41  Teste de Hipótese de Neyman-Pearson (Lema de Neyman-Pearson)

Estamos interessados em um teste \(\delta^*\) que produza menor \(\alpha_{\delta^*, \max}\) e maior poder possível. O Lema de Neyman-Pearson apresenta um teste que, dentre os testes de tamanho \(\alpha\), tem maior poder.

Seja \(\boldsymbol{X}_n\) a.a. de \(X \sim f_\theta, \theta \in \Theta = \{\theta_0, \theta_1\}\). Considere as hipóteses \[ \begin{cases} \mathcal{H}_0 : \theta = \theta_0 \\ \mathcal{H}_1 : \theta = \theta_1 \end{cases} \] nula e alternativa. A função teste \(\delta^* : \mathfrak{X}^{(n)} \rightarrow \{0,1\}\) tal que \[ \delta^*(\boldsymbol{x}_n) = \begin{cases} 0,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}(\theta_1) < \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}(\theta_0) \\ 1,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}(\theta_1) \geq \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}(\theta_0), \end{cases} \] em que \(\eta\) satisfaz \(\underbracket{P_{\theta_0}(\delta^*(\boldsymbol{X}_n) = 1)}_{\pi_{\delta^*}(\theta_0)} = \alpha\) para um \(\alpha\) fixado apriori, é o teste mais poderoso dentre todo os testes de tamanho \(\alpha\).

41.1 Exemplo (Normal)

Seja \(\boldsymbol{X}_n\) a.a. de \(X \sim N(\mu, \sigma^2)\) em que \(\theta = (\mu, \sigma^2) \in \{(0,1), (0,2)\}\). Considere \[ \begin{cases} \mathcal{H}_0 : \theta = (0,1) \\ \mathcal{H}_1 : \theta = (0,2) \end{cases} \] as hipóteses nula e alternativa.

Encontre o teste mais poderoso de tamanho \(\alpha = 5\%\).

41.1.0.1 Solução:

De acordo com o Lema de Neyman-Pearson, o teste mais poderoso de tamanho \(\alpha = 5\%\) é \[ \delta^*(\boldsymbol{x}_n) = \begin{cases} 0,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}((0,2)) < \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}((0,1)), \\ 1,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}((0,2)) \geq \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}((0,1)), \end{cases} \] em que \(\eta\) satisfaz \(\pi_{\delta^*}((0,1)) = 5\%\).

A função de verossimilhança é \[ \begin{aligned} \mathcal{L}_{\boldsymbol{x}_n}(\theta) &= \frac{1}{(\sqrt{2\pi\sigma^2})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum \frac{(x_i - \mu)^2}{\sigma^2}\right\}, \theta \in \{(0,1), (0,2)\} \\ &= \frac{1}{(\sqrt{2\pi\sigma^2})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum \frac{x_i^2}{\sigma^2} \right\}, \theta \in \{(0,1), (0,2)\} \\ \Rightarrow \mathcal{L}_{\boldsymbol{x}_n}(\theta_0) &= \frac{1}{(\sqrt{2\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum x_i^2 \right\} \\ \mathcal{L}_{\boldsymbol{x}_n}(\theta_1) &= \frac{1}{(\sqrt{4\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum \frac{x_i^2}{2} \right\} \\ \Rightarrow \mathcal{L}_{\boldsymbol{x}_n}((0,2)) &\geq \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}((0,1)) \\ \iff \frac{\mathcal{L}_{\boldsymbol{x}_n}((0,2))}{\mathcal{L}_{\boldsymbol{x}_n}((0,1))} &\geq \eta \\ \Rightarrow &\frac{\frac{1}{(\sqrt{4\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum \frac{x_i^2}{2} \right\}}{\frac{1}{(\sqrt{2\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum x_i^2 \right\}} \geq \eta \\ \iff &\frac{1}{2^{\frac{n}{2}}} \mathrm{e}^{\frac{1}{4}\sum x_i^2} \geq \eta \\ \iff &\mathrm{e}^{\frac{1}{4}\sum x_i^2} \geq 2^{n/2}\eta \\ \iff &\frac{1}{4}\sum x_i^2 \geq \ln(2^{n/2}\eta) \\ \iff &\sum x_i^2 \geq 4 \ln(2^{n/2}\eta). \end{aligned} \]

Logo, \(\sum x_i^2 \geq 4 \ln(2^{n/2}\eta)\) é equivalente a \(\frac{\frac{1}{(\sqrt{4\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum \frac{x_i^2}{2} \right\}}{\frac{1}{(\sqrt{2\pi})^n} \cdot \mathrm{Exp}\left\{ -\frac{1}{2} \sum x_i^2 \right\}}\).

