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STAT 801: Mathematical Statistics

Hypothesis Testing

Hypothesis testing: a statistical problem where you must choose, on the basis of data $X$, between two alternatives. We formalize this as the problem of choosing between two hypotheses: $H_o: \theta\in \Theta_0$ or $H_1: \theta\in\Theta_1$ where $\Theta_0$ and $\Theta_1$ are a partition of the model $P_\theta; \theta\in \Theta$. That is $\Theta_0 \cup \Theta_1 = \Theta$ and $\Theta_0 \cap\Theta_1=\emptyset$.

A rule for making the required choice can be described in two ways:

1. In terms of the set
   $$R=\{X:$$   we choose $\Theta_1$ if we observe $X$$$\}$$
   called the rejection or critical region of the test.

2. In terms of a function $\phi(x)$ which is equal to 1 for those $x$ for which we choose $\Theta_1$ and 0 for those $x$ for which we choose $\Theta_0$.

For technical reasons which will come up soon I prefer to use the second description. However, each $\phi$ corresponds to a unique rejection region $R_\phi=\{x:\phi(x)=1\}$.

Neyman Pearson approach treats two hypotheses asymmetrically. Hypothesis $H_o$ referred to as the null hypothesis (traditionally the hypothesis that some treatment has no effect).

Definition: The power function of a test $\phi$ (or the corresponding critical region $R_\phi$) is

$$\pi(\theta) = P_\theta(X\in R_\phi) = E_\theta(\phi(X))$$

Interested in optimality theory, that is, the problem of finding the best $\phi$. A good $\phi$ will evidently have $\pi(\theta)$ small for $\theta\in\Theta_0$ and large for $\theta\in\Theta_1$. There is generally a trade off which can be made in many ways, however.

Simple versus Simple testing

Finding a best test is easiest when the hypotheses are very precise.

Definition: A hypothesis $H_i$ is simple if $\Theta_i$ contains only a single value $\theta_i$.

The simple versus simple testing problem arises when we test $\theta=\theta_0$ against $\theta=\theta_1$ so that $\Theta$ has only two points in it. This problem is of importance as a technical tool, not because it is a realistic situation.

Suppose that the model specifies that if $\theta=\theta_0$ then the density of $X$ is $f_0(x)$ and if $\theta=\theta_1$ then the density of $X$ is $f_1(x)$. How should we choose $\phi$? To answer the question we begin by studying the problem of minimizing the total error probability.

Type I error: the error made when $\theta=\theta_0$ but we choose $H_1$, that is, $X\in R_\phi$.

Type II error: when $\theta=\theta_1$ but we choose $H_0$.

The level of a simple versus simple test is

$$\alpha = P_{\theta_0}($$We make a Type I error$$)$$

or

$$\alpha = P_{\theta_0}(X\in R_\phi) = E_{\theta_0}(\phi(X))$$

Other error probability denoted $\beta$ is

$$\beta= P_{\theta_1}(X\not\in R_\phi) = E_{\theta_1}(1-\phi(X)).$$

Minimize $\alpha+\beta$, the total error probability given by

$$\alpha + \beta = E_{\theta_0}(\phi(X))+E_{\theta_1}(1-\phi(X))$$

$$= \int[ \phi(x) f_0(x) +(1-\phi(x))f_1(x)] dx$$

Problem: choose, for each $x$, either the value 0 or the value 1, in such a way as to minimize the integral. But for each $x$ the quantity

$$\phi(x) f_0(x) +(1-\phi(x))f_1(x)$$

is between $f_0(x)$ and $f_1(x)$. To make it small we take $\phi(x) = 1$ if $f_1(x)> f_0(x)$ and $\phi(x) = 0$ if $f_1(x) < f_0(x)$. It makes no difference what we do for those $x$ for which $f_1(x)=f_0(x)$. Notice: divide both sides of inequalities to get condition in terms of likelihood ratio $f_1(x)/f_0(x)$.

Theorem: For each fixed $\lambda$ the quantity $\beta+\lambda\alpha$ is minimized by any $\phi$ which has

$$\phi(x) =\left\{\begin{array}{ll} 1 & \frac{f_1(x)}{f_0(x)} > \lambda \\ 0 & \frac{f_1(x)}{f_0(x)} < \lambda \end{array}\right.$$

Neyman and Pearson suggested that in practice the two kinds of errors might well have unequal consequences. They suggested that rather than minimize any quantity of the form above you pick the more serious kind of error, label it Type I and require your rule to hold the probability $\alpha$ of a Type I error to be no more than some prespecified level $\alpha_0$. (This value $\alpha_0$ is typically $0.05$ these days, chiefly for historical reasons.)

