# Computation of a Particular Convolution

Let $\xi_{1}, \xi_{2}, \xi_{3}$ be i.i.d. $N(0,1)$. I'm attempting to compute the density of $\max \{\xi_{1}, \xi_{2}\} + \xi_{3}$. I know the density of $\max \{\xi_{1}, \xi_{2}\}$ is $2\Phi(y) \phi(y)$ and the density of $\xi_{3}$ is $\phi(s)$ so that the density of interest is the convolution $\int_{\mathbb{R}} 2\Phi(y) \phi(y) \phi(z-y)dy$. Is there any way to get at this expression, say, express it in terms of $\Phi$?

-
Did you try integrating by parts? May be $\phi(y)$ as the second function? I am trying too, I'll post an answer if I succeed. – user21436 Apr 17 '12 at 9:26

The density of $\max\{\xi_1,\xi_2\}$ is given by $$g(t):=\frac 1{2\pi}e^{-t^2/2}\int_{-\infty}^te^{-s^2/2}ds$$ so we want to compute $$f(x):=\frac 1{2\pi}\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-t^2/2}\int_{-\infty}^te^{-s^2/2}ds\cdot e^{-(x-t)^2/2}dt.$$ We can write \begin{align*} e^{-t^2/2}e^{-(x-t)^2/2}&=\exp\left(-\frac 12(t^2+x^2-2xt+t^2)\right)\\\ &=\exp\left(-(t^2-xt)-\frac{x^2}2\right)\\\ &=\exp\left(-\frac{x^2}2\right)\exp\left(-\left(t-\frac x2\right)^2+\frac{x^2}4\right)\\\ &=\exp\left(-\frac{x^2}4\right)\exp\left(-\left(t-\frac x2\right)^2\right) \end{align*} hence \begin{align*}f(x)&=(2\pi)^{-3/2}\exp\left(-\frac{x^2}4\right)\int_{-\infty}^{+\infty}\exp\left(-\left(t-\frac x2\right)^2\right)\int_{-\infty}^te^{-s^2/2}dsdt\\\ &=(2\pi)^{-3/2}\exp\left(-\frac{x^2}4\right)\int_{-\infty}^{+\infty}\exp\left(-y^2\right)\int_{-\infty}^{y+x/2}e^{-s^2/2}dsdy. \end{align*} Put $h(x):=e^{x^2/4}(2\pi)^{3/2}f(x)$. We have \begin{align*} h'(x)&=\frac 12\int_{-\infty}^{+\infty}\exp\left(-y^2\right)e^{-(y+x/2)^2/2}dy \end{align*} and \begin{align*} \exp\left(-y^2\right)e^{-(y+x/2)^2/2}&=\exp\left(-\frac 32 y^2-xy-x^2/4\right)\\\ &=\exp\left(-\frac 32\left(y^2+\frac 23xy+\frac{x^2}6\right)\right)\\\ &=\exp\left(-\frac 32\left(\left(y+\frac 13x\right)^2-\frac{x^2}9+\frac{x^2}6\right)\right)\\\ &=\exp\left(-\frac{x^2}{12}\right)\exp\left(-\frac 32\left(y+\frac 13x\right)^2\right) \end{align*} hence \begin{align*} h'(x)&=\frac 12\exp\left(-\frac{x^2}{12}\right)\int_{-\infty}^{+\infty}e^{-3t^2/2}dt\\\ &=\frac 12\frac 1{\sqrt 3}\exp\left(-\frac{x^2}{12}\right)\int_{-\infty}^{+\infty}e^{-s^2/2}ds\\\ &=\frac 12\frac 1{\sqrt 3\sqrt{2\pi}}\exp\left(-\frac{x^2}{12}\right). \end{align*} We deduce that $$f(x)=\frac 12(2\pi)^{-3/2}\frac 1{\sqrt 3\sqrt{2\pi}}e^{-x^2/4}\int_{-\infty}^x\exp\left(-\frac{t^2}{12}\right)dt.$$

-
thank you, much appreciated – Red Rover Apr 18 '12 at 6:02
One question: I'm getting that your expression for $g$ is off by a constant of 2. Am I just calculating wrong or...? – Red Rover Apr 18 '12 at 6:22
Also, is there an easy way to generalize this for $\xi_{3}$ having variance $\sigma^{2}$? or will I have to perform this calculation again with the messy completion of squares...? – Red Rover Apr 18 '12 at 7:35
I missed the constant $2$, I will edit it. For the generalization, you can make the substitution $t':=\frac{x-t}{\sigma}$ in the expression of $f$; I think it will give you a nicer integral to compute. – Davide Giraudo Apr 18 '12 at 8:35