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This is problem # 5 from RSS's 2014 Graduate Diploma Module 2:

$$\mathbb{P}(X_j=k) = \begin{cases} (1-p)^3 & k=0\\ 3p(1-p) & k=1\\ p^3 & k=2\\ 0 & \text{otherwise} \end{cases}$$

$$Y_k = \sum_{i=1}^n I(X_i=k)$ for $k=0,1,2$$

It is straightforward to get an MLE estimate of $p$: $$\hat{p} = \frac{Y_1+3Y_2}{3Y_0+2Y_1+3Y_2}$$

Required to test: $H_0: p=p_0 \quad \text{versus}\quad H_1: p\neq p_0$.

I am stuck in the following part:

Using an asymptotic result, find the critical region of this test if its size is to be approximately 0.05

Attempt

Likelihood ratio(LR) test: $P(LR < k) = \alpha = 0.05$

\begin{align*} LR &= \frac{L(p|\mathbf{X})|_{p=p_0}}{L(p|\mathbf{X})|_{p=\hat{p}}}\\ &= \frac{(1-p_0)^{3Y_0}(3p_0(1-p_0))^{Y_1}(p_0^3)^{Y_2}}{(1-\hat{p})^{3Y_0}(3\hat{p}(1-\hat{p}))^{Y_1}(\hat{p}^3)^{Y_2}}\\ &= (\frac{p_0}{\hat{p}})^{Y_1+3Y_2} * (\frac{1-p_0}{1-\hat{p}})^{Y_1+3Y_0} \end{align*}

EDIT 1:

Using Wilk's theorem: $-2\log(LR) \longrightarrow^D \chi^2_1(0.95)$ in distribution Thus critical region $C$:

$C = \{\mathbf{Y}: (Y_1+3Y_2) \log{\frac{p_0}{\hat{p}}} + (Y_1+3Y_0) \log(\frac{1-p_0}{1-\hat{p}}) \geq \frac{-\chi^2_1(0.95)}{2} \}$

Is this further reducable since the final subpart of the question asks:

Show that in the case $Y_0 = Y_2$ this confidence interval can be written as $p(1-p) \geq 0.25 \exp{(\frac{-1.92}{Y_1+3Y_2})}$

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    $\begingroup$ @Slowloris, it is best not to edit someone's Q to add the [self-study] tag, but to ask them to add it themselves & read our policies. That way there is a greater chance they will be familiar w/ our policy. If they don't add the tag in a reasonable period of time, we can close the thread. There is example text you can use here. $\endgroup$ – gung May 12 '16 at 17:31
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    $\begingroup$ Have you come across Wilk's Theorem? $\endgroup$ – Greenparker May 12 '16 at 17:32
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    $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung May 12 '16 at 17:33
  • $\begingroup$ @Greenparker Thanks! That takes me a step closer to finding the rejection region. See edit # 1 $\endgroup$ – user239438 May 12 '16 at 17:56
  • $\begingroup$ Its a little confusing what confidence interval you are talking about, since $C$ is the rejection interval. Also, the "confidence interval" you have has $p$ instead of $p_0$, so again, it is unclear what you mean. In any case, if you plug in $Y_0 = Y_2$, you will notice that $\hat{p}$ turns out to be nice, and solving you get that very expression in terms of $p_0(1-p_0) \geq...$ $\endgroup$ – Greenparker May 12 '16 at 18:37
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$\chi^2_1(0.95) = 3.841$

$C = \{\mathbf{Y}: (Y_1+3Y_2) \log{\frac{p_0}{\hat{p}}} + (Y_1+3Y_0) \log(\frac{1-p_0}{1-\hat{p}}) \geq \frac{-\chi^2_1(0.95)}{2} \}$

When $Y_0 =Y_2$, $\hat{p} = 1/2$

Thus, $\log(4p_0(1-p_0)) \geq \frac{-3.841}{2(Y_1+3Y_2)} $ $\implies p_0(1-p_0) \geq 0.25 \exp{(\frac{-1.92}{Y_1+3Y_2})}$

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