Observe ainda que \[ \begin{aligned} \pi_{\delta^*}((0,1)) &= P_{(0,1)}(\delta^*(\boldsymbol{X}_n) = 1) \\ &\stackrel{\text{Def.}}{=} P_{(0,1)}\left(\frac{\mathcal{L}_{\boldsymbol{x}_n}((0,2))}{\mathcal{L}_{\boldsymbol{x}_n}((0,1))} \geq \eta\right) \\ &\stackrel{\text{Desenv.}}{=} P_{(0,1)}\left(\sum X_i^2 \geq \eta^*\right), \end{aligned} \] em que \(\eta^* = 4\ln(2^{n/2} \eta)\). Como, sob \(\mathcal{H}_0, \sum X_i^2 \sim \chi^2_n\), temos que \[ \pi_{\delta^*}((0,1)) = P_{(0,1)}(\chi^2_{n} \geq \eta^*) = 5\%. \]

Com \(n=2\), temos que \[ \pi_{\delta^*}((0,1)) = P_{(0,1)}(\chi^2_{n} \geq \eta^*) = 5\%. \Rightarrow \eta^* = 5.991. \]

Portanto, o teste mais poderoso é \[ \delta^*(\boldsymbol{x}_n) = \begin{cases} 0,&\ \text{se}\ \sum x_i^2 < 5.991 \\ 1,&\ \text{se}\ \sum x_i^2 \geq 5.991. \end{cases} \]

41.2 Exemplo (Bernoulli)

Seja \(\boldsymbol{X}_n\) a.a. de \(X \sim \mathrm{Ber}(\theta), \theta \in \{0.1, 0.9\}\). Considere \[ \begin{cases} \mathcal{H}_0 : \theta = 0.9 \\ \mathcal{H}_1 : \theta = 0.1 \end{cases} \] as hipóteses nula e alternativa e \(n=10\). Encontre o Teste Mais Poderoso (TMP) de tamanho \(\alpha = 10\%\).

41.2.1 Resposta

De acordo com o Lema de Neyman-Pearson, o teste mais poderoso de tamanho \(\alpha = 5\%\) é \[ \delta^*(\boldsymbol{x}_n) = \begin{cases} 0,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}(0.1) < \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}(0.9), \\ 1,&\ \text{se}\ \mathcal{L}_{\boldsymbol{x}_n}(0.1) \geq \eta \cdot \mathcal{L}_{\boldsymbol{x}_n}(0.9), \end{cases} \] em que \(\eta\) satisfaz \(\pi_{\delta^*}(0.9) = 10\%\).

Note que \[ \begin{aligned} \mathcal{L}_{\boldsymbol{x}_n}(\theta) &= \theta^{\sum x_i} (1-\theta)^{n - \sum x_i} \\ \Rightarrow \frac{\mathcal{L}_{\boldsymbol{x}_n}(0.1)}{\mathcal{L}_{\boldsymbol{x}_n}(0.9)} &= \frac{0.1^{\sum x_i} (0.9)^{n - \sum x_i}}{0.9^{\sum x_i} (0.1)^{n - \sum x_i}} \\ &= \frac{0.1^{\sum x_i} \cdot 0.9^{n} \cdot 0.1^{\sum x_i}}{0.9^{\sum x_i} \cdot 0.9^{\sum x_i} \cdot 0.1^n} \\ &= \left(\frac{0.1 \cdot 0.1}{0.9 \cdot 0.9}\right)^{\sum x_i} \cdot \left(\frac{0.9}{0.1}\right)^{n} \geq \eta \\ \iff &(\sum x_i) \ln \left(\frac{0.01}{0.81}\right) + n \ln (9) \geq \ln \eta \\ \iff &(\sum x_i) \ln \left(\frac{1}{81}\right) \geq \ln \eta - n \ln (9) \\ \iff &(-\sum x_i) \ln (81) \geq \ln \eta - n \ln (9) \\ \iff &-\sum x_i \geq \frac{\ln \eta - n \ln (9)}{\ln (81)} \\ \iff &\sum x_i \leq \frac{-\ln \eta + n \ln (9)}{\ln (81)} \end{aligned} \]

Logo, \(\left(\frac{0.1 \cdot 0.1}{0.9 \cdot 0.9}\right)^{\sum x_i} \cdot \left(\frac{0.9}{0.1}\right)^{n} \geq \eta \iff \sum x_i \leq \frac{-\ln \eta + n \ln (9)}{\ln (81)}\). \[ \pi_{\delta^*}(0.1) = P_{0.1}(\delta^*(\boldsymbol{X}_n) = 1) = P_{0.1}(\sum X_i \leq \eta^*) = \alpha = 10\% \]

Note que, sob \(\mathcal{H}_0, \sum^10_{i=1} X_i \sim \mathrm{Bin}(10, 0.9)\). Ademais, \[ P_{0.9}(\sum X \leq 7) = 0.0702,\ \ \ P_{0.9}(\sum X \leq 8) = 0.2639. \] Logo, não é possível encontrar um valor para \(\eta\) exato com esse tamanho do teste.