A family of pdfs or pmfs $\{g(t|\theta):\theta\in\Theta\}$ for a univariate random variable $T$ with real-valued parameter $\theta$ has a monotone likelihood ratio (MLR) if, for every $\theta_2>\theta_1$, $g(t|\theta_2)/g(t|\theta_1)$ is a monotone (nonincreasing or nondecreasing) function of $t$ on $\{t:g(t|\theta_1)>0\;\text{or}\;g(t|\theta_2)>0\}$. Note that $c/0$ is defined as $\infty$ if $0< c$.
Consider testing $H_0:\theta\leq \theta_0$ versus $H_1:\theta>\theta_0$. Suppose that $T$ is a sufficient statistic for $\theta$ and the family of pdfs or pmfs $\{g(t|\theta):\theta\in\Theta\}$ of $T$ has an MLR. Then for any $t_0$, the test that rejects $H_0$ if and only if $T >t_0$ is a UMP level $\alpha$ test, where $\alpha=P_{\theta_0}(T >t_0)$.
Example 1
To better understand the theorem, consider a single observation, $X$, from $\mathrm{n}(\theta,1)$, and test the following hypotheses:
$$
H_0:\theta\leq \theta_0\quad\mathrm{versus}\quad H_1:\theta>\theta_0.
$$
Then $\theta_1>\theta_0$, and the likelihood ratio test statistics would be
$$
\lambda(x)=\frac{f(x|\theta_1)}{f(x|\theta_0)}.
$$
And we say that the null hypothesis is rejected if $\lambda(x)>k$. To see if the distribution of the sample has MLR property, we simplify the above equation as follows:
$$
\begin{aligned}
\lambda(x)&=\frac{\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{(x-\theta_1)^2}{2}\right]}{\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{(x-\theta_0)^2}{2}\right]}\\
&=\exp
\left[-\frac{x^2-2x\theta_1+\theta_1^2}{2}+\frac{x^2-2x\theta_0+\theta_0^2}{2}\right]\\
&=\exp\left[\frac{2x\theta_1-\theta_1^2-2x\theta_0+\theta_0^2}{2}\right]\\
&=\exp\left[\frac{2x(\theta_1-\theta_0)-(\theta_1^2-\theta_0^2)}{2}\right]\\
&=\exp\left[x(\theta_1-\theta_0)\right]\times\exp\left[-\frac{\theta_1^2-\theta_0^2}{2}\right]
\end{aligned}
$$
which is increasing as a function of $x$, since $\theta_1>\theta_0$.
|
Figure 1. Normal Densities with $\mu=1,2$. |
By illustration, consider Figure 1. The plot of the likelihood ratio of these models is monotone increasing as seen in Figure 2, where rejecting $H_0$ if $\lambda(x)>k$ is equivalent to rejecting it if $T\geq t_0$.
|
Figure 2. Likelihood Ratio of the Normal Densities. |
And by factorization theorem the likelihood ratio test statistic can be written as a function of the sufficient statistics since the term, $h(x)$ will be cancelled out. That is,
$$
\lambda(t)=\frac{g(t|\theta_1)}{g(t|\theta_0)}.
$$
And by Karlin-Rubin theorem, the rejection region $R=\{t:t>t_0\}$ is a uniformly most powerful level-$\alpha$ test. Where $t_0$ satisfies the following:
$$
\begin{aligned}
\mathrm{P}(T>t_0|\theta_0)&=\mathrm{P}(T\in R|\theta_0)\\
\alpha&=1-\mathrm{P}(X\leq t_0|\theta_0)\\
1-\alpha&=\int_{-\infty}^{t_0}\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{(x-\theta_0)^2}{2}\right]\operatorname{d}x
\end{aligned}
$$
Hence the quantile of the $1-\alpha$ probability, which is $z_{\alpha}$ is equal to $t_0$, that is $z_{\alpha}=t_0$, and thus we reject $H_0$ if $T>z_{\alpha}$.
Example 2
Now consider testing the hypotheses, $H_0:\theta\geq \theta_0$ versus $H_1:\theta< \theta_0$ using the sample $X$ (single observation) from Beta($\theta$, 2), and to be more specific let $\theta_0=4$ and $\theta_1=3$. Can we apply Karlin-Rubin?
Of course! Visually, we have something like in Figure 3.
|
Figure 3. Beta Densities Under Different Parameters. |
Note that for this test, $\theta_1<\theta_0$, and so the likelihood ratio test statistics is simplified as follows:
$$
\begin{aligned}
\lambda(x)&=\frac{f(x|\theta_1=3, 2)}{f(x|\theta_0=4, 2)}=\frac{\displaystyle\frac{\Gamma(\theta_1+2)}{\Gamma(\theta_1)\Gamma(2)}x^{\theta_1-1}(1-x)^{2-1}}{\displaystyle\frac{\Gamma(\theta_0+2)}{\Gamma(\theta_0)\Gamma(2)}x^{\theta_0-1}(1-x)^{2-1}}\\
&=\frac{\displaystyle\frac{\Gamma(5)}{\Gamma(3)\Gamma(2)}x^{2}(1-x)}{\displaystyle\frac{\Gamma(6)}{\Gamma(4)\Gamma(2)}x^{3}(1-x)}=\frac{\displaystyle\frac{12\Gamma(3)}{\Gamma(3)\Gamma(2)}x^{2}(1-x)}{\displaystyle\frac{20\Gamma(4)}{\Gamma(4)\Gamma(2)}x^{3}(1-x)}\\
&=\frac{3}{5x},
\end{aligned}
$$
which is decreasing as a function of $x$, see the plot of this in Figure 4. And we say that $H_0$ is rejected if $\lambda(x) > k$ if and only if $T < t_0$. Where $t_0$ satisfies the following equations:
$$
\begin{aligned}
\mathrm{P}(T < t_0|\theta_0)&=\mathrm{P}(X < t_0|\theta_0)\\
\alpha&=\int_{0}^{t_0}\frac{\Gamma(\theta_0+2)}{\Gamma(\theta_0)\Gamma(2)}x^{\theta_0-1}(1-x)^{2-1}\operatorname{d}x\\
\alpha&=\int_{0}^{t_0}\frac{\Gamma(6)}{\Gamma(4)\Gamma(2)}x^{3}(1-x)\operatorname{d}x.
\end{aligned}
$$
Hence the quantile of the $\alpha$ probability, $x_{\alpha}=t_0$. And thus we reject $H_0$ if $T < x_{\alpha}$.
|
Figure 4. Likelihood Ratio of the Beta Densities. |
Reference
-
Casella, G. and Berger, R.L. (2001). Statistical Inference. Thomson Learning, Inc.
As reported by Stanford Medical, It's indeed the SINGLE reason women in this country get to live 10 years longer and weigh an average of 42 pounds less than we do.
ReplyDelete(By the way, it really has NOTHING to do with genetics or some secret diet and really, EVERYTHING around "HOW" they are eating.)
BTW, I said "HOW", and not "WHAT"...
TAP on this link to discover if this easy test can help you release your real weight loss potential