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Let $X$ be a random variable such that

  • $\mathbb{P}(X\in[0,10])=1$
  • $\mathbb{E}(X)=2$
  • $\mathbb{P}(X<2)\leq1/2$

Find the supremum of all possible values of $\text{var}(X)$.

I have some intuition, but I'm not sure if it's correct.

  1. Is it true that in order to maximize the variance of $X$ we should aim to assign the highest possible probabilities to $10$ and $0$? If so, why is that?
  2. The second condition tells us that the area above the cumulative distribution function of $X$ equals $2$. Since $\mathbb{P}(X<2)\leq1/2$, the area above the cdf, when restricted to $[0,2)$, is greater or equal to $1$. So the area above the cdf on $[2,10]$ cannot be greater than $1$. For our needs, we take that value to be $1$, right?
  3. To assign the highest probability to $10$ we make the cdf constant on $(2,10]$. Then $P(X=10)=1/8$. Is this the right approach?

Any ideas and explanations will be appreciated.

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Variance usually measure the spread of your distribution. If your distribution is concentrated around your mean, the variance is less bcoz $Var(X) = E(X-E(X))^2$. For maximizing the variance you should assign the highest probability to the value which is far away from the mean ( usually end of your interval)

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In order to find the maximum of $E(X^2)$, note that $P(X>2)\ge 0.5$, so the maximum is obtained by shifting all the mass to the right, while still satisfying the mean criterion. The maximum you could go towards the right is $X=4$ with a probability 0.5, and you could satisfy the mean by putting the remaining mass of 0.5 on $X=0$.

Thus $\sup_{p_X}Var(X)=4^2*0.5-4=4.$

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however to find the maximum of $E(X^2)$ we dont have to consider $P(X>2)$. am i right? – Learner Feb 28 '13 at 11:43
@Learner: In this problem, the bound $ℙ(X<2)≤1/2$ turns uninteresting, but it could become if, say, the interval spans both positive and negative values. – Bravo Feb 28 '13 at 11:52
@Shyam If $\mathbb{P}(X=10)=1$, then how does the assumption that $\mathbb{E}(X)=2$ remain fullfilled? – Johnny Westerling Feb 28 '13 at 12:34
@JohnnyWesterling: Thanks, I stand corrected. – Bravo Feb 28 '13 at 14:57
@Shyam I'm afraid your answer is wrong. The correct answer happens to be $10$. Check this: $P(X=10)=1/8$, $P(X=2)=3/8$, $P(X=0)=1/2$. Even though the probability assigned to $10$ is pretty small, by squaring $10$ and multiplying by that probability we're better off than by squaring $4$ and multiplying by the max. $1$. – czachur Mar 2 '13 at 11:55

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