Show that if $X$ is a random variable such that $P(a \le X \le b) = 1$, then $E(X)$ and $Var(X)$ exist, and $a \le E(X) \le b$ and $Var(X) \le \frac{(b-a)^2}{4}$.
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A bounded random variable has a finite expectation and variance. Therefore, $E(X)$ and $Var(X)$ exist. Show more…
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Key Concepts
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Let $X$ be a random variable that takes on values between 0 and $c$. That is, $P\{0 \leq X \leq c\}=1$. Show that $$ \operatorname{Var}(X) \leq \frac{c^{2}}{4} $$ HINT: One approach is to first argue that $$ E\left[X^{2}\right] \leq c E[X] $$ Then use this to show that $$ \operatorname{Var}(X) \leq c^{2}[\alpha(1-\alpha)] \quad \text { where } \quad \alpha=\frac{E[X]}{c} $$
Continuous Random Variables
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Let $X$ be a random variable with mgf $M(t),-h<t<h$. Prove that $$ P(X \geq a) \leq e^{-a t} M(t), \quad 0<t<h $$ and that $$ P(X \leq a) \leq e^{-a t} M(t), \quad-h<t<0 $$ Hint: Let $u(x)=e^{t x}$ and $c=e^{t a}$ in Theorem 1.10.2. Note: These results imply that $P(X \geq a)$ and $P(X \leq a)$ are less than or equal to their respective least upper bounds for $e^{-a t} M(t)$ when $0<t<h$ and when $-\bar{h}<t<0$.
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