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

Dec 3, 2017

$E \left(X\right) = \frac{\alpha}{3}$

#### Explanation:

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

$E \left(X\right) = {\int}_{a l l \text{ } x} x f \left(x\right) \mathrm{dx}$

$E \left(X\right) = {\int}_{- 1}^{1} x \left(\frac{1 + \alpha x}{2}\right) \mathrm{dx}$

$E \left(X\right) = \frac{1}{2} {\int}_{- 1}^{1} \left(x + \alpha {x}^{2}\right) \mathrm{dx}$

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

$E \left(X\right) = \frac{1}{2} \left\{{\left[{x}^{2} / 2 + \alpha {x}^{3} / 3\right]}^{1} - {\left[{x}^{2} / 2 + \alpha {x}^{3} / 3\right]}_{0}\right\}$

$E \left(X\right) = \frac{1}{2} \left\{\left(\cancel{\frac{1}{2}} + \frac{\alpha}{3}\right) - \left(\cancel{\frac{1}{2}} - \frac{\alpha}{3}\right)\right\}$

$E \left(X\right) = \frac{1}{2} \left(\frac{\alpha}{3} + \frac{\alpha}{3}\right) = \frac{1}{2} \times \frac{2 \alpha}{3}$

$E \left(X\right) = \frac{\alpha}{3}$

Dec 3, 2017

$E \left(X\right) = \frac{a}{3}$

#### Explanation:

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

$E \left(X\right) = {\int}_{- \infty}^{\infty} \setminus x f \left(x\right) \setminus \mathrm{dx}$

So given that $f \left(x\right) = \frac{1 + a x}{2}$ for $- 1 \le x \le 1$ then we have:

$E \left(X\right) = {\int}_{- 1}^{1} \setminus x \setminus \left(\frac{1 + a x}{2}\right) \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{- 1}^{1} \setminus x \left(1 + a x\right) \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{- 1}^{1} \setminus x + a {x}^{2} \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left[{x}^{2} / 2 + \frac{a {x}^{3}}{3}\right]}_{- 1}^{1}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \left\{\left(\frac{1}{2} + \frac{a}{3}\right) - \left(\frac{1}{2} - \frac{a}{3}\right)\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \left(\frac{1}{2} + \frac{a}{3} - \frac{1}{2} + \frac{a}{3}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \frac{2 a}{3}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{a}{3}$