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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.

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  • $\begingroup$ 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
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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

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

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

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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).

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