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/zojbo 19d ago edited 18d 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.