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Here's a hypothetical problem: assume that mean diameter of a tennis ball is 6.7 cm. Assume that the diameter is normally distributed with a standard deviation of 0.1 cm (I may have picked up a weird value but that's not the point). Now say I buy a billion tennis balls. What is the expected diameter of the largest ball? What's the expected diameter of the smallest ball?

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3 Answers 3

Hint: Since your diameters are independent and identically distributed (IID) according to a Gaussian distribution, you essentially need to find the distribution of two extreme order statistics - min and max. Consider the cumulative distribution function (CDF) for the max random variable:

\begin{align} P(max(X_1,\ldots,X_n) \leq z) &= P(X_1 \leq z, X_2 \leq z,\ldots, X_n \leq z) \\ &= \prod_{i=1}^n P(X_i \leq z) \end{align}

The last step is due to independence. Since you know the CDF for a Gaussian random variable, you know the CDF of the max random variable as well. This will help to compute the PDF, which will lead you to the mean. Proceed with the min random variable in a similar fashion.

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I think what you should look for is the asymptotic density of the max (min) of a collection of iid Gaussian random variables. The mean of that density will tell you the expected value of the largest of the set. (I am using the word expected in the sense of expectation of a random variable.)

This post might help: Expectation of the maximum of gaussian random variables

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Also for the minimum you use the fact that P[min X$_i$>x] =P{X$_1$>x, X$_2$>x..X$_n$>x]= ∏P[X$_i$>x]. Now if you are looking at the maximum of a sample of size 10. The distribution G(x) =F$^10$(x) and the density is G'(x)= 10 F$^9$(x) f(x) where f is the normal density and F is th cumulative normal. To get the expected value you integrate ∫x 10 F$^9$(x) f(x)dx integrating from-infinity to + infinity.

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at Michael and Kartik Audhkhasi Thanks for the answer, but both of these answers tell me the probability of the max X (or min X) being greater than (or less than) a certain value. What I was looking for is the expected maximum (or minimum) value in say a billion selections. –  Shashank Sawant Aug 29 '12 at 18:30
    
@ShashankSawant Since distribution is Gaussian therre is no upper or lower limit to the observed values and so without normalization the maximum goes to infinity and the minimum to -infinity with probability 1. Hence the expected values also go to infinity and -infinity respectively. –  Michael Chernick Aug 29 '12 at 18:56
    
Thanks for quick resp. But I don't think it's that simple. The number of selections matter in the case maximum expected value (let's focus on the maximum, the minimum can be calculated similarly). Here's what I mean: suppose you randomly select 10 balls from the factory and then note the maximum diameter of the ball. If I start noting the maximum value of a such selections of 10, the maximum noted values will form a distribution around a mean. This mean is the maximum expected value for a selection of 10. Now if I increase the selection size to 2 billion the mean(expectation)will be different. –  Shashank Sawant Aug 29 '12 at 19:06
    
@ShashankSawant So you want the expected value of the maximum of a sample of size 10. That is different. –  Michael Chernick Aug 29 '12 at 19:09
    
In my original question, the sample size was 1 billion. The size of 10 is just an example to emphasize that the size of the selection is a finite number (1 billion in my original question). –  Shashank Sawant Aug 29 '12 at 21:36
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