Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / O / O4

Ignoring the multinomial coefficient, the likelihood is $L(\theta )={\prod }_{i,j}{\theta }_{ij}^{n_{ij}}$. Maximize $\ell ={\sum }_{i,j}{n}_{ij}\log {\theta }_{ij}$ with $\sum {\theta }_{ij}=1$. A Lagrange multiplier gives $n_{ij}/{\theta }_{ij}=\lambda$; summing the resulting ${\theta }_{ij}=n_{ij}/\lambda$ gives $\lambda =N$. ThusBecause every count is positive, the log-likelihood is strictly concave on the interior simplex and this solution is its unique maximum.

Under the product constraint the log-likelihood separates:Maximizing the two simplex problems independently gives ${\hat{\alpha }}_i=n_{i.}/N$ and ${\hat{\beta }}_j=n_{.j}/N$. HenceThe fitted expected cell counts under independence are consequently $N{\hat{\theta }}_{ij}^0=n_{i.}{n}_{.j}/N$.

The generalized likelihood ratio compares the constrained maximum from part (b) with the unrestricted maximum from part (a). Taking minus twice its logarithm givesThe unrestricted $3\times 3$ model has $9-1=8$ free parameters; the independence model has $(3-1)+(3-1)=4$. Thus the asymptotic null distribution is ${\chi }_4^2$, with four degrees of freedom. The supplied value givesTherefore reject independence at the 5% level, but not at the 1% level. There is evidence of an association between the two classifications. The fitted expected counts are all about nine or larger, so the usual chi-squared approximation has no very small expected cells here.