The Neyman and Pearson approach is then to minimize $\beta$ subject to the constraint $\alpha \le \alpha_0$. Usually this is really equivalent to the constraint $\alpha=\alpha_0$ (because if you use $\alpha<\alpha_0$ you could make $R$ larger and keep $\alpha \le \alpha_0$ but make $\beta$ smaller. For discrete models, however, this may not be possible. Example: Suppose $X$ is Binomial$(n,p)$ and either $p=p_0=1/2$ or $p=p_1=3/4$.

If $R$ is any critical region (so $R$ is a subset of $\{0,1,\ldots,n\}$) then

$$P_{1/2}(X\in R) = \frac{k}{2^n}$$

for some integer $k$. If we want $\alpha_0=0.05$ with say $n=5$ for example we have to recognize that the possible values of $\alpha$ are $0, 1/32=0.03125, 2/32=0.0625$ and so on.

Possible rejection regions for $\alpha_0=0.05$:

Region	$\alpha$	$\beta$
$R_1=\emptyset$	0	1
$R_2=\{x=0\}$	0.03125	$1-(1/4)^5$
$R_3=\{x=5\}$	0.03125	$1-(3/4)^5$

So $R_3$ minimizes $\beta$ subject to $\alpha<0.05$.

Raise $\alpha_0$ slightly to 0.0625: possible rejection regions are $R_1$, $R_2$, $R_3$ and $R_4=R_2\cup R_3$.

The first three have the same $\alpha$ and $\beta$ as before while $R_4$ has $\alpha=\alpha_0=0.0625$ an $\beta=1-(3/4)^5-(1/4)^5$. Thus $R_4$ is optimal!

Problem: if all trials are failures ``optimal'' $R$ chooses $p=3/4$ rather than $p=1/2$. But $p=1/2$ makes 5 failures much more likely does $p=3/4$.

Problem: discreteness. Here's how we get around the problem. First we expand the set of possible values of $\phi$ to include numbers between 0 and 1. Values of $\phi(x)$ between 0 and 1 represent the chance that we choose $H_1$ given that we observe $x$; the idea is that we actually toss a (biased) coin to decide! This tactic will show us the kinds of rejection regions which are sensible. In practice we then restrict our attention to levels $\alpha_0$ for which the best $\phi$ is always either 0 or 1. In the binomial example we will insist that the value of $\alpha_0$ be either 0 or $P_{\theta_0} ( X\ge 5)$ or $P_{\theta_0} ( X\ge 4)$ or ... Smaller example: 4 possible values of $X$ and $2^4$ possible rejection regions. Here is a table of the levels for each possible rejection region $R$:

$R$	$\alpha$
$\emptyset$	0
{3}, {0}	1/8
{0,3}	2/8
{1}, {2}	3/8
{0,1}, {0,2}, {1,3}, {2,3}	4/8
{0,1,3}, {0,2,3}	5/8
{1,2}	6/8
{0,1,3}, {0,2,3}	7/8
{0,1,2,3}	1

Best level $2/8$ test has rejection region $\{0,3\}$, $\beta = 1-[(3/4)^3+(1/4)^3] = 36/64$. Best level $2/8$ test using randomization rejects when $X=3$ and, when $X=2$ tosses a coin with $P(H)=1/3$, then rejects if you get H. Level is $1/8+(1/3)(3/8) = 2/8$; probability of Type II error is $\beta =1-[(3/4)^3 +(1/3)(3)(3/4)^2(1/4)] = 28/64$.

Definition: A hypothesis test is a function $\phi(x)$ whose values are always in $[0,1]$. If we observe $X=x$ then we choose $H_1$ with conditional probability $\phi(X)$. In this case we have

$$\pi(\theta) = E_\theta(\phi(X))$$

$$\alpha = E_0(\phi(X))$$

and

$$\beta = 1- E_1(\phi(X))$$

Note that a test using a rejection region $C$ is equivalent to

$$\phi(x) = 1(x \in C)$$

The Neyman Pearson Lemma: In testing $f_0$ against $f_1$ the probability $\beta$ of a type II error is minimized, subject to $\alpha \le \alpha_0$ by the test function:

$$\phi(x) =\left\{\begin{array}{ll} 1 & \frac{f_1(x)}{f_0(x)} > \la... ...{f_0(x)} = \lambda \\ 0 & \frac{f_1(x)}{f_0(x)} < \lambda \end{array}\right.$$

where $\lambda$ is the largest constant such that

$$P_0( \frac{f_1(X)}{f_0(X)} \ge \lambda) \ge \alpha_0$$

and

$$P_0( \frac{f_1(X)}{f_0(X)}\le \lambda) \ge 1-\alpha_0$$

and where $\gamma$ is any number chosen so that

$$E_0(\phi(X)) = P_0( \frac{f_1(X)}{f_0(X)} > \lambda)$$

$$\quad +\quad \gamma P_0( \frac{f_1(X)}{f_0(X)} =\lambda)$$

$$= \alpha_0$$

The value of $\gamma$ is unique if $P_0( \frac{f_1(X)}{f_0(X)} = \lambda) > 0$.

