Question

Consider the following proability density function: f(x) = { kx. 0<= x< 2, k(4-x). 2<= x <=4, 0. otherwise. find the value of k for which f(x) is a valid probability density funcion

Answer 1

The value of K for which f(x) is a valid probability density function is 1/4.

How to solve for the value of K

$\int\limits^4_0 {fx(x)} \, dx =1$

$\int\limits^2_0 {Kx} \, dx +\int\limits^4_2 {K(4-x)} \, dx =1$

$K[\frac{2^2}{2} -0]+[K[4(4-2)-(\frac{4^2}{2} -\frac{2^2}{2} )]=1$

open the equation

$K\frac{4}{2}+K[8 - (\frac{16}{2} -\frac{4}{2} )] = 1$

$2K+K[\frac{4}{2} ]=1$

2K + 2K = 1

4K = 1

divide through by 4

K = 1/4

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