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Suppose that $X$ is a random variable for which $\mathbb E(X) = 10$, $\Pr(X \leq 7) = 0.2$, and $\Pr(X \geq 13) = 0.3$. Prove that $\operatorname{Var}(X) \geq 4.5$.

I'm not really sure how to start solving this; thanks for any help!

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  • $\begingroup$ For a little more of a challenge, show $\operatorname{Var}(X) \geq 4.68$ rather than $4.5$ $\endgroup$ – Henry Dec 5 '19 at 13:11
  • $\begingroup$ @Henry The Intuition that $\mathsf P(X=7)=0.2, \mathsf P(X=13)=0.3$ and $\mathsf P(X=9.4)=0.5$ minimizes the Variance is clear but I am not so sure if a formal proof is easy $\endgroup$ – Maximilian Janisch Dec 7 '19 at 11:13
  • $\begingroup$ @MaximilianJanisch: If you have $\Pr(X \le 7)=0.2,\Pr(7 \lt X \lt 13)=0.5, \Pr(13 \le X)=0.3$ then (a) for given conditional means in the three intervals, the variance is minimised when the probability is concentrated at those three means, and (b) for probability concentrated at three points with the overall mean in the central interval, the variance is minimised when both the extreme points are as close to the overall mean as possible, changing the third point to accommodate this $\endgroup$ – Henry Dec 7 '19 at 13:44
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The way to start solving this is to figure out that $|10-7|=|13-10|=3$ so, plugging into the definition of variance,

$$ \text{Var}(X) = E(X-E(X))^2\geq E\left[1_{|X-E(X)|\geq 3}\right] \cdot 3^2 = 9(P(X\leq 7)+P(X\geq 13))=\frac{9}{2}, $$

where $1_{|X-E(X)|\geq 3}$ is the indicator function of the event $|X-E(X)|\geq 3$, i.e. $X \leq 7$ or $X \geq 13$, then $E\left[1_{|X-E(X)|\geq 3}\right] = P(X \leq 7$ or $X \geq 13)$.

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Apply Chebyshev's inequality: $\mathbb{P}\left(|X-\mu| \geq k\sigma \right) \leq k^{-2}$

Here set $\mu=10$, $k\sigma = 3$ so

$\mathbb{P}\left(|X-10| \geq 3\right) = 0.2+0.3 = 0.5\leq k^{-2}$

Rearranging $0.5 \leq \left(\dfrac{3}{\sigma}\right)^{-2}$ and

$Var(X) = \sigma^2 \geq \dfrac{9}{2}$.

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The variance is the average of $(X-\mathbb E(X))^2=(X-10)^2$. Let's call that random variable $Y$. What you have is that 30% of the time, $Y>3^2$ (because $X>13$), and 20% of the time, $Y\geq3^2$ (because $X\leq7$). The rest of the time, $Y$ is between $0$ and $3^2$. In summary, $Y$ is greater than or equal to $9$ at least half of the time, and the rest of the time it's positive. What does that tell you about its average value?

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