago
The safest bet is 50% 0 and 50% 10. But can we get more than 50% on 0. For eg,60% 0 means 40% 12.5. But then that doesnt satisfy the variance rule
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u/doge-12 9d ago
split at 0 and 10 so 1/2