Example: Binomial$(n,p)$ with $p_0=1/2$ and $p_1=3/4$: ratio $f_1/f_0$ is

$$3^x 2^{-n}$$

If $n=5$ this ratio is one of 1, 3, 9, 27, 81, 243 divided by 32.

Suppose we have $\alpha = 0.05$. $\lambda$ must be one of the possible values of $f_1/f_0$. If we try $\lambda = 243/32$ then

$$P_0(3^X 2^{-5} \ge 243/32) = P_0(X=5)$$

$$= 1/32 < 0.05$$

and

$$P_0(3^X 2^{-5} \ge 81/32) = P_0(X \ge 4)$$

$$= 6/32 > 0.05$$

So $\lambda=81/32$.

Since

$$P_0(3^X 2^{-5} > 81/32) =P_0( X=5) =1/32$$

we must solve

$$P_0(X=5) + \gamma P_0(X=4) = 0.05$$

for $\gamma$ and find

$$\gamma = \frac{0.05-1/32}{5/32}= 0.12$$

NOTE: No-one ever uses this procedure. Instead the value of $\alpha_0$ used in discrete problems is chosen to be a possible value of the rejection probability when $\gamma=0$ (or $\gamma=1$). When the sample size is large you can come very close to any desired $\alpha_0$ with a non-randomized test.

If $\alpha_0=6/32$ then we can either take $\lambda$ to be 243/32 and $\gamma=1$ or $\lambda=81/32$ and $\gamma=0$. However, our definition of $\lambda$ in the theorem makes $\lambda=81/32$ and $\gamma=0$.

When the theorem is used for continuous distributions it can be the case that the cdf of $f_1(X)/f_0(X)$ has a flat spot where it is equal to $1-\alpha_0$. This is the point of the word ``largest'' in the theorem.

Example: If $X_1,\ldots,X_n$ are iid $N(\mu,1)$ and we have $\mu_0=0$ and $\mu_1 >0$ then

$$\begin{multline*} \frac{f_1(X_1,\ldots,X_n)}{f_0(X_1,\ldots,X_n)} \\ = \\ \exp\{\mu_1 \sum X_i -n\mu_1^2/2 - \mu_0 \sum X_i + n\mu_0^2/2\} \end{multline*}$$

which simplifies to

$$\exp\{\mu_1 \sum X_i -n\mu_1^2/2 \}$$

Now choose $\lambda$ so that

$$P_0(\exp\{\mu_1 \sum X_i -n\mu_1^2/2 \}> \lambda ) = \alpha_0$$

Can make it equal because $f_1(X)/f_0(X)$ has a continuous distribution. Rewrite probability as

$$\begin{multline*} P_0(\sum X_i > [\log(\lambda) +n\mu_1^2/2]/\mu_1) \\ = \\ 1-\Phi\left(\frac{\log(\lambda) +n\mu_1^2/2}{n^{1/2}\mu_1}\right) \end{multline*}$$

Let $z_\alpha$ be upper $\alpha$ critical point of $N(0,1)$; then

$$z_{\alpha_0} = [\log(\lambda) +n\mu_1^2/2]/[n^{1/2}\mu_1]\, .$$

Solve to get a formula for $\lambda$ in terms of $z_{\alpha_0}$, $n$ and $\mu_1$.

The rejection region looks complicated: reject if a complicated statistic is larger than $\lambda$ which has a complicated formula. But in calculating $\lambda$ we re-expressed the rejection region in terms of

$$\frac{\sum X_i}{\sqrt{n}} > z_{\alpha_0}$$

The key feature is that this rejection region is the same for any $\mu_1 >0$. [WARNING: in the algebra above I used $\mu_1 >0$.] This is why the Neyman Pearson lemma is a lemma! Definition: In the general problem of testing $\Theta_0$ against $\Theta_1$ the level of a test function $\phi$ is

$$\alpha = \sup_{\theta\in\Theta_0}E_\theta(\phi(X))$$

The power function is

$$\pi(\theta) = E_\theta(\phi(X))$$

A test $\phi^*$ is a Uniformly Most Powerful level $\alpha_0$ test if

1. $\phi^*$ has level $\alpha \le \alpha_o$

2. If $\phi$ has level $\alpha \le \alpha_0$ then for every $\theta\in\Theta_1$ we have
   $$E_\theta(\phi(X)) \le E_\theta(\phi^*(X))$$

Proof of Neyman Pearson lemma: Given a test $\phi$ with level strictly less than $\alpha_0$ we can define the test

$$\phi^*(x) = \frac{1-\alpha_0}{1-\alpha} \phi(x) + \frac{\alpha_0-\alpha}{1-\alpha}$$

has level $\alpha_0$ and $\beta$ smaller than that of $\phi$. Hence we may assume without loss that $\alpha=\alpha_0$ and minimize $\beta$ subject to $\alpha=\alpha_0$. However, the argument which follows doesn't actually need this.

