Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Phi$?

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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 $$f(x)=\frac 12(2\pi)^{-3/2}\frac 1{\sqrt 3\sqrt{2\pi}}e^{-x^2/4}\int_{-\infty}^x\exp\left(-\frac{t^2}{12}\right)dt.$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.