Question

The probability density function of a random variable X is f (x) = C | x |; Find C,

and the value of x_o such that ,

F_X(x_o)=3/4 where F is the CDF.

Answer 1

Solution:

$f(x)=C|x|, \forall |x|\le a \\f(x)=\{Cx, x\ge a\ge0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ -Cx,-a\le x<0$

Now, $\int_{-a}^0Cx \ dx+\int_0^{a}Cx \ dx=1$

$\Rightarrow C.\dfrac {x^2}2|_{-a}^0+C.\dfrac {x^2}2|_{0}^a=1$

$\Rightarrow C.\dfrac {a^2}2+C.\dfrac {a^2}2=1 \\ \Rightarrow C.a^2=1 \\ \Rightarrow C=\dfrac 1{a^2}$

We have to assume a value of $a$ as it is not given, say, $a=1$

Then, $C=1$

So, $f(x)=|x|, \forall|x|\le 1$

Now, CDF$=F_X(x_0)=\int_{-1}^{x_0}(-x)dx+\int_{{x_0}}^{1}(x)dx=\frac34$

$\dfrac{-x^2}{2}|_{-1}^{x_0}+[\dfrac{x^2}{2}|_{x_0}^{1}]=\dfrac34$

$\dfrac{-{x_0}^2}{2}+\dfrac12+\dfrac12-\dfrac{{x_0}^2}{2}=\dfrac34$

$-{x_0}^2=\dfrac34-1 \\-{x_0}^2=-\dfrac14 \\x_0=\pm\dfrac12$