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## Main Question or Discussion Point

My goal here is to at least approximately calculate the probability density function (PDF) given the moment generating function (MGF), [itex]M_X(t)[/itex].

I have managed to calculate the exact form of the MGF as an infinite series in [itex]t[/itex]. In principle, if I replace [itex]t[/itex] with [itex]it[/itex] and perform an inverse Fourier transform I should be able to obtain the PDF, [itex]\rho(x)[/itex], as in

[tex]

\rho(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}{e^{-ixt}M_X(it)dt}

[/tex]

I have looked into simply using the first few terms in the series expansion of [itex]M_X(it)[/itex] in the integrand of the above integral, but this hasn't yielded anything very significant. I should also mention that, since the random variable [itex]x[/itex] is bounded above and below, the actual transform has finite limits:

[tex]

\rho(x)=\frac{1}{2\pi}\int^{t_{\rm max}}_{t_{\rm min}}{e^{-ixt}M_X(it)dt}.

[/tex]

Any advice?

I have managed to calculate the exact form of the MGF as an infinite series in [itex]t[/itex]. In principle, if I replace [itex]t[/itex] with [itex]it[/itex] and perform an inverse Fourier transform I should be able to obtain the PDF, [itex]\rho(x)[/itex], as in

[tex]

\rho(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}{e^{-ixt}M_X(it)dt}

[/tex]

I have looked into simply using the first few terms in the series expansion of [itex]M_X(it)[/itex] in the integrand of the above integral, but this hasn't yielded anything very significant. I should also mention that, since the random variable [itex]x[/itex] is bounded above and below, the actual transform has finite limits:

[tex]

\rho(x)=\frac{1}{2\pi}\int^{t_{\rm max}}_{t_{\rm min}}{e^{-ixt}M_X(it)dt}.

[/tex]

Any advice?