4.4 Likelihood ratio tests
The likelihood ratio test for testing the composite hypothesis \(H_0: \theta \in \Omega _0\) against \(H_1: \theta \in \Omega - \Omega _0\) has critical region \(R = \left \{ x: \, \Lambda (x) = > C\right \}\)
where
\[\Lambda (x) = \frac {\sup _{\theta \in \Omega } f_{\theta }(x)}{\sup _{\theta \in \Omega _0} f_{\theta }(x)} = \frac {\sup _{\theta \in \Omega } L(\theta )}{\sup _{\theta \in \Omega _0} L(\theta )}\]
and \(C\) is determined by the size of the test.
Note. that \(\Lambda (x) = \frac {L\left (\widehat {\theta }\right )}{L\left (\widehat {\theta }_R\right )}\) where \(\widehat {\theta }\) is the usual \(ML\) estimate and \(\widehat {\theta }_R\) is the \(ML\) estimate calculated under the restriction \(H_0: \theta \in \Omega _0\).
Example 4.4.1. Let \(X_1, \, \cdots \, , X_n\) be a random sample from then \(N(\mu , \sigma ^2)\) distribution where \(\mu \) and \(\sigma ^2\) are unknown.
Consider a test of \(H_0: \mu = 0, \, 0 < \sigma ^2 < \infty \,\) against \(H_1:\, \mu \neq 0, \, 0 < \sigma ^2 < \infty \).
Show that the likelihood ration test of \(H_0\) against \(H_1\) has critical region
\[ R = \left \{ x:\, \frac {n\overline {x}^2}{s^2}\, > C\right \}.\]
Show that under \(H_0\) the statistic \(\, \frac {n\, \overline {X}^2}{S^2}\,\) has an \(F(1, n- 1)\) distribution and thus find a size \(\alpha = 0.05\) test for \(n = 20\).
Solution. \(H_0: \mu = 0, \, 0 < \sigma ^2 < \infty \hspace {0.3cm}\) against \(\hspace {0.3cm} H_1: \mu \neq 0, \, 0 < \sigma ^2 < \infty \,, \, \, \theta = (\mu ,\sigma ^2)\) \(\Omega = \left \{(\mu , \sigma ^2):\, -\infty < \mu < \infty , \, \sigma ^2 > 0\right \}\hspace {0.3cm}\) and \(\hspace {0.3cm}\Omega _0 = \left \{(0,\sigma ^2):\, \sigma ^2>0\right \}\) \[\Lambda (x) = \frac {\left (\widehat {\theta }\right )}{\left (\widehat {\theta }_R\right )}\]
\[L(\theta ) = \prod ^n_{i = 1} \frac {1}{\sqrt {2\pi \sigma ^2}} e^{-\frac {1}{2\sigma ^2}(x_i - \theta )^2} = \left (2\pi \sigma ^2\right )^{-\frac {n}{2}}e^{-\frac {1}{2\sigma ^2}\sum ^n_{i = 1} (x_i - \mu )^2}.\] From example 2.2.4 \[\widehat {\mu } = \overline {x}\,, \hspace {0.3cm} \widehat {\sigma }^2 = \frac {1}{n}\sum ^n_{i = 1} \left (X_i - \overline {X}\right )^2\] replacing we obtain \begin {align*} L\left (\widehat {\theta }\right ) & = \left (2\pi \widehat {\sigma }^2\right )^{-\frac {n}{2}}\, e^{-\frac {1}{2\widehat {\sigma }^2}\sum ^n_{i = 1} \left (x_i - \widehat {\mu }\right )^2}\\ & = \left (2\pi \widehat {\sigma }^2\right )^{-\frac {n}{2}}\,e^{-\frac {1}{2\widehat {\sigma }^2} \sum ^n_{i = 1} (x_i - \overline {x})^2}\\ & = \left (2\pi \widehat {\sigma }^2\right )^{-\frac {n}{2}}\,e^{-\frac {1}{2\widehat {\sigma }^2}\left (n\widehat {\sigma }^2\right )}\\ & = \left (2\pi \widehat {\sigma }^2\right )^{-\frac {n}{2}}\, e^{-\frac {n}{2}} \end {align*}
Under \(H_0:\, \mu = 0\, , \hspace {0.3cm} \sigma ^2 > 0\) \[L(\theta ) = \left (2\pi \sigma ^2\right )^{-\frac {n}{2}}\, e^{-\frac {1}{2\sigma ^2}\sum ^n_{i = 1} x_i^2}.\]
