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A random variable $X$ has variance $Var(X)=9$, $E(X) = 2$, $E(X^2) = 13$. The value of $X$ is never greater than $10$.

True of False, $Pr[X \leq 1] \leq \frac{8}{9}$ ?

It looks like Markov's Inequality can be used there but the $\leq$ must somehow change to $\geq$.

My book claims that $Pr[X \leq 1] = Pr[10 - X \geq 9]$ but I don't understand how they came about this. I was trying to take the contrapositive of the statement $X \leq 1$ but I was confused was to how they got to $10 - X \geq 9$.

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$X \leq 1 \Rightarrow X-10 \leq 1-10 \Rightarrow X-10 \leq -9 \Rightarrow 10-X\geq 9$

Using Markov inequality,

$P(X \leq 1) = P(10-X\geq9) \leq \frac{E(10-X)}{9} = \frac{10-E(X)}{9} = \frac{8}{9}$

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  • $\begingroup$ Thanks! Will accept when it lets me $\endgroup$ – Carpetfizz Jun 11 '17 at 5:38
  • $\begingroup$ By the way, how did you get $X-10$ to begin with? $\endgroup$ – Carpetfizz Jun 11 '17 at 5:39
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$$X \leq 1$$

is equivalent to

$$-X \geq -1$$

which is equivalent to

$$10-X \geq 10-1$$

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