SOLUTION: Suppose that x is a random variable with the possible values 1,2,and 3. Given that E(X^2)<=6, find the maxium possible value of E(X).

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Question 1034101: Suppose that x is a random variable with the possible values 1,2,and 3. Given that E(X^2)<=6, find the maxium possible value of E(X).
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let p(1) = p, p(2) = q, and p(3) = 1 - p - q.
Now E%28X%5E2%29+=+1%5E2%2Ap%2B2%5E2%2Aq%2B3%5E2%2A%281-p-q%29+=+p%2B4q%2B9%281-p-q%29+=+9-8p-5q.
Since E%28X%5E2%29+%3C=+6, ==> 9-8p-5q+%3C=6 ==> 8p%2B5q+%3E=+3.
Also, there are other restrictions, like 0+%3C=p%3C=1, 0+%3C=q%3C=1, and p%2Bq+%3C=1.
Now E%28X%29+=+1%2Ap%2B2%2Aq%2B3%2A%281-p-q%29+=+p%2B2q%2B3%281-p-q%29+=+3-2p-q.
This now becomes a linear programming problem, namely,
Maximize E+=+3-2p-q,
subject to
8p%2B5q+%3E=+3,
p%2Bq+%3C=1,
0+%3C=p%3C=1, and
0+%3C=q%3C=1.
Letting p be the horizontal axis, and q the vertical axis, using the graphical method, the region of feasibility is found to be a quadrilateral with the corner points (0,3/5), (0,1), (1,0), and (3/8,0).
Now...
E(0,3/5) = 2.4,
E(0,1) = 2,
E(1,0) = 1, and
E(3/8,0) = 2.25.
Therefore the maximum possible value of E(X) is highlight%282.4%29, and happens when p(1) = 0, p(2) = 3/5, and p(3) = 2/5.
(This answer is also consistent with the upper and lower bounds for E(X). Var%28X%29%3E=0 ==> E%28X%5E2%29+%3E=+%28E%28X%29%29%5E2 ==> 6%3E=+%28E%28X%29%29%5E2 ==> sqrt%286%29+%3E=+E%28X%29+%3E=+-sqrt%286%29. sqrt%286%29 is around 2.4495 to 5 significant figures.)