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In my notes this was left as an exercise and I am a bit rusty with my calculus. Starting with the definitions: $$\mathcal{L}_X(t) = \mathbb{E}[e^{-tX}] = \int_0^\infty e^{-Xt}f(t)dt \;\;\text{ and }\;\;X\sim\mathcal{N}(\mu,\sigma)\;\;\text{ i.e. }\;\;f(t) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}$$ so

$$\mathcal{L}_X(t) = \int_0^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} e^{-xt}dx$$

$$= \frac{1}{\sqrt{2\pi}\sigma} \int_0^\infty e^{-\frac{1}{2}\big(\frac{(x-\mu)^2}{\sigma^2}\big)-tx }dx$$ suppose I now say $u = \frac{x-\mu}{\sigma}$ so that $x = u\sigma+\mu$ I get the following but have run out of ideas for how to continue:

$$ = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\frac{\mu}{\sigma}}^\infty e^{-\frac{1}{2}u^2-t(u\sigma+\mu) }du$$

I considered trying to complete the square but don't think it helped: the exponent becomes $-\frac{1}{2}(u^2-2tu\sigma-2t\mu)$ giving $-\frac{1}{2}((u-t\sigma)^2 -t^2\sigma^2+2t\mu)$, which doesn't seem to be any simpler to integrate.

EDIT

After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)

So now I get $$ \mathcal{L}_X(t) = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty e^{-\frac{1}{2}u^2-t(u\sigma+\mu) }du$$

Now I try completing the square as above and get

$$ \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty e^{-\frac{1}{2}((u-t\sigma)^2 -t^2\sigma^2+2t\mu) }du = \frac{1}{\sqrt{2\pi}\sigma} e^{-t^2\sigma^2-2t\mu} \int_{-\infty}^\infty e^{-\frac{1}{2}(u-t\sigma)^2}du$$

Now I make a substitution to say $z = \frac{1}{\sqrt{2}}(u-t\sigma)$ and then we get $u = \sqrt{2}z+t\sigma$, and $du = \sqrt{2}dz$ so finally:

$$ \frac{1}{\sqrt{2\pi}\sigma}e^{-t^2\sigma^2-2t\mu} \int_{-\infty}^\infty e^{-z^2} \sqrt{2}dz = \frac{1}{\sigma}e^{-t^2\sigma^2-2t\mu}$$

OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.

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  • $\begingroup$ Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-\infty, \infty)$. Are you sure it is the one-tailed Laplace you need? $\endgroup$ – Nero Nov 22 '15 at 11:00
  • $\begingroup$ @Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided? $\endgroup$ – Luskentyrian Nov 22 '15 at 11:47
  • $\begingroup$ Yes, 2 sided is much easier because the range is unchanged by the transformation. $\endgroup$ – Nero Nov 22 '15 at 12:28
  • $\begingroup$ @Nero On second thoughts I think it should be double sided as the expectation would run from $-\infty$ to $\infty$. I have made a new attempt at this in an edit. $\endgroup$ – Luskentyrian Nov 30 '15 at 12:42
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Instead of doing the $u$-substitution:

$$ \begin{align} \mathcal{L}_X (t) & = \int_{-\infty}^\infty e^{-tx} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2-tx} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2tx\right)} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2t(x-\mu)+2t\mu\right)} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+2t\mu\right)} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+t^2\sigma^2-t^2\sigma^2+2t\mu\right)} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\left(\frac{x-\mu}{\sigma}\right)^2+2(t\sigma)\left(\frac{x-\mu}{\sigma}\right)+t^2\sigma^2\right)+\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2+\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\ & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2}e^{\frac{1}{2}t^2\sigma^2-t\mu} \mathrm{d}x \\ & = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}+t\sigma\right)^2} \mathrm{d}x \\ & = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{(x+t\sigma^2)-\mu}{\sigma}\right)^2} \mathrm{d}x \end{align} $$ Let $y=x+t\sigma^2$. $\mathrm{d}y=\mathrm{d}x$, $\lim_{x\to\infty}y\to\infty$, $\lim_{x\to-\infty}y\to-\infty$. $$ \begin{align} \mathcal{L}_X (t) & = e^{\frac{1}{2}t^2\sigma^2-t\mu} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2} \mathrm{d}y \\ & = e^{\frac{1}{2}t^2\sigma^2-t\mu} \end{align} $$

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