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I am trying to figure out what the nth central moment is for the exponential distribution. Here is the formula for the nth moment:

$$ \mathop{\mathbb{E}}{[x^n]} =\dfrac{n!}{\lambda^n} $$

My question: what is the continuous formula for the exponential distribution for the nth central moment?

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  • $\begingroup$ This seems like a self-study question. What have you tried? Where are you stuck? The wikipedia page for central moments seems to answer this question completely: just use the definition & binomial theorem. en.wikipedia.org/wiki/Central_moment What, specifically, is giving you trouble? $\endgroup$
    – Sycorax
    Commented Dec 30, 2022 at 4:40
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    $\begingroup$ Its straight substitution into the formula there. For a derivation just use the straight binomial expansion of $(a+b)^n$ with $a=X$ and $ b=-\mu$ and use the same substitution to get the exponential result $\endgroup$
    – Glen_b
    Commented Dec 30, 2022 at 4:55
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    $\begingroup$ Isn't this straightforward by calculating the integral $E(X^n) = \int_0^\infty x^n \lambda e^{-\lambda x}dx$? $\endgroup$
    – Zhanxiong
    Commented Dec 30, 2022 at 5:01
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    $\begingroup$ @AlexanderMills Can you post what you have tried and pointed out where the difficulty lies in deriving the final answer? $\endgroup$
    – Zhanxiong
    Commented Dec 30, 2022 at 5:30
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    $\begingroup$ For a direct formula, see the Properties section of the Exponential distribution page on Wikipedia, $\mu_n=...$ $\endgroup$
    – Glen_b
    Commented Dec 30, 2022 at 5:54

1 Answer 1

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Generally, when $X$ is any random variable with a defined positive integral moment $\mu_n(X) = E[X^n],$ then (a) $\mu_1 = \mu = E[X]$ is defined and (b) its central moment of order $n$ is given by the Binomial Theorem as a linear combination of moments

$$\mu^\prime_n(X) = E[(X - \mu)^n] = E\left[\sum_{i=0}^n \binom{n}{i}\mu^{n-i}X^i \right] = \sum_{i=0}^n \binom{n}{i}\mu^{n-i}\mu_i(X).$$

This formula can be inverted to express moments in terms of central moments, too. See Stuart & Ord, Kendall's Advanced Theory of Statistics, Volume 1.


Perhaps the easiest way to obtain the moments of an exponential random variable is through its moment generating function (mgf), here computed for a standard exponential distribution with density function $f_X = e^{-x}$ defined on the positive real numbers:

$$\begin{aligned} \phi_X(t) &=\sum_{i=0}^\infty \frac{(-1)^i\mu_i\,t^i}{i!} = \sum_{i=0}^\infty \frac{-(1)^iE[X^i]\,t^i}{i!} = E\left[\sum_{i=0}^\infty \frac{(-tX)^i}{i!}\right] = E\left[e^{-tX}\right]\\& = \int_0^\infty e^{-tx} e^{-x}\,\mathrm d x = \int_0^\infty e^{-(t+1)x}\,\mathrm dx = \frac{1}{-(t+1)}e^{-(t+1)x}\bigg|_0^\infty = (1+t)^{-1}\\ &= \sum_{j=0}^\infty (-1)^i t^i. \end{aligned}$$

The first line is the entire point of the mgf, the second line is the elementary calculation of the integral, and the last line applies the Binomial Theorem.

Comparing these absolutely convergent power series term by term gives

$$\mu_i = i!.$$

When $\lambda$ is a rate parameter, $1/\lambda$ is a scale parameter. The units calculus tells us to multiply $\mu_i$ by $(1/\lambda)^i$ in this case, producing the formula $\mu_i = i!/\lambda^i$ quoted in the question. You may plug this in for $\mu_i$ in the general formula for central moments, bearing in mind $\mu = \mu_1 = 1/\lambda.$

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  • $\begingroup$ well that looks nice $\endgroup$ Commented Dec 31, 2022 at 0:14
  • $\begingroup$ alright so for moment about the origin, for exponential it's n!/𝜆^n, I can't really tell what the answer is for central moments (about the mean)? In other words, I can't really tell what the answer is $\endgroup$ Commented Jan 2, 2023 at 2:14

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