On 2 Jun 2000, Wen-Feng Hsiao wrote:
> I have no question about the derivation of Expectation and Variance
> for the 1x2 case (finite population with 2 classes) of hypergeometric
> distribution.
Given the sequel, perhaps you should have.
> However, when the problem is augemented to 2x2 case, I have no clue
> to find its Expected value and Variance. Suppose we have the
> following 2x2 contingency table:
> Row Level
> Col. 1 2 Total
> 1 n11 n12 n1.
> 2 n21 n22 n2.
> Total n.1 n.2 n
>
> Where nij is the number for cell{i, j}, n is the total number (the finite
> population).
So far, so good.
> I have no problem in representing the probability for nij, since
>
> P(nij) = C(n1., n11) x C(n2., n21)/ C(n, n.1)
Ah. Now why is this? You merely assert it without showing a
derivation. As you point out below, this quantity does not depend on i
nor j, which is counterintuitive and probably wrong. Why should it
depend only on the first column and the row marginals?
> = n1.! n2.! n.1! n.2! / (n! n11! n12! n21! n22!).
>
> Interestingly, it seems that the probability for each cell is the same,
> since P(nij) does not depend on i or j.
Pursue the logic. If P(nij) = constant, how can the marginal totals
ni. and n.j ever be UNequal?
> So I feel strange how one can
> derive its corresponding expectation and variance? My book shows
>
> E{nij|H0} = ni. n.j/n
> V{nij|H0} = n1. n2. n.1 n.2 / (n^2 (n-1)),
>
> where H0 is the condition that there is no association between row and
> column.
>
> Could someone give me a hint?
Perhaps you could give us a hint first, as to what led to the equation
you assert above without proof. One presumes you did not just pull it
"out of the blue" -- but then, perhaps one is being presumptuous.
-- DFB.
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