Question

Consider the following probability density function:

f_X(x)=kx 0<=x<2,

=k(4-x), 2<=x<=4

=0, otherwise

a)   Find the value of k for which f_X(x) is a valid probability density function.

b)  Find the mean of X.

c)   Find and skecth the probability distiribution function F_X(x).

Answer 1

CORRECTED SOLUTION.

1) 1=$\int_0^4 f(x)dx = \int_0^2 kx dx + \int_2^4 k(4-x)dx =$

$= \frac{ kx^2}{2}|_0^2 + (4kx- \frac{kx^2}{2})|_2^4 = 4k$

Therefore, k=1/4.

2) E[X] = $\int_0^4 xf(x)dx = \int_0^2 \frac{x^2}{4} dx + \int_2^4 \frac{x(4-x)}{4} dx =$

$= \frac{x^3}{12}|_0^2 + (\frac{x^2}{2} - \frac{x^3}{12})|_2^4 =$ (8/12 - 0/12) + (16/2 - 64/12) - (4/2 - 8/12) = 2

3) $F_X(x) = \int_0^x f(t)dt$

If $0\leq x\leq 2$, then $F_X(x) = \int_0^x t/4dt = \frac{t^2}{8}|_0^x = x^2/8$

In particurlarly, F_X(2) = 2²/8 = 1/2.

If $2\leq x\leq 4$, then $F_X(x) = F_X(2) +\int_2^x (4-t)/4dt = 1/2 + (t-\frac{t^2}{8})|_2^x =$

= 1/2 + (x - x²/8) - (2 - 2²/8) = - x²/8 + x - 1.

The graph of the probability distiribution function F_X(x):