# What is the maximum expected value in a selection?

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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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.