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Suppose that X has the moment generating function of the form $M_X(t)=\frac{1}{1-4t}$, $t<\frac{1}{4}$, find probability density function.

Moment generating function is usually given by M(t)= E($e^{tx}$)= $\int_{-\infty}^{\infty} (e^{tx}*f(x) dx$

How am I supposed to find f(x) then? I just don't get the concept.

Beacon
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    In general, the concept will involve the inverse Fourier transform. But you are in luck, because it seems your MGF matches that of the exponential distribution. Have a look at it. Also see here : https://math.stackexchange.com/questions/343930/calculate-probability-density-function-from-moment-generating-function – Sarvesh Ravichandran Iyer Feb 12 '20 at 03:48
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    @астонвіллаолофмэллбэрг I think this is what you mean: https://en.wikipedia.org/wiki/Mellin_inversion_theorem

    Fourier transform corresponds to the characteristic function of a probability distribution.

     – Math1000 Feb 12 '20 at 05:37
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    @Math1000 Yes, you are right. The Fourier transform comes once we look at the characteristic function, although I think you get that by replacing $t$ by $it$. – Sarvesh Ravichandran Iyer Feb 12 '20 at 09:28

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Let $\lambda>0$. If $X$ has probability density function

$$f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \textrm{ if } x\geq0 \\ 0 & \textrm{ if } x<0 \end{cases}$$ then the moment generating function of $X$ is $$\begin{align} M_X(t) &= \int_0^\infty e^{tx}(\lambda e^{-\lambda x})dx \\ &= \lambda \int_0^\infty e^{-(\lambda - t)x}dx \\ &= \frac{\lambda}{\lambda-t} \textrm{ for } t<\lambda. \end{align}$$ Now choose $\lambda = \tfrac14$.

azif00
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