Answer to Question #261922 in Statistics and Probability for sam21ucd

Question

A function f(x) has the following form:

f(x)=kx 1<x<∞

and zero otherwise.

a)For what values of k is f(x) a pdf?

b)Find the CDF based on (a).

c)For what values of k does E(X) exist?

Answer 1

$$f(x) = \begin{cases} kx^{-(k+1)} &1<x<\infin \\ 0 & otherwise \end{cases}$$

a)

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

Then

$k\not=0$

$$\displaystyle\int_{1}^{\infin}kx^{-(k+1)}dx=\lim\limits_{A\to\infin}\bigg[\dfrac{k}{-(k+1)+1}x^{-(k+1)+1}\bigg]\begin{matrix} A \\ 1 \end{matrix}$$

$$=-\lim\limits_{A\to\infin}A^{-k}+1=1$$

$$-\lim\limits_{A\to\infin}A^{-k}=0=>k>0$$

Answer: $k>0$

b)

$$F(x)=\displaystyle\int_{-\infin}^{x}f(y)dy$$

$$F(x)=0, x<1$$

$$F(x)=\displaystyle\int_{1}^{x} ky^{-(k+1)}dy$$

$$=\bigg[\dfrac{k}{-(k+1)+1}y^{-(k+1)+1}\bigg]\begin{matrix} x \\ 1 \end{matrix}=-x^{-k}+1$$

Answer:

$$F(x) = \begin{cases} 0 &x<1 \\ 1-x^{-k} &1\leq x<\infin \\ \end{cases}, k>0$$

c)

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{1}^{\infin}kx^{-(k+1)+1}dx$$

$$=\lim\limits_{A\to\infin}\bigg[\dfrac{k}{-(k+1)+2}x^{-(k+1)+2}\bigg]\begin{matrix} A \\ 1 \end{matrix}$$

$$=-\dfrac{k}{k-1}\lim\limits_{A\to\infin}A^{-k+1}+\dfrac{k}{k-1}, k>1$$

If $k=1$

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{1}^{\infin}kx^{-1}dx$$

$$=\lim\limits_{A\to\infin}\big[k\ln(|x|)\big]\begin{matrix} A \\ 1 \end{matrix}=does\ not\ exist$$

Answer:

 $E(X)$ exists for $k>1.$