Past exam of the options I course of the University of Oxford / 2004 / O

For equal one-second observation windows, the sample mean estimates the Poisson distribution mean:This matches the expected mean count; it is also the maximum-likelihood estimator for independent Poisson distribution observations. With $N=12169$ windows, fitted expected frequencies are $E_j=N{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!$ for $j=0,\dots ,4$, and $E_{\ge 5}=N[1-{\sum }_{j=0}^4{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!]$.

Each last-column entry is $(O_j-E_j{)}^2/E_j$, and the Pearson statistic is $\boxed{X^2=\sum _{j\in \{0,1,2,3,4,\ge 5\}}(O_j-E_j{)}^2/E_j}$. The displayed expected frequencies and contributions are rounded, so their printed values need not reproduce every final digit of a calculation using unrounded probabilities.

Separate high-count classes have very small expected frequencies, making the chi-square approximation unreliable. Pooling the upper tail gives a class with expected count about $20.9$, comfortably above the customary small-cell threshold. The pooling protects the asymptotic test approximation, while retaining the tail count as part of the goodness-of-fit comparison.

There are six pooled classes, one normalization constraint, and one fitted Poisson distribution parameter. Thus the reference degrees of freedom are $\boxed{6-1-1=4}$. Under the null model the Pearson statistic has approximately the Chi-squared distribution with four degrees of freedom. Its supplied 5% upper critical value is $9.49$.

Because $\boxed{2.07<9.49}$, do not reject the Poisson count distribution at the 5% level. This is consistency with that distribution, not proof that it is exactly true. A test of the marginal count frequencies also does not by itself verify independence of successive time windows.

For nondegenerate design let $S_{xx}={\sum }_i{x}_i^2>0$. Maximizing the Gaussian likelihood over the mean parameters is equivalent to minimizing ${\sum }_i(Y_i-\alpha -\beta x_i{)}^2$. Since ${\sum }_i{x}_i=0$, the normal equations separate:Its error is the linear combination $\hat{\beta }-\beta ={\sum }_i{x}_i{ϵ}_i/S_{xx}$. Independence and normality therefore giveIf every design point is zero, the slope is not identifiable and no such slope estimator can be inferred; the standard regression answer assumes the stated nondegeneracy.

For $n>2$, the usual unbiased estimator of variance isThe fitted mean lies in the two-dimensional span of the constant design vector and $x$. Orthogonal projection of the Gaussian error onto its complement gives $n-2$ independent standardized normal coordinates. HenceThis residual projection is independent of the fitted-coefficient projection. If maximum likelihood for variance is intended instead, its estimator is ${\hat{\sigma }}_{\mathrm{ML}}^2=\mathrm{RSS}/n$, with $n{\hat{\sigma }}_{\mathrm{ML}}^2/{\sigma }^2\sim {\chi }_{n-2}^2$; the denominator differs, not the residual degrees of freedom.

Independence of the slope estimate and residual variance, together with their normal and chi-square laws, gives the Student pivotal quantityThus an exact Gaussian-model two-sided 95% confidence interval isUsing a normal quantile without estimating the variance uncertainty would not give this exact finite-sample coverage.