Integral of product of two normal distribution densities

Question

I want to compute the integral:

$\displaystyle \int^{\infty} _{-\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-x)^2}{2}} \frac{1}{\sqrt{2\pi}ab} e^{-\frac{x^2}{2(ab)^2}} dx$

Maybe we can use that for a normal distribution with mean $\mu$ and variance $\sigma^2$ we have

$\displaystyle \int^{\infty} _{-\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx = 1$

In an effort to write the integral in this form, I tried to take the exponents together. This gives:

$\displaystyle -\frac{(y-x)^2}{2} - \frac{x^2}{2(ab)^2} = \frac{-[(ab)^2 (y-x)^2 + x^2]}{2(ab)^2} = \frac{-[(ab)^2 (y^2 -2xy +x^2) + x^2]}{2(ab)^2}$

But this leads to nowhere. Any suggestions?

Answer 1

Here are general formulas for multivariate Gaussian distribution in $\mathbb{R}^D$ (derivation): $$\rho_{\mu, \Sigma}(x):= \frac{1}{\sqrt{|2\pi\Sigma|}} e^{-\frac 12 (x-\mu)^T\Sigma^{-1} (x-\mu)}$$

Integral of product of Gaussian distributions with covariance matrix $\Sigma$ and $\Gamma$, shifted by $\mu$ vector:

$$\int_{\mathbb{R}^D} \rho_{\mu, \Sigma}(x)\cdot\rho_{\mathbf{0},\Gamma}(x)\,dx=\frac{\exp\left(-\frac 12 (\mu^T\Sigma^{-1}(\Sigma^{-1}+\Gamma^{-1})^{-1} \Gamma^{-1}\mu) \right)} {\sqrt{(2\pi)^D |\Sigma||\Gamma||\Sigma^{-1}+\Gamma^{-1}|}}$$ For spherically symmetric $\Sigma=\sigma^2 \mathbf{I}$, $\Gamma=\gamma^2 \mathbf{I}$ shifted by any length $l$ vector it becomes: $$\frac{\exp \left(-\frac 12 \frac {l^2}{\sigma^2+\gamma^2} \right)} {\sqrt{2\pi \left(\sigma^{2}+\gamma^{2}\right)}^D}$$

Answer 2

Actually, you're on the right track. Just keep on going with $$\begin{align}(ab)^2(y^2-2xy+x^2)+x^2 & =(a^2b^2+1)x^2-2a^2b^2yx+a^2b^2y^2 \\ & =(a^2b^2+1)\left[x^2-2\frac{a^2b^2y}{a^2b^2+1}x\right]+a^2b^2y^2 \\ & =(a^2b^2+1)\left[\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2-\frac{a^4b^4y^2}{(a^2b^2+1)^2}\right]+a^2b^2y^2 \\ & =(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2+\left[a^2b^2-\frac{a^4b^4}{a^2b^2+1}\right]y^2 \\ & =(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2+\left(\frac{a^2b^2}{a^2b^2+1}\right)y^2 \\ \end{align}$$ Then we know that $$\int_{-\infty}^{\infty}e^{\frac{-(a^2b^2+1)\left(x-\frac{a^2b^2y}{a^2b^2+1}\right)^2}{2a^2b^2}}dx=\int_{-\infty}^{\infty}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\sqrt{2\pi}\sigma$$ Where $$\mu=\frac{a^2b^2y}{a^2b^2+1}$$ $$\sigma=\frac{ab}{\sqrt{a^2b^2+1}}$$ So now you have $$\begin{align}\int_{-\infty}^{\infty}\frac1{\sqrt{2\pi}}e^{-\frac{(y-x)^2}{2}}\frac1{\sqrt{2\pi}ab}e^{-\frac{x^2}{2(ab)^2}}dx & =\frac1{\sqrt{2\pi}}\frac1{\sqrt{2\pi}ab}\frac{\sqrt{2\pi}ab}{\sqrt{a^2b^2+1}}e^{-\frac{y^2}{2\left(\sqrt{a^2b^2+1}\right)^2}} \\ & = \frac1{\sqrt{2\pi}\sqrt{a^2b^2+1}}e^{-\frac{y^2}{2\left(\sqrt{a^2b^2+1}\right)^2}} \end{align}$$ So the means add, $0+0=0$, as do the variances, $1+a^2b^2=\left(\sqrt{a^2b^2+1}\right)^2$, just like they are supposed to.

Answer 3

Since the given integral is a convolution an alternative solution could be based on Fouriertransformation.

We know that $$ e^{-x^{2}}\overset{\mathcal F}{\longmapsto} \sqrt{\pi}e^{-{\xi}^{2}/4}. $$ where ${\mathcal F}$ is the Fouriertransformation.

It is a basic fact that $$ f(x)\overset{\mathcal F}{\longmapsto} F(\xi) \Longrightarrow f(ax)\overset{\mathcal F}{\longmapsto} \dfrac{1}{|a|}F\left(\dfrac{\xi}{a}\right). $$ where $a$ is a non-zero real number. Now the Fourier transformation of the given integral yields that $$ \dfrac{1}{\sqrt{2\pi}}\sqrt{2\pi}e^{-2\xi^{2}/4}\dfrac{1}{\sqrt{2\pi}ab}\sqrt{2\pi}abe^{-2(ab)^{2}\xi^{2}/4}= e^{-2(1+(ab)^{2})\xi^{2}/4}. $$ Finally we receive the integral via inverse Fourier transformation $$ \dfrac{1}{\sqrt{2\pi}\sqrt{1+a^{2}b^{2}}}e^{-y^{2}/(2(1+a^{2}b^{2}))}. $$