\[\mathcal {L}(\theta ) = -\frac {1}{2\sigma ^2} \sum ^n_{i =1} x_i^2 - \frac {n}{2}\log \left (2\pi \sigma ^2\right ).\]
\[S(\theta ) = \frac {\partial \mathcal {L}}{\partial \sigma ^2} = \frac {1}{2(\sigma ^2)^2}\sum ^n_{i = 1} x_i^2 - \frac {n}{2}\cdot \frac {2\pi }{2\pi \sigma ^2}= \frac {1}{2(\sigma ^2)^2}\sum ^n_{i = 1} x_i^2 - \frac {n}{2\sigma ^2}.\] \(S(\theta ) = 0\) \[\frac {n}{2\sigma ^2} = \frac {1}{2(\sigma ^2)^2}\sum ^n_{ i = 1}x_i^2 \, \implies \, \widehat {\sigma }_R^2 = \frac {1}{n}\sum ^n_{i = 1} x_i^2.\]
\[L\left (\widehat {\theta }_R\right ) = \left (2\pi \widehat {\sigma }^2_R\right )^{-\frac {n}{2}}\, e^{-\frac {1}{2\widehat {\sigma }^2_R}\sum ^n_{i = 1} x_i^2} = \left (2\pi \widehat {\sigma }^2_R\right )^{-\frac {n}{2}}\, e^{-\frac {1}{2\widehat {\sigma }^2_R}\cdot n\widehat {\sigma }^2_R} = \left (2\pi \widehat {\sigma }^2_R\right )^{-\frac {n}{2}}\, e^{-\frac {n}{2}} \]
\begin {align*} \frac {L\left (\widehat {\theta }\right )}{L\left (\widehat {\theta }_R\right )} & = \frac {\left (2\pi \widehat {\sigma }^2\right )^{-\frac {n}{2}}\, e^{-\frac {n}{2}}}{\left (2\pi \widehat {\sigma }^2_R\right )^{-\frac {n}{2}}\, e^{-\frac {n}{2}}} = \frac {\left (\widehat {\sigma }_R^2\right )^{\frac {n}{2}}}{\left (\widehat {\sigma }^2\right )^{\frac {n}{2}}} = \left (\frac {\widehat {\sigma }^2_R}{\widehat {\sigma }^2}\right )^{\frac {n}{2}}\\\\ & = \left (\frac {\frac {1}{n}\sum ^n_{i = 1} x_i^2}{\frac {1}{n}\sum ^n_{i = 1} (x_i - \overline {x})^2}\right )^{\frac {n}{2}}\\\\ & = \left (\frac {\sum ^n_{i = 1} \left ((x_i - \overline {x})^2 + \overline {x}(x_i - \overline {x}) + \overline {x}^2\right )}{\sum ^n_{i = 1} (x_i - \overline {x})^2}\right )^{\frac {n}{2}} \end {align*}
\begin {align*} & = \left (\frac {\sum ^n_{i = 1} (x_i - \overline {x})^2 + n\overline {x}^2}{\sum ^n_{i = 1} (x_i - \overline {x})^2}\right )^{\frac {n}{2}}\\\\ & =\left (\frac {(n-1)S^2 + n\overline {x}^2}{(n-1) S^2}\right )^{\frac {n}{2}} = \left (1 + \frac {n\overline {x}^2}{(n -1)S^2}\right )^{\frac {n}{2}}\\\\ & = \left (1 + \frac {1}{n - 1}\cdot \frac {n\overline {x}^2}{S^2}\right )^{\frac {n}{2}} \end {align*}
\(\left (1 + \frac {1}{n - 1}\cdot \frac {n\overline {x}^2}{S^2}\right )^{\frac {n}{2}} > K\,\) is equivalent to \(\frac {n\overline {x}^2}{S^2} > C\) where \(C\) is determined by the size of the test \[\therefore \,\, R = \left \{ x:\, \frac {n\overline {x}^2}{S^2} > C\right \}.\]
Under \(H_0:\, \mu = 0\) \[T = \frac {n\, \overline {X}^2}{S^2} = \frac {\left (\dfrac {\overline {X} - 0}{\sigma /\sqrt {n}}\right )^2}{\dfrac {(n - 1) S^2/\sigma ^2}{n - 1}}\]
\[\left (\frac {\overline {X} - 0}{\frac {\sigma }{\sqrt {n}}}\right )^2 = Z^2 \, \thicksim \, \chi ^2_{(1)}\hspace {0.5cm}\text {and}\hspace {0.5cm}\frac {(n - 1)S^2}{\sigma ^2} \thicksim \chi ^2_{n - 1}\] \[T = \frac {\dfrac {\chi ^2_{(1)}}{1}}{\dfrac {\chi ^2_{n - 1}}{n - 1}}\, \thicksim \, F(1, n-1).\]
