Answer to Question #268404 in Statistics and Probability for Rinah

Question

A continuous random variable X has the following pdf f (x) = kxe− 1 x ; k is a constant.x > 0

 X2

(a) Find the value of k for fX(x) to be a valid probability density function.

(b) Find the cumulant generating function of X.

(b) Using the cumulant generating function or otherwise, find the mean and variance of X.

Answer 1

(a)

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{\infin}kxe^{-x}dx$$

$$=k\lim\limits_{A\to \infin}\displaystyle\int_{0}^{A}xe^{-x}dx$$

$$=k\lim\limits_{A\to \infin}[-xe^{-x}]\begin{matrix} A \\ 0 \end{matrix}+k\lim\limits_{A\to \infin}\displaystyle\int_{0}^{A}e^{-x}dx$$

$$=0-k\lim\limits_{A\to \infin}[e^{-x}]\begin{matrix} A \\ 0 \end{matrix}$$

$$=k=1=>k=1$$ $$f(x) = \begin{cases} xe^{-x} &x>0 \\ 0 & x\leq0 \end{cases}$$

(b)

$$M_x(t)=E(e^{xt})=\displaystyle\int_{-\infin}^{\infin}e^{xt}f(x)dx$$

$$=\lim\limits_{A\to \infin}\displaystyle\int_{0}^{A}xe^{x(t-1)}dx$$

$$\int xe^{(t-1)x}dx=\dfrac{1}{t-1}xe^{(t-1)x}-\dfrac{1}{t-1}\int e^{(t-1)x}dx$$

$$=\dfrac{1}{t-1}xe^{(t-1)x}-\dfrac{1}{(t-1)^2}e^{(t-1)x}+C$$

$M_x(t)$ exists for $t<1$

$$M_x(t)=\lim\limits_{A\to \infin}\displaystyle\int_{0}^{A}xe^{x(t-1)}dx$$

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

$$=-0-0-(-0-\dfrac{1}{(t-1)^2})=\dfrac{1}{(t-1)^2}, t<1$$

$$K_x(t)=\ln(M_x(t))=-2\ln(1-t), t<1$$

c)

$$K_x'(t)=\dfrac{2}{1-t}$$

$$K_x''(t)=\dfrac{2}{(1-t)^2}$$

$$mean=E(X)=K_x'(0)=\dfrac{2}{1-0}=2$$

$$Var(X)=K_x''(0)=\dfrac{2}{(1-0)^2}=2$$