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I need to compute the inverse Laplace transform of the function

$$ M(t)=e^{\frac{t^2}{2}} $$ Now, I know that this is a normal distribution with mean zero and variance 1, but how the computations are done? The formula for the inverse Laplace transform doesn't help me either... It would be great if somebody could help!

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  • $\begingroup$ You use $t$ for the Inverse Laplace Transform (so you want from $t$ to ...)? Usually we use: $\mathcal{L}_{s}^{-1}\left[F_{(s)}\right]_{(t)}=f_{(t)}$ $\endgroup$ – Jan Nov 14 '15 at 17:55
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The inverse Laplace transform can be computed using the standard formula (see this page), which states that if $F$ is the Laplace transform of $f$, then $f$ can be recovered via the line integral $$f(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT} e^{tz}F(z)dz$$ where $\gamma$ is suitably chosen. In our case, $F(z)=e^{z^2/2}$. This function is smooth over the whole complex plane, so (as explained on the wikipedia page) we may take $\gamma=0$, and the line integral is along the imaginary axis from $-iT$ to $iT$. Define a parameterisation of this path in the natural way i.e. $z:[-T,T]\rightarrow\mathbb{C}$ maps $s$ to $is$. Substituting $z(s)=is$ into the line integral gives $$\int_{-iT}^{iT} e^{tz}F(z)dz=\int_{-T}^{T}e^{ist}e^{-s^2/2}z'(s)ds=i\int_{-T}^{T}e^{ist}e^{-s^2/2}ds$$ Next, write $e^{ist}=\cos(st)+i\sin(st)$. Sine is an odd function so the sine term will evaluate to zero. We are therefore left with $$f(t)=\frac{1}{2\pi i}i\int_{-\infty}^{\infty}\cos(ts)e^{-s^2/2}ds=\frac{2i}{2\pi i}\int_0^\infty \cos(\sqrt2 tx)e^{-x^2}\sqrt2dx$$ where we have used the fact that cosine is an even function, and the substitution $x=s/\sqrt2$. This integral is well known - see this question/answer. Evaluating the integral, we finally get $$f(t)=\frac{2i}{2\pi i}\sqrt2 \frac{\sqrt\pi}{2}e^{-2t^2/4}=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$$

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  • $\begingroup$ The problem is that the Laplace transform of $t\mapsto \mathrm e^{-t^2/2}/\sqrt{2\pi}$ is $s\mapsto\frac12\mathrm e^{s^2/2}\operatorname{erfc}(s/\sqrt2)$. $\endgroup$ – Tom-Tom Dec 7 '15 at 10:28
  • $\begingroup$ @Tom-Tom That's true if you use a one sided Laplace transform i.e. integrate from $0$ to $\infty$. In the context of probability, a two sided definition is normally used i.e. you integrate from $-\infty$ to $\infty$, which gives $s\mapsto e^{s^2/2}$. $\endgroup$ – S. Catterall Reinstate Monica Dec 7 '15 at 11:03
  • $\begingroup$ In that case I agree (and the inverse is elementary). $\endgroup$ – Tom-Tom Dec 7 '15 at 14:43
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The Laplace function $l_X (u)$ of a random variable $X$ (with real values) is defined by $$l_X (u) = \mathbf{E}[e^{uX}]$$ for each $u\in\mathbf{R}$ for which the expectation makes sense. (You will see soon that it makes sense for any real number $u$.) Suppose $X$ has the law of a normal random variable with variance $1$ and mean $0$. It's density is $\varphi(t)=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$ so that $$l_X (u) = \mathbf{E}[e^{uX}] = \int_{-\infty}^{+\infty} e^{ut} \frac{1}{\sqrt{2\pi}}e^{-t^2/2} dt = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{ut} e^{-t^2/2} dt $$ and the integral makes sense as $t\mapsto e^{ut} e^{-t^2/2}$ is integrable for any $u$. Now you can write $$u t - \frac{t^2}{2} = -\frac{1}{2}\left( t^2 - 2ut \right) = -\frac{1}{2}\left( t^2 - 2ut + u^2\right) + \frac{u^2}{2} = -\frac{1}{2} (t-u)^2 + \frac{u^2}{2}$$ and plugging this in the last integral gives $$l_X (u) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2} (t-u)^2 + \frac{u^2}{2}} dt = e^{\frac{u^2}{2}} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2} (t-u)^2} dt$$ by taking out the $e^{\frac{u^2}{2}}$ and as any integral over $\mathbf{R}$ is invariant by translation we have $$\int_{-\infty}^{+\infty} e^{-\frac{1}{2} (t-u)^2} dt = \int_{-\infty}^{+\infty} e^{-\frac{1}{2} t^2} dt$$ so that $$l_X (u) = e^{\frac{u^2}{2}} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2} t^2} dt$$. Finally, the well know equality $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2} t^2} dt = 1$ allows to conclude that $$l_X (u) = e^{\frac{u^2}{2}}.$$

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  • $\begingroup$ I know that stuff, but how can one pass from the MGF to the PDF using the inverse laplace transform? $\endgroup$ – james42 Dec 4 '15 at 20:03
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The moment generating function of a probability distribution function is $$\phi_p(t)=\mathbb E\left[\mathrm e^{t X}\right] =\int\mathrm e^{tx}p(x)\mathrm dx.$$ In other words, if $p$ is defined on $\mathbb R$, it is the two-sided Laplace transform of $p$ taken at the point $-t$.

If you know $\phi_p$ and are looking for $p$, you have to compute $$\frac1{2\pi\mathrm i}\lim_{T\to\infty}\int_{\gamma-\mathrm iT}^{\gamma+\mathrm iT} \mathrm e^{zx}\phi_p(-z)\mathrm dz\tag1$$ this is justified by the following relation $$\frac1{2\pi\mathrm i}\lim_{T\to\infty}\int_{\gamma-\mathrm iT}^{\gamma+\mathrm iT} \mathrm e^{zx}\phi_p(-z)\mathrm dz= \frac1{2\pi\mathrm i}\lim_{T\to\infty}\int_{\gamma-\mathrm iT}^{\gamma+\mathrm iT} \mathrm e^{zx}\int\mathrm e^{zy}p(-y)\mathrm dy\mathrm dz$$ that gives after integration over $z$ (by a standard relation for Fourier transforms) $$\int\delta(x-y)p(y)\mathrm dy=p(x).$$ It happens that (1) is precisely the inverse Laplace transform of $\phi_p$. I hope this answers your question.

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