Re: expectation of the minimum

From: Jack Tomsky (jtomsky_at_ix.netcom.com)
Date: 03/25/05 

Date: Fri, 25 Mar 2005 11:09:12 EST

> I did a simulation on Excel by generating 10,000
> pairs of independent N(0, 1) and getting x and y with
> the appropriate variances and correlation by linear
> combinations. For various r, I came up with the
> following expected minimums. It appears that your
> intuition is correct.
>
> r E(min)
> 1.0 0.000
> 0.5 -0.421
> 0.0 -0.583
> -0.5 -0.704
> -1.0 -0.799
>
> Jack

Here's a more complete table. For r = -1, I replaced the simulated E(min) by the analytic value, E(min) = -E(|x|) = -sqrt(2/pi) = -0.798.

 r E(min)
1.0 0.000
0.9 -0.199
0.8 -0.274
0.7 -0.331
0.6 -0.379
0.5 -0.421
0.4 -0.459
0.3 -0.493
0.2 -0.525
0.1 -0.555
0.0 -0.583
-0.1 -0.609
-0.2 -0.634
-0.3 -0.658
-0.4 -0.681
-0.5 -0.704
-0.6 -0.725
-0.7 -0.745
-0.8 -0.765
-0.9 -0.784
-1.0 -0.798

   Jack

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