Question

Let X be a continuous random variable with density function below.

$x = \begin{cases} \frac{x}{2}, &\text{for } 0 ≤ x ≤ 2 \\ 0, &\text{otherwise } \end{cases}$

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

Answer 1

Solution:

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

$E[X]=\int_0^2x(\dfrac x2)dx=\int_0^2(\dfrac {x^2}2)dx \\=[\dfrac{x^3}6]_0^2 \\=\dfrac 86-0 \\=\dfrac 43$

Now, $E[|X-E[X]|]=E[|X-\dfrac 43|]$

$=\int_0^2|x-\dfrac 43|(\dfrac x2) dx \\=\int_0^{\frac 43}(\dfrac 43-x)(\dfrac x2) dx+\int_{\frac 43}^2(x-\dfrac 43)(\dfrac x2) dx$

$=\int_0^{\frac 43}(\dfrac 23x-\dfrac {x^2}2) dx+\int_{\frac 43}^2(-\dfrac 23x+\dfrac {x^2}2) dx$

$=[(\dfrac {x^2}3-\dfrac {x^3}6)]_0^{\frac 43}+[(-\dfrac {x^2}3+\dfrac {x^3}6)]_{\frac 43}^2$

$=[(\dfrac {({\frac 43})^2}3-\dfrac {({\frac 43})^3}6)-0]+[(-\dfrac {2^2}3+\dfrac {2^3}6)]-[(-\dfrac {({\frac 43})^2}3+\dfrac {({\frac 43})^3}6)]$

$=[(\dfrac {{\frac {16}9}}3-\dfrac {{\frac {64}{27}}}6)]+[-\dfrac {4}3+\dfrac {8}6]-[-\dfrac {{\frac {16}9}}3+\dfrac {{\frac {64}{27}}}6]$

$=[\dfrac {16}{27}-\dfrac {32}{81}]+[0]-[-\dfrac {16}{27}+\dfrac {32}{81}] \\=2[\dfrac {16}{27}-\dfrac {32}{81}] \\=\dfrac {32}{81}$