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Say for $t<1$:

$$M(t) = \frac{1}{(1-t)^2}$$

How to find the p.d.f of the random variable?

$$M(t) = E(e^{tx})=\int_{-\infty}^{+\infty}e^{tx}f(x)dx$$

How do we find:

$f(x) = xe^{-x}$ on $(0,+\infty)$ and zero elsewhere? Is there a general method?

George
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  • http://math.stackexchange.com/questions/343930/calculate-probability-density-function-from-moment-generating-function –  Aug 06 '15 at 01:03
  • The easiest thing to do is to compute the MGF of your $f(x)$ and verify it equals what the question claims. Otherwise, lookup the inverse of a two-sided laplace transform. – Alex R. Aug 06 '15 at 01:09

1 Answers1

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Replace $t$ with $it$ to get the CF, then compute the inverse Fourier transform of the CF: that gives the density. For short, the density is the inverse Laplace transform of the moment generating function. It is useful to exploit integration by parts and some well-known (inverse) transform, together with the fact that the Fourier transform of a product is a convolution of two Fourier transforms.

Jack D'Aurizio
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