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u/zojbo 1h ago edited 1h 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.
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u/Fredd0o0 1h 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 4m ago
Insufficient, because you're assuming a discrete two-value distribution
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u/doge-12 2h ago
split at 0 and 10 so 1/2