\(\alpha = P_{H_0}(X\in R)\) \[0.05 = P(T>C) = P(F_{1,19} > C)\] Therefore \(\hspace {0.2cm} C = 4.35\). □
Problem 4.4.1. Suppose \(X\thicksim GAM(2,\beta _1)\) and \(Y\thicksim GAM(2,\beta _2)\) independently. Show that the likelihood ratio statistic for
testing the hypothesis \(H_0: \beta _1 = \beta _2\,\) against the alternative \(\, H_1:\beta _1\neq \beta _2\) is a function of the statistic
\[T = \frac {X}{X + Y}.\]
Find the distribution of \(T\) under \(H_0\). Find the critical region for a size \(\alpha = 0.01\) test.
Hint: If \(X\thicksim GAM(a,1)\) and \(Y\thicksim GAM(b,1)\) are independent random variable then \[\frac {X}{X + Y} \thicksim BETA(a,b).\]
Show solution
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.
Note. \(\Lambda \) is largest when \(t=\tfrac 12\) and falls away symmetrically, so rejecting for small \(\Lambda \) means rejecting when \(T\) is far from \(\tfrac 12\) in either direction — exactly what a two-sided test of \(\beta _1=\beta _2\) should do.
Problem 4.4.2. Suppose \(X_1, \, \cdots \, , X_n\) are independent \(EXP(\lambda )\) random variables and independently \(Y_1, \, \cdots \, , \, Y_n\) are
independent \(EXP(\mu )\) random variables.
Show that the likelihood ratio statistic for testing the hypothesis \(H_0: \lambda = \mu \) against the alternative \(H_1: \lambda \neq \mu \) is
a function of
\[T = \frac {\sum ^n_{i = 1} X_i}{\sum ^n_{i =1} X_i + \sum ^n_{i = 1} Y_i}.\]
Show solution
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.
Problem 4.4.3. Suppose \(X_1, \, \cdots \, , X_n\) is a random sample from a regular statistical model \(\{f_{\theta }(x); \theta \in \Omega \}\) with \(\Omega \) being an open set in a \(k\)-dimensional Euclidean space. Consider a subset of \(\Omega \) defined by \[\Omega _0 = \{\theta (\eta ):\, \eta \in \, \text { open subset of a}\, q-\text {dimensional Euclidean space}\}.\] Then the likelihood ratio statistic defined by \[\Lambda _n(x) = \frac {\sup _{\theta \in \Omega } \prod ^n_{i = 1} f_{\theta }(x_i)}{\sup _{\theta \in \Omega _0} \prod ^n_{i =1} f_{\theta }(x_i)} = \frac {\sup _{\theta \in \Omega }L(\theta )}{\sup _{\theta \in \Omega _0} L(\theta )}\] is such that, under \(H_0: \theta \in \Omega _0\), \[2\, \log \Lambda _n(x) \, \underset {D}{\longrightarrow } \, W\thicksim \chi ^2_{(k - q)}.\]
Show solution
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 \[2\log \Lambda _n \underset {D}{\longrightarrow } W \sim \chi ^{2}_{(k-q)} .\]
Note. The degrees of freedom are the number of restrictions — the dimension the null hypothesis removes — and not the dimension of either space on its own. That is the practical content of the theorem: to use it, count constraints. The regularity conditions matter too: \(\Omega _0\) must be a smooth surface in the interior of \(\Omega \), which is why the result fails when the null puts a parameter on the boundary, as in testing a variance component equal to zero.
