How do I find all even moments (and odd moments) for $f_X(x)=\frac{1}{2}e^{-|x|}$?

Question

I was asked to find a formula for all even moments of the form $E(X^{2n})$ and all odd moments of the form $E(X^{2n+1})$ using a mgf. Can you help me find the even moments? I will attempt to solve for the odd moments using similar techniques.

Let $X$ be a continuous random variable having the density $f_X(x)=\frac{1}{2}e^{-|x|}$, where $-\infty < x < \infty$.

My work: I was able to obtain the following moment-generating function: $M_X(t)=(1-t^2)^{-1}.$ However, I am unsure of where to go from here. Thank you for your help.

Answer 1

One way to make your mgf approach to the problem easier is to use the power series

$$(1-t^2)^{-1}=\sum_{j=0}^{\infty} t^{2j}$$

Differentiating the rhs repeatedly is much easier than differentiating the lhs. (note this only applies for $|t|<1$ but as you are differentiating at $t=0$ it still works). You should see that it will just be a factorial. However is it probably even simpler to just evaluate the expectation directly

$$E(X^{2n})=\int_{-\infty}^{\infty}\frac{x^{2n}}{2}\exp(-|x|)dx$$

The result should be a gamma function (ie factorial). This also becomes clear from the mgf power series.

update regarding the mgf approach, to evaluate $E(X^{2n})$ we need to differentiate the rhs $2n$ times. The following result will be useful

$$\frac{\partial^k x^{r}}{\partial x^k} =\left( \begin{matrix}  \frac{r!}{(r-k)!}x^{r-k} & r=k,k+1,\dots \\ 0 & r=1,\dots,k-1 \end{matrix} \right) $$

Now if you apply this to the term $t^{2j}$ and take the "2n-th" derivative, we have $r=2j$ and $k=2n$ and $x=t$. Then we get $$\frac{\partial^{2n} t^{2j}}{\partial t^{2n}} =\left( \begin{matrix}  \frac{(2j)!}{(2j-2n)!}t^{2j-2n} & 2j=2n,2n+2,2n+4,2n+6,\dots \\ 0 & 2j=0,2,\dots,2n-2 \end{matrix} \right) $$

This means when we add up over all the terms we can write this as $$\sum_{j=0}^{\infty}\frac{\partial^{2n} t^{2j}}{\partial t^{2n}}=0+\dots+0+(2n)!+t^2\frac{(2n+2)!}{2!}+t^4\frac{(2n+4)!}{4!}+\dots$$

only the single tertm $(2n)!$ is not $0$ and also not a multiple of $t$. So setting $t=0$ leaves only that term.

Answer 2

Hint: This is an example of a probability density function that is symmetric about zero:

$$f_X(0+x) = f_X(0-x) \quad \quad \quad \text{for all } x \in \mathbb{R}.$$

Visually, this means that the distribution is reflected around the zero line, and is the same on both sides. See if you can use this property to figure out (and then prove) what the odd moments would be. The even moments are a bit trickier, but see if you can use this property to reduce those moments to a simpler form.

Answer 3

Since the OP seems to be having difficulty with the various hints in the comments and the other answers, here is a heuristic method that yields the right answer in this instance. \begin{align} E[\exp(tX)] &= E\left[1 + tX + \frac{(tX)^2}{2!} + \frac{(tX)^3}{3!} + \cdots\right]\\ &= 1 + tE[X] + \frac{t^2}{2!}E[X^2] + \frac{t^3}{3!}E[X^3] + \cdots\tag{1}\\ \end{align} and so if we have a power series for $E[\exp(tX)] = M_X(t)$ that we have managed to find by hook or by crook and without looking at the displayed equation above since that will just serve to confuse us, then we can just look at the coefficient of $t^n$ in the power series that we have and say "Hey Ma! I think that the coefficient of $t^n$ is just $\frac{1}{n!}E[X^n]$ and so I can find $E[X^n]$ by multiplying the coefficient that I already have by $n!$.

For the OP's specific case, he has already found that $$M_X(t) = \frac{1}{1-t^2} = 1 + t^2 + t^4 + \cdots + t^{2n} + \cdots \tag{2}$$ (that's the "by hook or by crook" part) and so he can compare $(1)$ and $(2)$ to figure out what the moments are. I will leave it to the OP to tell Ma that $E[X^n]=0$ whenever $n$ is odd. Whether he wishes to tell Ma what $E[X^n]$ is for even $n$ is a matter for him to decide. The $E[X^4]=12$ that the OP claims (in a comment) to have calculated doesn't sound right to me (I think it should be $E[X^4] = 4! = 24$), but who am I to interfere in the sacred relationship between mother and child?

