0
$\begingroup$

I was wondering whether it was possible to derive the kth moment of an exponential random variable $X \sim \text{EXP}(\theta)$ without using the MGF and then finding the kth derivative.

I came up with this derivation:

Let $X \sim \operatorname{EXP}(\theta)$

Then the PDF of $X$ is:

$$f(x) = \begin{cases} \frac{1}{\theta} e^{\frac{-x}{\theta}} & x \gt 0 \\ 0 & \text{otherwise} \\ \end{cases}$$

The definition of the kth moment is $\operatorname{E}(X^k)$, so

$$\begin{align} \operatorname{E}(X^k) & = \int_{-\infty}^\infty x^k \; f(x) \; dx \\ & = \int_{0}^\infty x^k \; \frac{1}{\theta} e^{\frac{-x}{\theta}} \; dx \\ & = \frac{\theta^k}{\theta^k} \int_{0}^\infty x^k \; \frac{1}{\theta} e^{\frac{-x}{\theta}} \; dx\\ & = \theta^k \int_{0}^\infty x^k \; \frac{1}{\theta^{k+1}} e^{\frac{-x}{\theta}} \; dx\\ & = \frac{\Gamma(k+1)}{\Gamma(k+1)} \theta^k \int_{0}^\infty x^k \; \frac{1}{\theta^{k+1}} e^{\frac{-x}{\theta}} \; dx\\ & = \Gamma(k+1) \theta^k \int_{0}^\infty x^k \; \frac{1}{\Gamma(k+1) \theta^{k+1}} e^{\frac{-x}{\theta}} \; dx \quad (*) \\ \end{align}$$

Since $\int_{0}^\infty x^k \; \frac{1}{\Gamma(k+1) \theta^{k+1}} e^{\frac{-x}{\theta}} \; dx$ is the PDF of $X \sim \Gamma(k+1, \theta)$, it equals one, so:

$$ \begin{align} \operatorname{E}(X^k) = \cdots & = \Gamma(k+1) \theta^k \int_{0}^\infty x^k \; \frac{1}{\Gamma(k+1) \theta^{k+1}} e^{\frac{-x}{\theta}} \; dx \quad (*) \\ & = \Gamma (k+1) \theta^k \quad \square \end{align}$$

Is this derivation correct?

$\endgroup$
1
$\begingroup$

There are no mistakes in the derivation but that's all. Things can be done much easyer.

First let me note that it is common to write $\lambda e^{-\lambda x}$ for the PDF of exponential distribution on positive axis. Apparantly you go out from $\theta=\frac{1}{\lambda}$. There is nothing wrong with that, but it must be mentioned to avoid confusion.

Secondly you could save yourself a lot of work by stating that your $X$ has the same distribution as $\theta Y$ where $Y$ has exponential distribution with parameter $1$.

Then $\mathbb EX^k=\theta^k\mathbb EY^k$ so based on this observation it is enough to prove that $\mathbb EY^k=\Gamma(k+1)$. Parameters are annoying and if you do without then the probability on mistakes is much less.

You ended with stating that "$\int_0^{\infty}\cdots$ is the PDF of $X\sim\Gamma(k+1,\theta)$, so it equals $1$".

But it is a better idea to start with equivalent statement for $\theta=1$ that: $$\Gamma(k+1)=\int_0^{\infty}x^ke^{-x}dx$$

It tells you immediately that $\mathbb EY^k=\Gamma(k+1)$ and consequently that $\mathbb EX^k=\theta^k\Gamma(k+1)$.

$\endgroup$

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.