2

I've got two distributions

$$ p_1(x) = \cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot e^{-x^2/2\sigma_1^2} $$

and similarly

$$ p_2(x) = \cfrac{1}{\sqrt{2\pi}\sigma_2}\cdot e^{-x^2/2\sigma_2^2} $$

I'm told that using the convolution theorem is the way to go so we'll start from there.

I know that $F(p_1(x)) = \int\limits_{-\infty}^\infty p_1(x) e^{2\pi ikx} dx$

so $$ \begin{align} F(p_1(x))F(p_2(x)) &= \int\limits_{-\infty}^\infty \cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx \cdot \int\limits_{-\infty}^\infty \cfrac{1}{\sqrt{2\pi}\sigma_2}\cdot e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\ &= \cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot \cfrac{1}{\sqrt{2\pi}\sigma_2} \int\limits_{-\infty}^\infty e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx \cdot \int\limits_{-\infty}^\infty e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\ &= \cfrac{1}{2\pi\sigma_1 \sigma_2} \int\limits_{-\infty}^\infty e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx \cdot \int\limits_{-\infty}^\infty e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\ \end{align} $$

And then I'm not sure what to do from here... I know that I'll take the inverse Fourier transform at the end and that will reveal the final gaussian distribution but I'm not sure how to evaluate these next couple steps... I've never used Fourier before and haven't taken a class that uses integrals in a complex space.

financial_physician
  • 813

1 Answers1

3

Let's be even more general by computing the convolution of $f_j:=\frac{1}{\sigma_j\sqrt{2\pi}}\exp\frac{-(x-\mu_j)^2}{2\sigma_j^2},\,j\in\{1,\,2\}$. First note $f_j$ has Fourier transform$$\begin{align}\int_{\Bbb R}\frac{1}{\sigma_j\sqrt{2\pi}}\exp\frac{-(x-\mu_j)^2+4\pi\sigma_j^2ikx}{2\sigma_j^2}dx&=\Bbb E[\exp2\pi ikX|X\sim N(\mu_j,\,\sigma_j^2)]\\&=\exp(2\pi ik\mu_j-2\pi^2k^2\sigma_j^2).\end{align}$$Now use$$\begin{align}f_1\ast f_2&=\mathcal{F}^{-1}(\mathcal{F}f_1\cdot\mathcal{F}f_2)\\&=\mathcal{F}^{-1}\exp(2\pi ik(\mu_1+\mu_2)-2\pi^2k^2(\sigma_1^2+\sigma_2^2))\\&=\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_2^2)}}\exp\frac{-(x-\mu_1-\mu_2)^2}{2(\sigma_1^2+\sigma_2^2)}.\end{align}$$

J.G.
  • 118,053
  • Thanks for the help! Would yoou mind explaining to me how you got for the indefinite integral to the Expectation? Since the Gaussian is symmetric, I can see that the first term in the numerator goes away. I can see that the expression within the Expectation is just the second term in the numerator divided by the denominator, where did the $1/(\sigma_j\sqrt{2\pi})$ go? I do recognize the evaluation of the expectation though so that's good.

    The other part that I don't quite understand is how you took the inverse Fourier. Could you try coloring that in a bit more for me?

     – financial_physician Sep 24 '20 at 22:10
  • In fewer words, I'd love a little help with 1) understanding how the Fourier transform of the distribution is what you have as the expectation and 2) how the inverse fourier transform of that expression is equal to that final pdf. Thanks you! – financial_physician Sep 24 '20 at 22:12
  • 1
    @financial_physician I started by evaluating the characteristic function at $t=2\pi k$, which is how you've defined the Fourier transform of PDFs (although your definition isn't limite to PDFs). The convolutions' multiplication clearly adds means, and also adds variances. Finally, we use $\mathcal{F}(f_1\ast f_2)=\mathcal{F}f_1\cdot\mathcal{F}f_2$, a standard result. – J.G. Sep 25 '20 at 06:00