If the probability density function of $X$ $X$ is $f (x) = \frac{1 + a x}{2}, - 1 \leq x \leq 1, - 1 \leq a \leq 1$ $f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1$ then the expectation of $X$ $X$ is? (a) $\frac{6}{a}$ $6/a$ (b) $\frac{a}{3}$ $a /3$ (c) $\frac{a}{2}$ $a /2$ (d) $\frac{3}{a}$ $3/a$

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If the probability density function of $X$ $X$ is $f (x) = \frac{1 + a x}{2}, - 1 \leq x \leq 1, - 1 \leq a \leq 1$ $f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1$ then the expectation of $X$ $X$ is? (a) $\frac{6}{a}$ $6/a$ (b) $\frac{a}{3}$ $a /3$ (c) $\frac{a}{2}$ $a /2$ (d) $\frac{3}{a}$ $3/a$

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Answer 1 · 2017-12-03T17:09:28

$E (X) = \frac{α}{3}$ $E(X)=alpha/3$

Explanation:

for a pdf $f (x)$ $f(x)$ the expectation of X

$E (X) = \int_{a l l x} x f (x) d x$ $E(X)=int_(all" "x)xf(x)dx$

$E (X) = \int_{- 1}^{1} x (\frac{1 + α x}{2}) d x$ $E(X)=int_(-1)^1x((1+alphax)/2)dx$

$E (X) = \frac{1}{2} \int_{- 1}^{1} (x + α x^{2}) d x$ $E(X)=1/2int_(-1)^1(x+alphax^2)dx$

$E (X) = \frac{1}{2} {[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}_{- 1}^{1}$ $E(X)=1/2[x^2/2+alphax^3/3]_(-1)^1$

$E (X) = \frac{1}{2} {{[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}^{1} - {[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}_{0}}$ $E(X)=1/2{[x^2/2+alphax^3/3]^1-[x^2/2+alphax^3/3]_0}$

$E(X)=1/2{(cancel(1/2)+alpha/3)-(cancel(1/2)-alpha/3)}$

$E(X)=1/2(alpha/3+alpha/3)=1/2xx(2alpha)/3$

$E(X)=alpha/3$

Answer 2 · 2017-12-03T17:34:02

$E(X) =a/3$

making (b) the correct answer

Explanation:

By definition if $f(x)$ is a continuous probability density function then:

$E(X) =int_(-oo)^(oo) \ xf(x) \ dx$

So given that $f(x) = (1+ax)/2$ for $-1 le x le 1$ then we have:

$E(X) = int_(-1)^(1) \ x \ ((1+ax)/2) \ dx$

$\ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x (1+ax) \ dx$

$\ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x+ax^2 \ dx$

$\ \ \ \ \ \ \ \ = 1/2 [ x^2/2+(ax^3)/3 ]_(-1)^(1)$

$\ \ \ \ \ \ \ \ = 1/2 {(1/2+a/3)-(1/2-a/3)}$

$\ \ \ \ \ \ \ \ = 1/2 (1/2+a/3-1/2+a/3)$

$\ \ \ \ \ \ \ \ = 1/2 (2a)/3$

$\ \ \ \ \ \ \ \ = a/3$

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If the probability density function of $X$ $X$ is $f (x) = \frac{1 + a x}{2}, - 1 \leq x \leq 1, - 1 \leq a \leq 1$ $f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1$ then the expectation of $X$ $X$ is? (a) $\frac{6}{a}$ $6/a$ (b) $\frac{a}{3}$ $a /3$ (c) $\frac{a}{2}$ $a /2$ (d) $\frac{3}{a}$ $3/a$

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If the probability density function of XXX is f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1f(x)=1+ax2,−1≤x≤1,−1≤a≤1f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1 then the expectation of XXX is? (a) 6/a6a6/a (b) a /3a3a /3 (c) a /2a2a /2 (d) 3/a3a3/a

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Explanation:

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If the probability density function of $X$ $X$ is $f (x) = \frac{1 + a x}{2}, - 1 \leq x \leq 1, - 1 \leq a \leq 1$ $f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1$ then the expectation of $X$ $X$ is? (a) $\frac{6}{a}$ $6/a$ (b) $\frac{a}{3}$ $a /3$ (c) $\frac{a}{2}$ $a /2$ (d) $\frac{3}{a}$ $3/a$