0

I'm trying to find a formula to find the pdf from the distribution's moments.

I did the following:

$$ f_X(x) = \int_{0}^{\infty} M_X(t)\text{e}^{-tx} dt = \int_{0}^{\infty} \sum_{k=0}^{\infty} \frac{t^k}{k!} E \left[ X^k\right]\text{e}^{-tx} dt = \sum_{k=0}^{\infty} E \left[ X^k\right] \int_{0}^{\infty}\frac{t^k}{k!} \text{e}^{-tx} dt \nonumber\\ \therefore f_X(x) = \sum_{k=0}^{\infty} \frac{1}{x^{k+1}} E \left[ X^k\right] $$

But this doesn't seem quite right. If I substitute the raw moments from the Rayleigh distribution:

$$ E \left[ X^k\right] = \sigma^k 2^{\frac{k}{2}} \Gamma\left(1+\frac{k}{2}\right) \nonumber \\$$

I do not get the Rayleigh distribution.

mjw
  • 8,647
  • 1
  • 8
  • 23
Felipe
  • 334
  • 2
  • 15

1 Answers1

1

You've attempted to identify $f_X$ with the Laplace transform of $M_X$, which would imply $M_X$ is the inverse Laplace transform of $f_X$, whereas by definition it isn't. Explicitly, it would be wrong to claim $M_X(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{tx}f_X(x)dx$, because actually $M_X(t)=\int_{-\infty}^\infty e^{tx}f_X(x)dx$.

J.G.
  • 115,835
  • So its not possible to recover the pdf from the MGF by a inverse transform? – Felipe Feb 17 '20 at 18:17
  • @Felipe You can, but it's unclear that the calculation would be easy. – J.G. Feb 17 '20 at 18:18
  • So, its possible to find a infinite sum of moments calculating the right inverse of $t^k$? – Felipe Feb 17 '20 at 18:55
  • @Felipe Since the distribution has support $[0,,\infty)$, $M_X(t)=\int_0^\infty f_X(x)e^{tx}dx$, so $f_X(x)=\frac{1}{2\pi i}\int_{\gamma+i\Bbb R}M_X(t)e^{-tx}dt$. This integration range is unsuited to your strategy. – J.G. Feb 17 '20 at 19:06