4
$\begingroup$

The $n$-th moment ($n \geq 1$) of a random variable $X$ is given by: $m_n = \frac{2^n}{n+1}$. Find the probability distribution of $X$.

Here's my attempt at a solution: I expand the moment generating function, $\psi (t) = E[e^{tX}]=\int_{-\infty}^{\infty}dxf(x)e^{tx}$, as a Taylor series about $0$: $$\psi (t) = 1+ \sum_{n=0}^{\infty}\frac{\psi^{(n)}(0)}{n!}t^n = 1 +\sum_{n=0}^{\infty}\frac{2 ^n}{(n+1)!}t^n = 1 + \frac{e^{2t}}{2t}.$$

Now, I have the identity $\psi (-t) = \int_{-\infty}^{\infty}dxf(x)e^{-tx}$ which is apparently a bilateral Laplace transform. So to find $f(x)$ I should invert it right? But the resulting integral seems not to converge at all. I haven't done Laplace transforms at all yet and I have no idea how to solve this problem. Is there perhaps a simpler way? In the lecture notes I couldn't see any formula relating the probability density function to the moment generating function.

$\endgroup$
1
$\begingroup$

Apparently one needs to consider the characteristic function, $\phi (t) = \psi(it)$, then the identity to use is

$$f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty}dt \ e^{-itx}\phi(t) =\frac{1}{2\pi} \int_{-\infty}^{\infty}dt \ \frac{e^{it(2-x)}-e^{-itx}}{2it} = \frac{1}{4} (sgn(2-x)+sgn(x)) = \frac{1}{2} {\chi_{[0,2]}}$$

So it's a uniform distribution on $[0,2]$. God knows how to get the integral result with the signum function, but Wolfram was able to do it with extended computation time, thankfully. If anyone knows how to do it 'by hand' I'd love to see it, cause it looks like a useful result.

$\endgroup$
  • $\begingroup$ The signum looks to me like a standard fourier transform result. $\endgroup$ – Pat May 9 '13 at 11:20

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.