Question

Probability density function of random variable X is given as:

f(x)=0.4&*(x+4)+0.2[u(x+3)-u(x+2)]+A*e^(-x)*u(x)

where u(x) is the unit step function.

a) Find and sketch probability distribution function F_X(x).

b) Calculate the following probabilities.

P(X=3)

P(-4 ≤ X < -2)

P(-4 < X < -2)

P(X >-4)

P(X < 0)

Answer 1

a)

$$f(x)=0.4e^{-(x+4)}u(x+4)$$ $$+0.2(u(x+3)-u(x+2))+Ae^{-x}u(x)$$

where $u(x)$ is the unit step function.

$$\displaystyle\int_{-\infin}^\infin f(x)dx=1$$

$$\displaystyle\int_{-4}^\infin 0.4e^{-(x+4)}dx+0.2(1)+\displaystyle\int_{0}^\infin Ae^{-x}dx=1$$

$$0.4(1)+0.2+A(1)=1$$

$$A=0.4$$

$$f(x)=0.4e^{-(x+4)}u(x+4)$$ $$+0.2(u(x+3)-u(x+2))+0.4e^{-x}u(x)$$

where $u(x)$ is the unit step function.

$F_X(x)=\begin{cases} 0 & x<-4 \\ 0.4(1-e^{-x-4}) &-4\leq x<-3 \\ 0.4(1-e^{-x-4})+0.2(x+3) & -3\leq x<-2 \\ 0.4(1-e^{-x-4})+0.2 & -2\leq x<0 \\ 0.4(1-e^{-x-4})+0.2+0.4(1-e^{-x}) & x\geq0\\ \end{cases}$

b)

$$P(X=3)=0$$

$$P(-4\leq X<-2)=F(-2)-F(-4)$$

$$=0.4(1-e^{-(-2)-4})+0.2-0=0.6-e^{-2}$$

$$P(-4< X<-2)=P(-4\leq X<-2)=0.6-e^{-2}$$

$$P(X>-4)=1$$

$$P(X<0)=F(0)=0.6-e^{-4}$$