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I'm working on a few problems in Reed & Simon, and I ran across this problem.

Compute the Fourier transform of $f(x) = e^{-\alpha x^2/2}$ via the following steps. (a) Prove that $-\lambda \hat{f}(\lambda)= \alpha \frac{d}{d\lambda}\hat{f}(\lambda)$ and conclude that $\hat{f}(\lambda) = ce^{-\lambda^2/(2\alpha)}$.

This is what I have so far (with help from first comment):

\begin{align*} -\frac{\alpha}{\lambda} \frac{d}{d\lambda} \hat{f}(\lambda) &= -\frac{\alpha}{\lambda \sqrt{2\pi}} \int \frac{d}{d\lambda} e^{-i\lambda x}e^{-\alpha x^2/2} dx \\ &= \frac{\alpha i}{\lambda\sqrt{2\pi}} \int x e^{-\alpha x^2/2} e^{\frac{2i\lambda}{\alpha}} dx \\ &= \frac{\alpha i}{\lambda\sqrt{2\pi}} \Big[-\frac{i\lambda}{\alpha} e^{-\frac{\lambda^2}{2\alpha}}\sqrt{\frac{2}{\alpha\pi}}\Big] \\ &=\frac{1}{\sqrt{\alpha}\pi}e^{-\frac{\lambda^2}{2\alpha}} \end{align*}

I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.

Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^{-x^2/2}$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.

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    $\begingroup$ Try integrating $\int e^{i \lambda x} (x e^{-\alpha x^2/2}) \,\mathrm{d}x$ by parts. $\endgroup$ – stochasticboy321 Jul 28 '18 at 0:48
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    $\begingroup$ Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ \int_{\mathbb{R}} e^{-i\lambda x} (x e^{-\alpha x^2/2}) \,\mathrm{d}x = \left. e^{-i \lambda x} \frac{- e^{-\alpha x^2 /2}}{\alpha}\right|_{-\infty}^{+ \infty} - i \frac{\lambda}{\alpha}{\int_{\mathbb{R}} e^{-i\lambda x} e^{-\alpha x^2/2} \,\mathrm{d}x } = -i \frac{ \lambda \sqrt{2\pi}}{\alpha} \hat{f}(\lambda).$$ $\endgroup$ – stochasticboy321 Jul 28 '18 at 2:16
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    $\begingroup$ So, from the second equality in the question, $$ -\frac{\alpha}{\lambda} \frac{\mathrm{d} }{\mathrm{d} \lambda} \hat{f} (\lambda) = \frac{i \alpha}{\lambda \sqrt{2\pi} } \cdot -i \frac{ \lambda \sqrt{2\pi}}{\alpha} \hat{f}(\lambda) = \hat{f} (\lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument. $\endgroup$ – stochasticboy321 Jul 28 '18 at 2:16
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Theorem: The characteristic function of a standard normal random variable $X$ is $ \varphi_{X}(t)=e^{-t^{2}/2} $.

Proof. Note that \begin{align*} \varphi_{X}(t)=\mathbb{E}\left[e^{itX}\right] & =\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}e^{itx}dx\\ & =\frac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^{0}e^{-\frac{x^{2}}{2}}e^{itx}dx+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{itx}dx\right)\\ & =\frac{1}{\sqrt{2\pi}}\left(\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{-itx}dx+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}e^{itx}dx\right)\\ & =\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx. \end{align*} Now, take the derivative with respect to $t$ to get $$ \varphi_{X}^{\prime}(t)=-\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}x\sin(tx)dx. $$ Integrate by parts to get \begin{align*} \varphi_{X}^{\prime}(t) & =\frac{2}{\sqrt{2\pi}}\left(e^{-\frac{x^{2}}{2}}\sin(tx)\mid_{0}^{\infty}-t\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx\right)\\ & =-\frac{2t}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}\cos(tx)dx\\ & =-t\varphi_{X}(t). \end{align*} Note that $$ \varphi_{X}^{\prime}(t)=-t\varphi_{X}(t) $$ is an ODE with solution $$ \varphi_{X}(t)=ce^{-t^{2}/2}. $$ We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $\varphi_{X}(0)=1$ so that $c=1$, as desired.

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  • $\begingroup$ This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you. $\endgroup$ – mathishard.butweloveit Jul 28 '18 at 1:11
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    $\begingroup$ Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $\varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept. $\endgroup$ – parsiad Jul 28 '18 at 1:16
  • $\begingroup$ I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks $\endgroup$ – mathishard.butweloveit Jul 28 '18 at 1:22

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