Alexander Day

2022-02-13

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

Javon Ross

Answer: $E\left(X\right)=\frac{\alpha }{3}$
Explanation: for a pdf f(x) the expectation of X
$E\left(X\right)={\int }_{\text{all}x}xf\left(x\right)dx$
$E\left(X\right)={\int }_{-1}^{1}x\left(\frac{1+\alpha x}{2}\right)dx$
$E\left(X\right)=\frac{1}{2}{\int }_{-1}^{1}\left(x+\alpha {x}^{2}\right)dx$
$E\left(X\right)=\frac{1}{2}{\left[\frac{{x}^{2}}{2}+\alpha \frac{{x}^{3}}{3}\right]}_{-1}^{1}$
$E\left(X\right)=\frac{1}{2}\left\{{\left[\frac{{x}^{2}}{2}+\alpha \frac{{x}^{3}}{3}\right]}^{1}-{\left[\frac{{x}^{2}}{2}+\alpha \frac{{x}^{3}}{3}\right]}_{0}\right\}$
$E\left(X\right)=\frac{1}{2}\left\{\left(\text{⧸}\frac{1}{2}+\frac{\alpha }{3}\right)-\left(\text{⧸}\frac{1}{2}-\frac{\alpha }{3}\right)\right\}$
$E\left(X\right)=\frac{1}{2}\left(\frac{\alpha }{3}+\frac{\alpha }{3}\right)=\frac{1}{2}×\frac{2\alpha }{3}$
$E\left(X\right)=\frac{\alpha }{3}$

Hannah Escobar

Explanation:
By definition if f(x) is a continuous probability density function then:
$E\left(X\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf\left(x\right)dx$
So given that then we have:
$E\left(X\right)={\int }_{-1}^{1}x\left(\frac{1+ax}{2}\right)dx$
$=\frac{1}{2}{\int }_{-1}^{1}x\left(1+ax\right)dx$
$=\frac{1}{2}{\int }_{-1}^{1}x+a{x}^{2}dx$
$=\frac{1}{2}{\left[\frac{{x}^{2}}{2}+\frac{a{x}^{3}}{3}\right]}_{-1}^{1}$
$=\frac{1}{2}\left\{\left(\frac{1}{2}+\frac{a}{3}\right)-\left(\frac{1}{2}-\frac{a}{3}\right)\right\}$
$=\frac{1}{2}\left(\frac{1}{2}+\frac{a}{3}-\frac{1}{2}+\frac{a}{3}\right)$
$=\frac{1}{2}\frac{2a}{3}$
$=\frac{a}{3}$

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