Let $X \sim \operatorname{Exp}(\lambda).$

Find $E[X^n]$, for all $n \in N$.

I'm having difficulty figuring out how to do this problem and not sure quite how to solve this. I know that I'm supposed to apply the exponential distribution here, but still can't figure this out. Thanks in advance.

  • $\begingroup$ Probably you want to do all the work at once by computing the moment generating function. $\endgroup$ – André Nicolas Apr 25 '16 at 22:00

The moment generating function of a random variable $X$ is defined by

$$ M_X(t) = E(e^{tX}) = \begin{cases} \sum_i e^{tx_i}p_X(x_i), & \text{(discrete case)} \\ \\ \int_{-\infty}^{\infty} e^{tx}f_X(x)dx, & \text{(continuous case)} \end{cases} $$

If we express $e^{tX}$ formally and take expectation

$$M_X(t) = E(e^{tX}) = 1 + tE(X) + \frac{t^2}{2!}E(X^2)+ \ldots + \frac{t^n}{n!}E(X^n)+ \ldots$$

Hence the $n$th moment of $X$ is given by

$$E(X^n) = M_X^{(n)}(0) \:\:\:\:\:\:n = 1, 2 \ldots$$

$$M_X^{(n)}(0) = \frac{d^n}{dt^n} M_X(t) |_{t=0}$$


Write $I_n = E[X^n]$. You have

$$ I_n = \int_{-\infty}^{\infty} x^n \cdot f_X(x) \, dx $$

where $f_X$ is the probability density function of $X$. In your case,

$$ I_n = \int_0^{\infty} x^n \cdot (\lambda e^{-\lambda x}) \, dx. $$

Applying integration by parts with $u = x^n$ and $dv = \lambda e^{- \lambda x} \, dx$ (so $v = -e^{- \lambda x}$), you have

$$ I_n = \int_0^{\infty} u \, dv = [-x^n e^{-\lambda x}]^{x = \infty}_{x = 0} - \int_{0}^{\infty} v \, du = n \int_0^{\infty} x^{n-1} e^{-\lambda x} \, dx = \frac{n}{\lambda}I_{n-1}.$$

This is a recursion for $I_n$ whose solution is $I_n = \frac{n!}{\lambda^n} \cdot I_0 = \frac{n!}{\lambda^{n}}$ (as $I_0 = 1$ since $f_X$ is a probability density function).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.