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We have, for a positive random variable X:

$$ M_X(t) = E \left[\text{e}^{tX}\right] = \sum_{k=0}^{\infty} \frac{t^k}{k!} E \left[ X^k\right] $$

Then, the pdf is given by:

$$ f_X(x) = \frac{1}{2\pi i} \lim_{T\to\infty} \int_{0}^{\gamma + iT} \text{e}^{-tx} M_X(t) dt = \sum_{k=0}^{\infty}\frac{E \left[ X^k\right]}{2\pi i} \lim_{T\to\infty} \int_{0}^{\gamma + iT} \text{e}^{-tx} \frac{t^k}{k!} dt $$

This is correct? If it is, how can I calculate the inverse laplace transform of $t^k$? I just found the solution when k is non-integer and negative. Thanks!

Felipe
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    Can you not use the derivative $\rightarrow$ multiplication property from here? https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems – user27182 Feb 21 '20 at 23:20
  • How I can use this property? thanks – Felipe Feb 22 '20 at 14:45
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    Are you not taking inverse laplace of polynomials on the right hand side? – user27182 Feb 23 '20 at 00:01
  • Yes. I was thinking of using the derivative property. But the pdf will be expanded with the derivative of delta dirac functions – Felipe Feb 24 '20 at 15:28
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    Is it $\int_{\gamma - \mathrm{i}T}^{\gamma + \mathrm{i}T}$ in the integral? – River Li Feb 28 '20 at 17:23
  • the random variable is positive, but no problem in consider this interval – Felipe Feb 29 '20 at 00:39
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    @Felipe Do you think that $\int_{\gamma - \mathrm{i}T}^{\gamma + \mathrm{i}T} \mathrm{e}^{-tx} t^k \mathrm{d}t$ does not exist, and so we can not obtain $f_X(x)$ in this way? – River Li Feb 29 '20 at 16:23
  • I'm using $\int_{0}^{\gamma + iT} $ because my Va is positive. But when i tried, with fourier transform, I get a series with derivatives of delta dirac. In reality, I'm trying to obtain the pdf from the moments of the Random variable. But, it seems to broke somewhere – Felipe Mar 05 '20 at 15:15
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    @Felipe (You may use @ otherwise I don't know your reply.) Consider the exponential distribution with PDF $f(x) = \mathrm{e}^{-x}1_{x > 0}$. We have $F(s) = \int_0^\infty f(x) \mathrm{e}^{sx} \mathrm{d} x = \int_0^\infty \mathrm{e}^{-(1-s)x} \mathrm{d} x = \frac{1}{1-s}$ when $\mathrm{Re}(1-s) > 0$. We have for $x > 0$, $ \frac{1}{2\pi \mathrm{i}}\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \mathrm{e}^{-sx}F(s) ds = \frac{1}{2\pi \mathrm{i}}\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \mathrm{e}^{-sx} \frac{1}{1-s} ds = \mathrm{e}^{-x}. $ – River Li Mar 08 '20 at 09:47
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    @Felipe Alternatively, with the substitution $s = \gamma + \mathrm{i}u$, then $\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \cdots \mathrm{d} s = \int_{-\infty}^\infty \cdots \mathrm{d}u$, we get the same result. – River Li Mar 08 '20 at 09:53
  • @RiverLi I was trying to obtain an expansion of the pdf $f_X(x)$ from its moments. But that last integral give the derivatives of delta function, but it dont make sense for me. Note that i'm not trying to obtain the pdf from MGF with laplace transform – Felipe Mar 08 '20 at 17:23
  • @RiverLi The reasoning is the following: if I have the pdf constructed from its moments, I can estimate the behavior of the pdf from the estimated moments. But it seems not so trivial – Felipe Mar 08 '20 at 17:25
  • @Felipe In my opinion, $\int_{\gamma - \mathrm{i}T}^{\gamma + \mathrm{i}T} \mathrm{e}^{-tx} t^k \mathrm{d}t$ does not exist, you can not obtain $f_X(x)$ in this way. – River Li Mar 08 '20 at 23:56
  • @Felipe Actually, you want to interchange the order of integration and summation, i.e. $\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \sum_{k=0}^\infty \frac{1}{2\pi \mathrm{i}}\mathrm{e}^{-tx}\frac{t^k}{k!}\mathrm{E}[X^k] \mathrm{d}t \overset{?} = \sum_{k=0}^\infty \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{2\pi \mathrm{i}}\mathrm{e}^{-tx}\frac{t^k}{k!}\mathrm{E}[X^k] \mathrm{d}t. $ I think that the answer is negative. – River Li Mar 08 '20 at 23:57
  • @RiverLi, Yes, the moments don't integrate. You know why this equality is broken? Thanks for the answers – Felipe Mar 09 '20 at 17:14
  • @Felipe The existence of the integral for each $k$ may be crucial. – River Li Mar 10 '20 at 04:06

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