Question

A continuous random variable x has the pdf given by

f(x)={ e^-x for x>0

{0,ellsewhere

Find the mean and the standard deviation of

a)x

b)y=e^(3/4)x

Answer 1

a)

$$E[X]=\displaystyle\int_{0}^{\infin}xe^{-x}dx$$

$$\int xe^{-x}dx$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=e^{-x}dx, v=\int e^{-x}dx=-e^{-x}$$

$$\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx=-xe^{-x}-e^{-x}+C_1$$

$$E[X]=\displaystyle\int_{0}^{\infin}xe^{-x}dx=\lim\limits_{t\to\infin}\displaystyle\int_{0}^{t}xe^{-x}dx$$

$$=\lim\limits_{t\to\infin}[-xe^{-x}-e^{-x}]\begin{matrix} t \\ 0 \end{matrix}=-0-0+0+1=1$$

$$E[X^2]=\displaystyle\int_{0}^{\infin}x^2e^{-x}dx$$

$$\int x^2e^{-x}dx$$

$$\int udv=uv-\int vdu$$

$$u=x^2, du=2xdx$$

$$dv=e^{-x}dx, v=\int e^{-x}dx=-e^{-x}$$

$$\int x^2e^{-x}dx=-x^2e^{-x}+2\int xe^{-x}dx$$

$$=-x^2e^{-x}-2xe^{-x}-2e^{-x}+C_2$$

$$E[X^2]=\displaystyle\int_{0}^{\infin}x^2e^{-x}dx=\lim\limits_{t\to\infin}\displaystyle\int_{0}^{t}x^2e^{-x}dx$$

$$=\lim\limits_{t\to\infin}[-x^2e^{-x}-2xe^{-x}-2e^{-x}]\begin{matrix} t \\ 0 \end{matrix}$$

$$=-0-0-0+0+0+2=2$$

$$Var(X)=\sigma^2=E[X^2]-(E[X])^2=2-(1)^2=1$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{2}$$

b)

$$E[Y]=\displaystyle\int_{0}^{\infin}e^{(3/4)x}e^{-x}dx=\lim\limits_{t\to\infin}\displaystyle\int_{0}^{t}e^{(-1/4)x}dx$$ $$=\lim\limits_{t\to\infin}[-4e^{(-1/4)x}]\begin{matrix} t \\ 0 \end{matrix}=-0+4=4$$

$$E[Y^2]=\displaystyle\int_{0}^{\infin}e^{2(3/4)x}e^{-x}dx=\lim\limits_{t\to\infin}\displaystyle\int_{0}^{t}e^{(1/2)x}dx$$

$$=\lim\limits_{t\to\infin}[2e^{(1/2)x}]\begin{matrix} t \\ 0 \end{matrix}=\infin$$

$$Var(Y)=\sigma^2=E[Y^2]-(E[Y])^2=\infin$$

$$\sigma=\sqrt{\sigma^2}=\infin$$