Chapter 5: Q 5.8 (page 215)

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
$V a r (X) \leq \frac{c^{2}}{4}$
Hint: One approach is to first argue that
localid="1646883602992" $E [X^{2}] \leq c E [X]$
and then use this inequality to show that
$V a r (X) \leq c^{2} [α (1 - α)] w h e r e α = \frac{E [X]}{c}$

Short Answer

Expert verified

Hence, proved that $V a r (X) \leq \frac{c^{2}}{4}$ .

Step by step solution

Given information:

Let $X$ be a random variable that takes on values between $0$ and $c$ . That is, $P \{0 \leq X \leq c\} = 1$

Hint: One approach is to first argue that

$E [X^{2}] \leq c E [X]$

and then use this inequality to show that

$V a r (X) \leq c^{2} [α (1 - α)] w h e r e α = \frac{E [X]}{c}$

Explanation:

Since $X$ be, a random variable that takes on values between $0$ and $c$ , we know that

$f (x) = 0 i f x \leq 0 o r x > 0$

Thus,

$E [X^{2}] = \int_{0}^{c} x^{2} f (x) d x \leq \int_{0}^{c} c x f (x) d x = c E [X]$

Explanation:

Hence,

$V a r (X) = E [X^{2}] - E {[X]}^{2} \leq c E [X] - E {[X]}^{2} = c^{2} (α - α^{2})$

Where $α = \frac{E [X]}{c}$ we know $α - α^{2}$ attains its maximum value of $\frac{1}{4} a t α = \frac{1}{2}$ so

$V a r (X) \leq c^{2} {(α - α^{2})}_{α = \frac{1}{2}} = c^{2} (\frac{1}{2} - \frac{1}{4}) = \frac{c^{2}}{4}$

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Let Xbe a random variable that takes on values between0andc. That isP0≤X≤c=1.Show thatVar(X)≤c24Hint: One approach is to first argue thatlocalid="1646883602992" EX2≤cEXand then use this inequality to show thatVar(X)≤c2[α(1-α)]whereα=E[X]c

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