Answer to Question #210150 in Statistics and Probability for Ri Jin

Question

Let X be a continuous random variable with density function $f(x) = \begin{cases} \frac{x}{2} &\text{for } 0 \le x \le 2 \\ 0 &\text{, } otherwise \end{cases}$

Find E[|X-E[X]|].

Answer 1

$f(x) = \begin{cases} \frac{x}{2} &\text{for } 0 \le x \le 2 \\ 0 &\text{, } otherwise \end{cases} ​$

$E[X]= \int_{-\infty}^\infty xf(x)dx$

Here,

$E[X]=\int _0^2x\times \frac{x}{2}dx=\int_0^2\frac{x^2}{2}dx=\dfrac{x^3}{6}]_0^2=\dfrac{4}{3}$

Now,

$E[|X-E[X]|]=E[|X-\frac{4}{3}|]$

So,

$E[|X-\frac{4}{3}|]=\int_0^2|x-\frac{4}{3}|f(x)dx=\int_0^2|x-\frac{4}{3}|(\frac{x}{2})dx=\int_0^2|\frac{x^2}{2}-\frac{2x}{3}|dx$

$\implies \dfrac{x^3}{6}]_0^2\ \ -\ \dfrac{x^2}{3}]_0^2=\dfrac{8}{6}-\dfrac{4}{3}=0$

Hence, $E[|X-E[X]|]=0$