Answer 4

If you don't care to do unnecessary calculation, it is convenient to view your distribution as an equal mixture of an Exponential and its negative:

$$\frac{1}{2} e^{-|x|} = \frac{1}{2} e^{-x}\,\mathcal{I}(x\gt 0) + \frac{1}{2} e^{x}\,\mathcal{I}(x \lt 0).$$

Because $((-1)^n + (1)^n)/2$ is either $-1+1=0$ or $(1+1)/2 = 2/2 = 1$ as $n$ is odd or even, respectively, the odd moments of your distribution are zero and the even moments are the same as those of the Exponential. But, by definition, the even Exponential moments are

$$\mu_{2k} = \int_0^\infty x^{2k} e^{-x}\mathrm{d}x = \Gamma(2k+1) = (2k)!$$

Because this required no calculation more difficult than $1+1=2,$ it's hard to imagine a simpler solution.

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If instead you wish to pursue the moment generating approach, begin by noting that the mgf of the Exponential distribution is $$\phi(t)=\int_0^\infty e^{tx} e^{-x}\mathrm{d}x = \int_0^\infty e^{-(1-t)x}\mathrm{d}x = \frac{1}{1-t}.$$ Thus the mgf of this mixture is

$$(\phi(t) + \phi(-t))/2 = \frac{1}{2}\left(\frac{1}{1-t} + \frac{1}{1-(-t)}\right) = \frac{1}{1-t^2}.$$

This is analytic near $0$ and therefore equals the power series expansion given by the Binomial Theorem as

$$\frac{1}{1-t^2} = (1 + (-t^2))^{-1} = \sum_{k=0}^\infty \binom{-1}{k} (-t^2)^k = \sum_{k=0}^\infty t^{2k} = \sum_{k=0}^\infty \color{red}{(2k)!}\, \color{gray}{\frac{t^{2k}}{(2k)!}},$$

from which you may read off the moments as the coefficients of $t^n/n!:$ again we see they are zero for the odd moments and $(2k)!$ for the even moments.

Answer 5

So I calculated the first, second, third, and fourth derivatives. I got $E(X^1)=0$, $E(X^2)=2$, $E(X^3)=0$, and $E(X^4)=12$. These derivatives are quite long to compute at this point, so I m wondering if there is an easier way to go about this to obtain a formula for the evens.

You could use the Taylor series expansion:

$$\frac{1}{1-t^2} = \sum_{k=0}^\infty t^{2k}$$

However, this is a bit of a circular reasoning since the Taylor series expansion is itself derived by computing the derivatives. In that case you can just as well look up directly a formula for the higher order moments of the Laplace distribution.

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You might find that Taylor series expansion indirectly - not using $f(x) = \sum_{n=0}^\infty f^{(n)}/n! t^k $ - by instead using the formula for a geometric series.

However, you could also 'manually' derive the derivatives (that means straight forward computation using chain rule and product rule) and when you look at the pattern of the terms then you will find that many of the terms become zero and a regular pattern emerges.

Say we substitute $u = t^2$ then the derivation it looks simpler:

$$ \frac{\text{d}^n}{\text{d}u^n} \frac{1}{(1-u)} = \frac{n!}{(1-u)^n}$$

Now use Faà di Bruno's formula (chain rule but then applied several times):

$$ \frac{\text{d}^n}{\text{d}t^n} \frac{1}{(1-u)} = \sum_{k=1}^n \frac{k!}{(1-u)^k} \cdot B_{n,k}(2t,2,0,...,0)$$

where $B_{n,k}$ refers to Bell polynomials. Most of the terms will be zero and you get

$$ \frac{\text{d}^{2n}}{\text{d}t^{2n}} \frac{1}{(1-t^2)} = \sum_{k=0}^n c_{nk} \frac{t^{2k}}{(1-t^2)^{1+n+k}}$$

with

$$c_{nk} = 2^{2k} \frac{(2n)! \cdot (n+k)!}{(n-k)! \cdot (2k)!} $$

and for the value at $t=0$ you have

$$ \frac{\text{d}^{2n}}{\text{d}t^{2n}} \frac{1}{(1-t^2)} = c_{n0} = (2n)! $$