Determine $E(X^3)$ if $X$ is $\exp(2)$

Question

I want to calculate $E(X^3)$ knowing that $X\sim\exp(2).$ I have tried using the definition of expected value to calculate $E(X^3)$ and got the integral $$ \int_0^\infty x^3 2e^{-2x} \,dx.$$ Is this correct or is there an easier way to do it that I am missing?

Answer 1

Looks right to me.

You can calculate that integral by doing integration by parts three times. Or if you already know $E(X^2),$ once is enough, because integration by parts gives you an expression in terms of $E(X^2).$

That’s how I would expect you to do the question (at least, if you don’t know the integral already.)

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A more advanced trick lets you calculate all $E(X^n)$ at once, using the Moment Generating Function:

$$\begin{align}\sum_{n=0}^\infty \frac{E(X^n)}{n!}t^n&=E\left(e^{tX}\right)\\&=\int_0^\infty 2e^{tx}e^{-2x}\,dx\\ &=\frac{2}{t-2}e^{(t-2)x}\Bigg|_0^\infty\\ &=\frac{1}{1-t/2}\quad\text{Assuming }t<2 \end{align}$$

This shows that $$E(X^n)=\frac{n!}{2^n}$$

Answer 2

You can use the Gamma r.v. density function, i.e., if $X\sim Gamma(n, \lambda)$, then $$ \int_{\mathbb{R}_+} \frac{\lambda ^ n}{ \Gamma(n)}e^{-\lambda x} x^{n-1} dx = 1 $$ In your case $$ \int x ^ 3 \cdot 2e ^{-2x} dx = 2\cdot \frac{3!}{2^4}\int\frac{2^4}{3!}x ^ 3 e ^{-2x}dx = \frac{3!}{2^3} $$

where $\Gamma(n) = (n-1)!$ for $n\in \mathbb{N}$.

Answer 3

You can compute $E[X^n]$ by taking derivative of mgf $n$ times and evaluate at $t=0$.

$M_X(t) = \frac{\lambda}{\lambda - t} = \frac{2}{2 - t}$, then $\frac{d^nM_X}{d t^n} = \frac{2n!}{(2-t)^{n+1}}$ and at $t = 0$, we get $E[X^n] = \frac{n!}{2^n}$