Note. The number of degrees of freedom is the difference between the number of parameters that need to be estimated in the general model and the number of parameters to be estimated under the restrictions imposed by \(H_0\).
Example 4.4.2. Suppose \(X_1, \, \cdots \, , X_n\) are independent \(POI(\lambda )\) random variables and independently \(Y_1, \, \cdots \, , \, Y_n\) are independent \(POI(\mu )\) random variables. Find the likelihood ratio test of \(H_0:\lambda = \mu \) against \(H_1: \lambda \neq \mu \).
Solution. \(H_0: \lambda = \mu \hspace {0.3cm}\) against \(\hspace {0.3cm} H_1: \lambda \neq \mu \). \[\Omega = \{(\lambda , \mu ):\, \lambda , \, \mu > 0\} \hspace {0.5cm}\text {and}\hspace {0.5cm} \Omega _0 = \{(\lambda ,\lambda ):\, \lambda > 0\}.\]
\[L(\theta ) = \prod ^n_{i = 1} f_{\lambda }(x_i)\, \prod ^n_{i = 1} f_{\mu }(y_i)= \prod ^n_{i = 1} \frac {e^{-\lambda }\,\lambda ^{x_i}}{x_i!}\,\prod ^n_{i = 1} \frac {e^{-\mu }\,\mu ^{y_i}}{y_i!} = \frac {e^{-n\lambda }\, \lambda ^{\sum ^n_{i=1} x_i}}{\prod ^n_{i = 1} x_i!}\,\frac {e^{-n\mu }\, \mu ^{\sum ^n_{i=1} y_i}}{\prod ^n_{i = 1} y_i!}.\]
\[\mathcal {L}(\theta ) = -n\lambda + \sum ^n_{i = 1} x_i \, \log \lambda - \log \prod ^n_{i = 1} x_i! - n\mu + \sum ^n_{i = 1} y_i \log \mu - \log \prod ^n_{i = 1} y_i!.\]
\[S(\theta ) = \begin {pmatrix} -n + \frac {\sum ^n_{i =1} x_i}{\lambda }\\\\ -n + \frac {\sum ^n_{i = 1}y_i}{\mu }\\ \end {pmatrix}\] \(S(\theta ) = 0\) \[\implies \hspace {0.3cm} n = \frac {\sum ^n_{i =1} x_i}{\lambda }\hspace {0.5cm} , \hspace {0.5cm} n = \frac {\sum ^n_{i = 1} y_i}{\mu }\] \[\widehat {\lambda } = \frac {1}{n}\sum ^n_{i = 1} x_i = \overline {x}\hspace {0.5cm} , \hspace {0.5cm} \widehat {\mu } = \frac {1}{n}\sum ^n_{i = 1} y_i = \overline {y}.\]
\begin {align*} L\left (\widehat {\theta }\right ) & = \frac {e^{-n \widehat {\lambda }}\, \widehat {\lambda }^{\sum ^n_{i = 1}x_i}}{\prod ^n_{i = 1} x_i!}\,\cdot \, \frac {e^{-n \widehat {\mu }}\, \widehat {\mu }^{\sum ^n_{i = 1}y_i}}{\prod ^n_{i = 1} y_i!} =\frac {e^{-n \overline {x}}\, \overline {x}^{\sum ^n_{i = 1}x_i}}{\prod ^n_{i = 1} x_i!}\,\cdot \, \frac {e^{-n \overline {y}}\, \overline {y}^{\sum ^n_{i = 1}y_i}}{\prod ^n_{i = 1} y_i!}\\\\ & = \frac {e^{-n(\overline {x} + \overline {y})}\, \overline {x}^{n\overline {x}}\, \overline {y}^{n\overline {y}}}{\prod ^n_{i = 1} x_i!\ \prod ^n_{i = 1} y_i!}\\\\ & = \frac {e^{-n(\overline {x} + \overline {y})}\, \left (\overline {x}^{\overline {x}}\right )^n\, \left (\overline {y}^{\overline {y}}\right )^n}{\prod ^n_{i = 1} x_i!\ \prod ^n_{i = 1} y_i!}. \end {align*}
Under \(H_0\) \[L(\theta ) = \frac {e^{-n\lambda }\, \lambda ^{\sum ^n_{i = 1} x_i}}{\prod ^n_{i = 1}x_i!}\, \cdot \, \frac {e^{-n\lambda }\, \lambda ^{\sum ^n_{i = 1} y_i}}{\prod ^n_{i = 1}y_i!} = \frac {e^{-2n\lambda }\, \lambda ^{\sum ^n_{i = 1} x_i + \sum ^n_{i = 1} y_i}}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1} y_i!}.\]
