Let $X_1...X_n\sim N(\mu,\sigma)$ be normal random variables. Find the expected value of $\max_i(X_i)$ and $\min_i(X_i)$.

The sad truth is I don't have any good idea how to start and I'll be glad for a hint.

up vote 8 down vote accepted

Presumably the $X_i$ are independent. If $\Phi$ is the standard normal cdf, $$P(\max_i X_i < \mu + t \sigma) = \prod_i P(X_i < \mu + t \sigma) = \Phi(t)^n$$ so $$ E[\max_i X_i] = \mu + \sigma \int_{-\infty}^\infty t \dfrac{d}{dt} \Phi(t)^n\ dt $$ Maple tells me that for $n=2$ the integral is $1/\sqrt{\pi}$, and for $n=3$ it is $3/(2\sqrt{\pi})$. It doesn't know of closed forms for $n > 3$, and neither do I. The numerical values for $n=2$ to $10$ are $\matrix{.56418958354775628695,& .84628437532163443042,& 1.0293753730039641321,\cr 1.1629644736405196128,& 1.2672063606114712976,& 1.3521783756069043992,\cr 1.4236003060452777531,& 1.4850131622092370063,& 1.5387527308351728560\cr}$

  • Why you took $P(max<\mu+t\sigma)?$ – user65985 Aug 22 '13 at 1:47
  • 1
    Scaling and translation: $X_i = \mu + \sigma Z_i$ where $Z_i$ are standard normal, and $\max_i X_i = \mu + \sigma \max_i Z_i$, so this takes care of the dependence on $\mu$ and $\sigma$. – Robert Israel Aug 22 '13 at 4:07
  • Probably unhelpful, but the median of this distribution is $\sqrt{2} \text{erf}^{-1}\left(2^{\frac{n-1}{n}}-1\right)$ – barrycarter Mar 17 '16 at 15:26

As Robert Israel says, you only need to consider standard $N(0,1)$s. If $E_n$ is the expected value of the maximum of $n$ independent $N(0,1)$s, then the expected value of $\max_i(X_i)$ is $\mu+\sigma E_n$ and the expected value of $\min_i(X_i)$ is $\mu-\sigma E_n$.

Actually this is quite a cute problem, because there is a closed form up to $n=5$ given by $E_1=0$, $E_2=\pi^{-1/2}$, $E_3=(3/2)\pi^{-1/2}$, $E_4=3\pi^{-3/2}\cos^{-1}(-1/3)$, $E_5=(5/2)\pi^{-3/2}\cos^{-1}(-23/27)$.

But I don't think you will get a neat closed form for general $n$. Depending on your application, maybe an approximation formula would be useful, or maybe a computer program would be good. The first approximation formula you might use is $E_n\sim\sqrt{2\log(n)}$, but there are more accurate (and elaborate) ones if that is what you need (in which case, please say so).

To show the first five values of $E_n$ are as above, let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ be the standard normal density and $\Phi(x)$ the standard cdf. Then \begin{equation}\begin{split} E_n&=\int_{-\infty}^\infty x\dfrac{d}{dx}\Phi(x)^ndx\\ &=n\int_{-\infty}^\infty x\phi(x)\Phi(x)^{n-1}dx\\ &=n(n-1)\int_{-\infty}^\infty \phi(x)^2\Phi(x)^{n-2}dx\quad\text{(by parts)}\\ &=n(n-1)\int_{-\infty}^\infty \phi(x)^2\left(A(x)+\tfrac12\right)^{n-2}dx\quad\text{(where $A(x)=\Phi(x)-\tfrac12$),}\\ &=\frac{n(n-1)}{2\pi}\sum_{r=0}^{[n/2]-1}(\tfrac12)^{n-2-2r}\binom{n-2}{2r}\int_{-\infty}^\infty e^{-x^2}A(x)^{2r}dx\quad\text{(using antisymmetry of $A(x)$)}. \end{split}\end{equation}

So the $E_n$ come in pairs, and $E_{2n}$ and $E_{2n+1}$ can be written in terms of $\int_{-\infty}^\infty e^{-x^2}A(x)^{2r}dx$ for $r=0, \ldots, n-1$. We can get a closed form for this integral for $r=0$ and also, slightly surprisingly, for $r=1$, giving $E_1, \ldots, E_5$ above.

The only way I know how to do this is by introducing $$I_n(s)=\int_{-\infty}^\infty e^{-sx^2/2}A(x)^{2n}dx.$$ Then $I_n(s)$ satisfies a strange reduction rule $$I_{n+1}(s)=\frac{2n+1}{2\pi\sqrt{s}}\int_0^{1/\sqrt{s}}\left(1+y^2\right)^{-1}I_{n}\left(1+s(1+y^2)\right)dy.$$ As a consequence, \begin{equation}\begin{split} I_0(s)&=\sqrt\frac{2\pi}{s},\\ I_1(s)&=\frac{1}{\sqrt{2\pi s}}\tan^{-1}\left(\frac{1}{\sqrt{s(s+2)}}\right),\\ %I_2(s)&=3(2\pi)^{-3/2}s^{-1/2}\int_0^{s^{-1/2}}\left(1+y^2\right)^{-1}\left(3+2y^2\right)^{-1/2}\tan^{-1}\left(\left((3+2y^2)(5+2y^2)\right)^{-1/2}\right)dy \end{split}\end{equation} and the above formulae for $E_1, \ldots, E_5$ follow from the values of $I_0(2)$ and $I_1(2)$, using standard trig equivalences.

To prove the reduction rule, \begin{equation}\begin{split} \int_0^{1/\sqrt{s}}\left(1+y^2\right)^{-1}&I_{n}\left(1+s(1+y^2)\right)dy\\ &=\int_0^{1/\sqrt{s}}\int_{-\infty}^\infty\left(1+y^2\right)^{-1}e^{-(1+s(1+y^2))x^2/2}A(x)^{2n}dxdy\\ &=\int_0^{1/\sqrt{s}}\int_{-\infty}^\infty\left(1+y^2\right)^{-1}e^{-s(1+y^2)x^2/2}\sqrt{2\pi}\phi(x)A(x)^{2n}dxdy\\ &=\int_0^{1/\sqrt{s}}\int_{-\infty}^\infty\left(1+y^2\right)^{-1}e^{-s(1+y^2)x^2/2}\frac{\sqrt{2\pi}}{2n+1}\dfrac{d}{dx}A(x)^{2n+1}dxdy\\ &=s\frac{\sqrt{2\pi}}{2n+1}\int_0^{1/\sqrt{s}}\int_{-\infty}^\infty xe^{-s(1+y^2)x^2/2}A(x)^{2n+1}dxdy\quad\text{(by parts in $x$)}\\ &=s\frac{\sqrt{2\pi}}{2n+1}\int_{-\infty}^\infty xe^{-sx^2/2}A(x)^{2n+1}\left(\int_0^{1/\sqrt{s}} e^{-sy^2x^2/2}dy\right)dx\\ &=s\frac{\sqrt{2\pi}}{2n+1}\int_{-\infty}^\infty xe^{-sx^2/2}A(x)^{2n+1}\left(\int_0^x \frac{e^{-y^2/2}}{\sqrt{s}x}dy\right)dx\\ &=\sqrt{s}\frac{2\pi}{2n+1}\int_{-\infty}^\infty e^{-sx^2/2}A(x)^{2n+2}dx\\ &=\sqrt{s}\frac{2\pi}{2n+1}I_{n+1}(s). \end{split}\end{equation}

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.