Answer to Question #272651 in Statistics and Probability for Stone

Question

A random variable X has density function given by:

f(x)= {3e^-3x

x is greater than or equal to 0, otherwise. Find the first four moments.

Answer 1

$E(X)=\int^{\infin}_0xf(x)dx=3\int^{\infin}_0xe^{-3x}dx=-\frac{3(3x+1)e^{-3x}}{9}|^{\infin}_0=1/3$

$E(X^2)=\int^{\infin}_0x^2f(x)dx=3\int^{\infin}_0x^2e^{-3x}dx=-\frac{3(9x^2+6x+2)e^{-3x}}{27}|^{\infin}_0=2/9$

$E(X^3)=\int^{\infin}_0x^3f(x)dx=3\int^{\infin}_0x^3e^{-3x}dx=-\frac{3(9x^3+9x^2+6x+2)e^{-3x}}{27}|^{\infin}_0=2/9$

$E(X^4)=\int^{\infin}_0x^4f(x)dx=3\int^{\infin}_0x^4e^{-3x}dx=-\frac{3(27x^4+36x^3+36x^2+24x+8)e^{-3x}}{27}|^{\infin}_0=8/27$

it was used integrating by parts:

$\int fg'=fg-\int f'g$

$\int xe^{-3x}dx=-xe^{-3x}/3-\int (-e^{-3x}/3)dx=-xe^{-3x}/3-e^{-3x}/9$

$\int x^2e^{-3x}dx=-x^2e^{-3x}/3-\int 2x(-e^{-3x}/3)dx=-x^2e^{-3x}/3+\frac{2}{3}(xe^{-3x}/3+e^{-3x}/9)$

$\int x^3e^{-3x}dx=-x^3e^{-3x}/3-\int 3x^2(-e^{-3x}/3)dx=$

$=-x^3e^{-3x}/3-x^2e^{-3x}/3+\frac{2}{3}(xe^{-3x}/3+e^{-3x}/9)$

$\int x^4e^{-3x}dx=-x^4e^{-3x}/3-\int 4x^3(-e^{-3x}/3)dx=$

$=-x^4e^{-3x}/3+\frac{4}{3}(-x^3e^{-3x}/3-x^2e^{-3x}/3+\frac{2}{3}(xe^{-3x}/3+e^{-3x}/9))$