Answer to Question #185254 in Statistics and Probability for mufafa

Question

. Let the probability distribution function of X be given by f(x) = 0.25 exp(−0.25x), x > 0 Show that E(X) = 4 and V ar(X) = 16

Answer 1

$$E(X)=\displaystyle\int_{0}^{\infin}x(0.25e^{-0.25x})dx$$

$$\int0.25xe^{-0.25x}dx$$

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

$u=0.25x, du=0.25dx$

$v=-4e^{-0.25x}$

$$\int0.25xe^{-0.25x}dx=-xe^{-0.25x}+\int e^{-0.25x}dx$$

$$=-xe^{-0.25x}-4e^{-0.25x}+C$$

$$E(X)=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}x(0.25e^{-0.25x})dx$$

$$=\lim\limits_{t\to \infin}[-xe^{-0.25x}-4e^{-0.25x}]\begin{matrix} t \\ 0 \end{matrix}=4$$

$$\int0.25x^2e^{-0.25x}dx$$

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

$u=0.25x^2, du=0.5xdx$

$v=-4e^{-0.25x}$

$$\int0.25x^2e^{-0.25x}dx=-x^2e^{-0.25x}+2\int xe^{-0.25x}dx$$

$$=-x^2e^{-0.25x}-8xe^{-0.25x}-32e^{-0.25x}+C$$

$$E(X^2)=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}x^2(0.25e^{-0.25x})dx$$

$$=\lim\limits_{t\to \infin}[-x^2e^{-0.25x}-8xe^{-0.25x}-32e^{-0.25x}]\begin{matrix} t \\ 0 \end{matrix}$$

$$=32$$

$$Var(X)=E(X^2)-(E(X))^2$$

$$=32-(4)^2=16$$