Expected Values of Max(X1, X2) and Min(X1, X2)?

[qquito]

Dear Everyone:

X1 and X2 are two independent samples from the standard normal
distribution, N(0, 1). What are the expected values of Max(X1, X2) and
Min(X1, X2)?

I know that these are the order statistics when the number of samples
is 2, and their CDFs are, respectively, [F(z)]^2 and 1-[1-F(z)]^2, but
how do you proceed from here to evaluate their *expected* values?

Thank you for reading and replying!

--Roland

[Jack Tomsky]

The only way to get those expected values is numerically. We certainly have E(Max(X1, X2)) = -E(Min(X1, X2)).

The following paper has tables which includes your case of n = 2.

H. Leon. Harter (1961),"Expected values of normal order statistics", Biometrika, pp. 151-165.

Jack
www.tomskystatistics.com

[Ray Koopman]

E[max[X1,X2]] = Integrate 2 x f[x] F[x] from -infinity to infinity,

where f and F are the standard normal pdf and cdf.

Mathematica gets it symbolically as 1/sqrt[pi] = .564190

E[min[X1,X2]] = -E[max[X1,X2]]

[dvsarwate]

On Nov 13, 5:26 pm, Ray Koopman <koop...@sfu.ca> wrote:

>
> E[max[X1,X2]] = Integrate 2 x f[x] F[x] from -infinity to infinity,
>
> where f and F are the standard normal pdf and cdf.
>
> Mathematica gets it symbolically as 1/sqrt[pi] = .564190
>
> E[min[X1,X2]] = -E[max[X1,X2]]

In fact, it is also straightforward to evaluate
the integral of 2 x f[x] F[x] from -infinity
to infinity analytically to get the result
obtained by Mathematica. Since the antiderivative
of x f[x] is -f[x], integration by parts gives

E[max[X1,X2]] = (1/pi) integral exp(-x^2)

and since the pdf of a N(0, 1/2) random variable
is (1/sqrt(pi))exp(-x^2), the expected value is
1/sqrt(pi).

--Dilip Sarwate

[qquito]

Thank you both, Jack and Ray, very much for the helpful and useful
information! --Roland

[qquito]

Thank you for the reply! I searched a bit more and found the following
paper

Jones, H.L., 1948. Exact Lower Moments of Order Statistics in Small
Samples from A Normal Distribution. The Annals of Mathematical
Statistics, 19, 270-273.

And this paper gives some exact moments for n=2, 3, 4. For examples,
for n=2, the E[max(X1, X2)] = 1/sqrt(pi); for n=3, E[max(X1, X2, X3)]
= 3/[2sqrt(pi)].