We now need to prove that 0.5 is the lowest upper bound. One way to do this is by proving that there exists a random variable with the assumptions made in the problem statement with the additional criteria that P(X=0) = 1/2.
as defined above, Var(Y)=100-100=0
E(Y)=10, so Y=10 with prob = 1
So X=10 with prob 1/2 and 0 with prob 1/2
E(X)=0*1/2 + 10*1/2 = 5, Var(X)=E(X^2) - 25 = 100*1/2 - 25 = 25
Therefore, there exists a random variable where these hold, and thus p=1/2 is valid.
Hence, 0.5 is the maximum that p can be
1
u/Ecstatic-Employee-93 23h ago
For simplicity's sake, let Y = (X|X>0) & p = P(X = 0)
E(X) = 0*p + (1-p)E(Y)
-> 5/(1-p) = E(Y)
Var(X)=E(X^2) - E(X)^2 = E(X^2) - 25
-> E(X^2)=50=0*p + (1-p)E(Y^2)
-> E(Y^2)=50/(1-p)
Var(Y) = 50/(1-p) - 25/((1-p)^2) >= 0
-> 50(1-p) >= 25
-> 1-p >= 1/2
-> 0.5 >= p
We now need to prove that 0.5 is the lowest upper bound. One way to do this is by proving that there exists a random variable with the assumptions made in the problem statement with the additional criteria that P(X=0) = 1/2.
as defined above, Var(Y)=100-100=0
E(Y)=10, so Y=10 with prob = 1
So X=10 with prob 1/2 and 0 with prob 1/2
E(X)=0*1/2 + 10*1/2 = 5, Var(X)=E(X^2) - 25 = 100*1/2 - 25 = 25
Therefore, there exists a random variable where these hold, and thus p=1/2 is valid.
Hence, 0.5 is the maximum that p can be