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The probability that $X \in (-1, 3)$ equals $0$ ;

Expectation: $\mathbb{E} X=0$

How to find minimal variance?

I suppose that the answer is 3. Is that right? And how to prove it?

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up vote 4 down vote accepted

We should use Markov's inequality.Note that $$P(-2<(X-1)<2)=0$$ Consequently, $$P((X-1)^2<4)=0$$ Next, Markov inequality for $Y=(X-1)^2$: $$P(Y \geq 4) \leq E(Y)/4 = E(X^2-2X+1)/4 = (E(X^2)+1)/4$$ Because $E(X) = 0$. Then, $D(X)=E(X^2)$. So, $$E(X^2)\geq P(Y \geq 4)-1=3$$

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Hint: the smallest variance will occur when the entire probability mass is at $−1$ and $3$ and distributed so that the mean is $0$.

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Why not? if expactation 0, then P(x=-1)=3/4 and P(x=3)=1/4; E(X^2)=1*3/4+9*1/4=3 – Xxx Dec 4 '12 at 17:20
@Xxx You are correct. I used $(-1,2)$ instead of $(-1,3)$. I will edit my answer. Thanks. – Dilip Sarwate Dec 4 '12 at 18:14

Let $X$ be a random variable such that $\mathbb{E}X=0$, $\mathbb{P}[X \in (-1,3)]=0$. You want to solve

$$\mathbb{E}((X-\mathbb{E}X)^2)=\mathbb{E}(X^2) = \int_{[X \leq -1]} X^2 \, d\mathbb{P}+ \int_{[X \geq 3]} X^2 \, d\mathbb{P} \to \min$$

One can easily see that the minimum is attained for $X \sim p_1 \cdot \delta_{-1}+p_2 \cdot \delta_{3}$. From the condition $\mathbb{E}X=0$ and $p_1+p_2=1$ you obtain $p_1$, $p_2$.

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$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\mathbb E}$

Let $$X=\begin{cases} 3 & \text{with probability }1/4, \\ -1 & \text{with probability }3/4. \end{cases}$$ This puts the values that $X$ can assume as close to $0$ as they can get without violating the constraints you've set. We get $\E(X)=0$ and $\var(X) = 3$.

Later edit: Suppose $Y$ is some other r.v. satisfying the constraints. Then $$ \E(Y) = \E(Y\mid Y\ge3)\Pr(Y\ge3)+\E(Y\mid Y\le-1)\Pr(Y\le-1)=0, $$ and \begin{align} \var(Y) & = \E(\var(Y\mid 1_{Y\ge3}))+\var(E(Y\mid 1_{Y\ge3})) \\[12pt] & = \overbrace{\var(Y\mid Y\ge3)\Pr(Y\ge3)+\var(Y\mid Y\le-1)\Pr(Y\le-1)}^\text{unexplained component of the variance} \\[4pt] &{}\quad {}+ \overbrace{\underbrace{(\E(Y\mid Y\ge3)-\E(Y\mid Y\le-1))^2}_\text{square}\ \underbrace{\Pr(Y\ge3)\Pr(Y\le-1)}_\text{product}}^\text{explained component of the variance}. \end{align} The "unexplained" component of the variance will be $0$ precisely in the case ofthe r.v. $X$ defined above.

The "product" is of the form $p(1-p)$ and gets bigger as $p$ gets closer to $1/2$ and smaller as $p$ gets closer to either endpoint. The "square" is minimized subect to the required constraints by the variable $X$ defined above. So our only hope of making $\var(Y)$ smaller than $3=\var(X)$ must lie in making the "product" small. That requires making $p$ closer to $1$ or to $0$.

Probably I'll continue this further later . . . . . . .

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and now we need to show that other variance will be bigger, don't we? – Xxx Dec 4 '12 at 17:30
Yes.... Maybe I'll expand this further. If one takes this $X$ and writes $Y=X+\varepsilon$, where $\varepsilon$ is always nonnegative when $X=3$ and and nonpositive when $X=-1$, then it's easy to show that $\operatorname{var}(Y)\ge\operatorname{var}(X)$, and if the distribution of $\varepsilon$ is suitably chosen, then $\mathbb E(Y)=0$. But then there's the question of altering the weights $1/4$ and $3/4$ and compensating to keep the expectation of $Y$ equal to $0$ by altering the distribution of $\varepsilon$. – Michael Hardy Dec 4 '12 at 18:22

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