Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for some insight into a problem:

Consider a group of $T$ persons, and let $a_1, a_2, ..., a_T$ denote the height of these $T$ persons. Suppose that $n$ are selected from this group at random without replacement, and let $X$ denote the sum of the heights of these $n$ persons. I'm supposed to find the mean and the variance.

This is in the section of hypergeometric distributions for my probability textbook, so I am trying to see how to turn this into an "urn" problem. Normally, I think one would split $X = X_1+...+X_T$ where $X_i = 1$ is chosen, $0$ otherwise, where the probability in this problem would be $Pr(X_i = 1) = \frac{1}{T}$, but I am confused because each $X_i$ is "weighted" by the height. I was having trouble finding other useful examples online.

EDIT: Here is how the book explains hypergeometric distributions:

Assume that $n$ balls are selected at random without replacement from a box containing $A$ red balls and $B$ blue balls. The expected number of red balls is $E(X) = \frac{nA}{A+B}$ and the variance is $Var(X) = \frac{nAB}{(A+B)^2}\cdot\frac{A+B-n}{A+B-1}$

I am trying to understand how to use these formulas for this particular problem. I am pretty sure I can calculate each straight from the definitions.

share|improve this question
    
this is a problem from "Probability and Statistics" by Degroot and Schervish, 5.3.8 –  Tyler Mar 23 '11 at 1:16

2 Answers 2

up vote 1 down vote accepted

It seems to me that the average is very intuitive and just what you'd expect (g). So try starting with:
$$X = \Sigma_{i=1}^T a_i\;X_i$$

share|improve this answer
    
It seems to me that will give us $\frac{a_1 + ... + a_T}{T}$, the average height of everyone, so we would just multiply by $n$ to give us $E(X)$, is this correct? –  Tyler Mar 23 '11 at 1:46
    
@Tyler; I didn't do the math, but it's hard for me to imagine any other result. Now, do you have a formula for the variance? –  Carl Brannen Mar 23 '11 at 1:47
    
Well, I have $Var(X) = E((X-\mu)^2) = E(X^2) - E(X)^2$, but I am trying to use one specifically for hypergeometric distributions, and I'm not sure how to apply it. The text explains this distribution as: given A red balls and B blue balls, select n at random without replacement. I feel like my problem has "Select n balls at random without replacement, and each ball has a number, then add up the numbers." I am having problems trying to figure out how to go between the two concepts. –  Tyler Mar 23 '11 at 2:01
    
@Tyler; so it boils down to you need to calculate $E(X^2) = $ $E(\Sigma_i\Sigma_j a_ia_jX_iX_j)$. Hmmm. Seems like the math might be similar to how your textbook did the calculation for the hypergeometric variance. –  Carl Brannen Mar 23 '11 at 2:32
    
Haha thanks... Yikes. I was hoping for a more elegant answer, but I guess the authors just wanted us to get our hands (very) dirty. –  Tyler Mar 23 '11 at 2:36

To compute the variance, you need to compute $E(X^2)$; the rest is simple. For simplicity I'll sketch this in the case when $n = 3$. So you have

$$ E(X^2) = E((X_1 + X_2 + X_3)^2) = E(X_1^2 + X_2^2 + X_3^2 + 2X_1 X_2 + 2X_1 X_3 + 2X_2 X_3)$$.

By linearity of expectation this is

$$ E(X_1^2) + E(X_2^2) + E(X_3^2) + 2E(X_1 X_2) + 2(X_1 X_3) + 2(X_2 X_3) $$

and some of these terms are equal to each other by symmetry; then you can write each of them in terms of the $a_i$. This works similarly (but is a bit harder to notate) for larger $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.