Question #261922

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?


1
Expert's answer
2021-11-09T16:50:16-0500
f(x)={kx(k+1)1<x<0otherwisef(x) = \begin{cases} kx^{-(k+1)} &1<x<\infin \\ 0 & otherwise \end{cases}

a)

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


Then

k0k\not=0

1kx(k+1)dx=limA[k(k+1)+1x(k+1)+1]A1\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}

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

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

Answer: k>0k>0

b)


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

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

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

=[k(k+1)+1y(k+1)+1]x1=xk+1=\bigg[\dfrac{k}{-(k+1)+1}y^{-(k+1)+1}\bigg]\begin{matrix} x \\ 1 \end{matrix}=-x^{-k}+1


Answer:

F(x)={0x<11xk1x<,k>0F(x) = \begin{cases} 0 &x<1 \\ 1-x^{-k} &1\leq x<\infin \\ \end{cases}, k>0

c)


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

=limA[k(k+1)+2x(k+1)+2]A1=\lim\limits_{A\to\infin}\bigg[\dfrac{k}{-(k+1)+2}x^{-(k+1)+2}\bigg]\begin{matrix} A \\ 1 \end{matrix}

=kk1limAAk+1+kk1,k>1=-\dfrac{k}{k-1}\lim\limits_{A\to\infin}A^{-k+1}+\dfrac{k}{k-1}, k>1

If k=1k=1


E(X)=xf(x)dx=1kx1dxE(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{1}^{\infin}kx^{-1}dx

=limA[kln(x)]A1=does not exist=\lim\limits_{A\to\infin}\big[k\ln(|x|)\big]\begin{matrix} A \\ 1 \end{matrix}=does\ not\ exist


Answer:

 E(X)E(X) exists for k>1.k>1.



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