Solution. With \(X\sim \) GAM\((2,\beta _1)\) and \(Y\sim \) GAM\((2,\beta _2)\) independent, \[L(\beta _1,\beta _2) = \frac {x\,e^{-x/\beta _1}}{\beta _1^{2}}\cdot \frac {y\,e^{-y/\beta _2}}{\beta _2^{2}} .\] Unrestricted, the maximum likelihood estimates are \(\widehat {\beta }_1 = x/2\) and \(\widehat {\beta }_2 = y/2\). Under \(H_0:\beta _1=\beta _2=\beta \) the total \(x+y\) has shape \(4\), so \(\widehat {\beta } = (x+y)/4\). Substituting, \[\Lambda = \frac {\sup _{H_0}L}{\sup L} = \frac {\left [(x+y)/4\right ]^{-4}e^{-4}} {\left (x/2\right )^{-2}\left (y/2\right )^{-2}e^{-2}e^{-2}} = \frac {\left (x/2\right )^{2}\left (y/2\right )^{2}}{\left [(x+y)/4\right ]^{4}} = \frac {64\,x^{2}y^{2}}{(x+y)^{4}} .\] Writing \(t = x/(x+y)\), so that \(x=t(x+y)\) and \(y=(1-t)(x+y)\), the factors of \((x+y)\) cancel and \[\Lambda = 64\,t^{2}(1-t)^{2} = 64\left [t(1-t)\right ]^{2},\] a function of \(T\) alone.

The distribution of \(T\) under \(H_0\)

Under \(H_0\), \(X\) and \(Y\) are independent gamma variables with the same scale and shapes \(2\) and \(2\), so \[T = \frac {X}{X+Y} \ \sim \ \text {BETA}(2,2),\] with density \(6t(1-t)\) on \((0,1)\). The common scale \(\beta \) cancels, which is what makes the null distribution free of the nuisance parameter and the test usable.

Solution. Write \(S=\sum _{i=1}^{n}X_i\) and \(U=\sum _{i=1}^{n}Y_i\). With the exponential parameterised by its mean, \[L(\lambda ,\mu ) = \lambda ^{-n}e^{-S/\lambda }\,\mu ^{-n}e^{-U/\mu },\] so unrestricted \(\widehat {\lambda }=S/n\), \(\widehat {\mu }=U/n\), while under \(H_0:\lambda =\mu \) the pooled estimate is \(\widehat {\lambda }_0=(S+U)/(2n)\). Hence \[\Lambda = \frac {\left [(S+U)/2n\right ]^{-2n}e^{-2n}} {\left (S/n\right )^{-n}\left (U/n\right )^{-n}e^{-n}e^{-n}} = \frac {\left (S/n\right )^{n}\left (U/n\right )^{n}}{\left [(S+U)/2n\right ]^{2n}} = 2^{2n}\,\frac {S^{n}U^{n}}{(S+U)^{2n}} .\] Putting \(T = S/(S+U)\) gives \[\Lambda = 2^{2n}\left [T(1-T)\right ]^{n},\] so the statistic is a function of \(T\) alone — the same shape as the previous problem with \(n\) in place of \(1\).

Under \(H_0\) the sums \(S\) and \(U\) are independent GAM\((n,\lambda )\) variables with a common scale, so \[T = \frac {S}{S+U}\ \sim \ \text {BETA}(n,n),\] free of \(\lambda \). The test rejects for small \(\Lambda \), that is when \(T\) departs from \(\tfrac 12\) in either direction, with the critical values taken from the BETA\((n,n)\) distribution.

Solution. This is Wilks’ theorem. A full proof belongs to asymptotic theory, but the argument is short enough to give in outline, and it explains where the degrees of freedom come from.

Let \(\widehat {\theta }\) maximise the likelihood over \(\Omega \) and \(\widehat {\theta }_0\) over \(\Omega _0\), and expand \(\log L\) about \(\widehat {\theta }\) to second order. The first-order term vanishes because \(\widehat {\theta }\) is a stationary point, leaving \[2\log \Lambda _n = 2\left [\log L\left (\widehat {\theta }\right ) - \log L\left (\widehat {\theta }_0\right )\right ] \approx \left (\widehat {\theta }-\widehat {\theta }_0\right )^{t} J\left (\theta _0\right ) \left (\widehat {\theta }-\widehat {\theta }_0\right ),\] since \(-\partial ^{2}\log L/\partial \theta \partial \theta ^{t}\) converges to \(J(\theta _0)\).

Under \(H_0\) the large-sample theory of the previous section gives \[J\left (\theta _0\right )^{1/2}\left (\widehat {\theta }-\theta _0\right ) \underset {D}{\longrightarrow } Z \sim N_k(0,I),\] so the quadratic form above is asymptotically \(Z^{t}PZ\), where \(P\) is the orthogonal projection onto the orthogonal complement of the tangent space of \(\Omega _0\) at \(\theta _0\). That tangent space has dimension \(q\), so \(P\) has rank \(k-q\). A quadratic form \(Z^{t}PZ\) in a standard normal vector, with \(P\) an orthogonal projection of rank \(r\), has the \(\chi ^{2}_{(r)}\) distribution. Hence \[2\log \Lambda _n \underset {D}{\longrightarrow } W \sim \chi ^{2}_{(k-q)} .\]