\[\mathcal {L}(\theta ) = -2n\lambda + \left (\sum ^n_{i = 1} x_i + \sum ^n_{i = 1} y_i \right ) \log \lambda - \log \prod ^n_{i = 1} x_i! - \log \prod ^n_{i = 1} y_i!.\]
\[S(\theta ) = \frac {\partial \mathcal {L}}{\partial \lambda } = -2n + \frac {\sum ^n_{i = 1} x_i + \sum ^n_{i = 1} y_i}{\lambda }.\] \(S(\theta ) = 0\) \[2n\lambda = \sum ^n_{i = 1} + \sum ^n_{i = 1} y_i\]
\[\widehat {\lambda }_R = \frac {\sum ^n_{i = 1} x_i + \sum ^n_{i = 1} y_i}{2n} = \frac {\overline {x} + \overline {y}}{2}.\] \begin {align*} L\left (\widehat {\theta }_R\right ) & = \frac {e^{-2n\widehat {\lambda }_R}\, \widehat {\lambda }_R^{\sum ^n_{i = 1} x_i + \sum ^n_{i = 1} y_i}}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1}y_i!}\\\\ & = \frac {e^{-2n\left (\frac {\overline {x} + \overline {y}}{2}\right )}\, \, \, \left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{n\overline {x} + n\overline {y}}}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1}y_i!} = \frac {e^{-n\left (\overline {x} + \overline {y}\right )}\, \, \, \left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{n(\overline {x} + \overline {y})}}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1}y_i!} \end {align*}
\begin {align*} \Lambda (x,y) & = \frac {L\left (\widehat {\theta }\right )}{L\left (\widehat {\theta }_R\right )} = \frac {\dfrac {e^{-n(\overline {x} + \overline {y})}\, \left (\overline {x}^{\overline {x}}\right )^n \, \left (\overline {y}^{\overline {y}}\right )^n}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1} y_i!}}{\dfrac {e^{-n\left (\overline {x} + \overline {y}\right )}\, \, \, \left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{n(\overline {x} + \overline {y})}}{\prod ^n_{i = 1} x_i!\, \prod ^n_{i = 1}y_i!}} =\frac {\left (\overline {x}^{\overline {x}}\right )^n\, \, \left (\overline {y}^{\overline {y}}\right )^n}{\left [\left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{\overline {x} + \overline {y}}\right ]^n} = \left (\frac {\overline {x}^{\overline {x}}\, \, \overline {y}^{\overline {y}}}{\left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{\overline {x} + \overline {y}}}\right )^n. \end {align*}
\[\therefore \,\,\, R = \left \{(x,y):\, \frac {\overline {x}^{\overline {x}}\, \overline {y}^{\overline {y}}}{\left (\dfrac {\overline {x} + \overline {y}}{2}\right )^{\overline {x} + \overline {y}}}\, > C\right \}\]
where \(C\) is determined by the size of the test.
Under \(H_0\), \[2\, \log \Lambda (X, Y) \, \underset {D}{\longrightarrow }\, W\, \thicksim \, \chi ^2_{(2-1)} = \chi ^2_{(1)}.\] □
Questions on this section
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