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Let $g_t(x)=\frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$, in some probability (stochastic processes) papers and exercises I saw that this function satisfies the (very useful) property: $$ (g_t * g_s)(x)=g_{t+s}(x) $$

Writing down the integrals, this is equivalent to:

$$ \frac{1}{\sqrt{2\pi(t+s)}}e^{-\frac{x^2}{2(t+s)}}=\int_{\mathbb{R}} \frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}} \frac{1}{\sqrt{2\pi s}}e^{-\frac{y^2}{2s}} \, dy $$

I tried out various basic substitutions with various coefficients (like $y\sqrt{t+s}=u$) and also tried to write something as an integral and then switch integration order (Fubini-Tonelli), but I couldn't turn it into a form I can handle yet. May someone give me some hints? Thanks in advance.

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Note that by completing the square we have

$$\begin{align} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}} \frac{1}{\sqrt{2\pi s}}e^{-\frac{y^2}{2s}} \, dy&=\frac{1}{\sqrt{4\pi^2st}}\int_{-\infty}^\infty e^{-\frac1{2st}\left(s(x-y)^2+ty^2\right)}\,dy\\\\ &=\frac{1}{\sqrt{4\pi^2st}}\int_{-\infty}^\infty e^{-\frac1{2st}\left((s+t)\left(y-\frac{sx}{s+t}\right)^2+\frac{st}{s+t}x^2\right)}\,dy\\\\ &=\frac{1}{\sqrt{4\pi^2st}}e^{-\frac{x^2}{s+t}}\underbrace{\int_{-\infty}^\infty e^{-\frac{s+t}{2st}y^2}\,dy }_{=\sqrt{\frac{\pi(2st)}{s+t}}}\\\\ &=\frac{1}{\sqrt{2\pi(s+t)}}\,e^{-\frac{x^2}{s+t}} \end{align}$$

as was to be shown!

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    I can't believe..tried out circus substitutions while it was just some clear algebra. Kind of overestimated it. By the way, fast and neat solution, thank you :) – 2017-02-09
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    You're welcome. My pleasure. -Mark – 2017-02-09
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Do some algebra including completing the square: $$ \frac{(x-y)^2}{2t} + \frac{y^2}{2s} = \frac{(y-[{\cdots}])^2}{2(\cdots)}. $$

If you have $\displaystyle\int_{-\infty}^\infty e^{-(y-[\cdots])^2/(2[\cdots])} \, dy,$ what do you get?