Lagrange Multipliers

Suppose you want to minimize $f(x)$ subject to $g(x) = 0$. Consider first the function

$$h_\lambda(x) = f(x) + \lambda g(x)$$

If $x_\lambda$ minimizes $h_\lambda$ then for any other $x$

$$f(x_\lambda) \le f(x) +\lambda[ g(x) - g(x_\lambda)]$$

Now suppose you can find a value of $\lambda$ such that the solution $x_\lambda$ has $g(x_\lambda) = 0$. Then for any $x$ we have

$$f(x_\lambda) \le f(x) +\lambda g(x)$$

and for any $x$ satisfying the constraint $g(x) = 0$ we have

$$f(x_\lambda) \le f(x)$$

This proves that for this special value of $\lambda$ the quantity $x_\lambda$ minimizes $f(x)$ subject to $g(x) = 0$.

Notice that to find $x_\lambda$ you set the usual partial derivatives equal to 0; then to find the special $x_\lambda$ you add in the condition $g(x_\lambda) = 0$.

Return to proof of NP lemma

For each $\lambda> 0$ we have seen that $\phi_\lambda$ minimizes $\lambda\alpha+\beta$ where $\phi_\lambda=1(f_1(x)/f_0(x) \ge \lambda)$.

As $\lambda$ increases the level of $\phi_\lambda$ decreases from 1 when $\lambda=0$ to 0 when $\lambda = \infty$. There is thus a value $\lambda_0$ where for $\lambda < \lambda_0$ the level is less than $\alpha_0$ while for $\lambda > \lambda_0$ the level is at least $\alpha_0$. Temporarily let $\delta=P_0(f_1(X)/f_0(X) = \lambda_0)$. If $\delta = 0$ define $\phi=\phi_\lambda$. If $\delta > 0$ define

$$\phi(x) =\left\{\begin{array}{ll} 1 & \frac{f_1(x)}{f_0(x)} > \la... ...(x)} = \lambda_0 \\ 0 & \frac{f_1(x)}{f_0(x)} < \lambda_0 \end{array}\right.$$

where $P_0(f_1(X)/f_0(X) < \lambda_0)+\gamma\delta = \alpha_0$. You can check that $\gamma\in[0,1]$.

Now $\phi$ has level $\alpha_0$ and according to the theorem above minimizes $\alpha+\lambda_0\beta$. Suppose $\phi^*$ is some other test with level $\alpha^* \le \alpha_0$. Then

$$\lambda_0\alpha_\phi+ \beta_\phi \le \lambda_0\alpha_{\phi^*} + \beta_{\phi^*}$$

We can rearrange this as

$$\beta_{\phi^*} \ge \beta_\phi +(\alpha_\phi-\alpha_{\phi^*})\lambda_0$$

Since

$$\alpha_{\phi^*} \le \alpha_0 = \alpha_\phi$$

the second term is non-negative and

$$\beta_{\phi^*} \ge \beta_\phi$$

which proves the Neyman Pearson Lemma.

Example application of NP: Binomial$(n,p)$ to test $p=p_0$ versus $p_1$ for a $p_1>p_0$ the NP test is of the form

$$\phi(x) =1(X>k)+\gamma 1(X=k)$$

where we choose $k$ so that

$$P_{p_0}(X>k) \le \alpha_0 < P_{p_0}(X \ge k)$$

and $\gamma\in[0,1)$ so that

$$\alpha_0 = P_{p_0}(X>k) + \gamma P_{p_0}(X =k)$$

This rejection region depends only on $p_0$ and not on $p_1$ so that this test is UMP for $p=p_0$ against $p>p_0$. Since this test has level $\alpha_0$ even for the larger null hypothesis it is also UMP for $p\le p_0$ against $p>p_0$.

Application of the NP lemma: In the $N(\mu,1)$ model consider $\Theta_1=\{\mu>0\}$ and $\Theta_0=\{0\}$ or $\Theta_0=\{\mu \le 0\}$. The UMP level $\alpha_0$ test of $H_0: \mu\in\Theta_0$ against $H_1:\mu\in\Theta_1$ is

$$\phi(X_1,\ldots,X_n) = 1(n^{1/2}\bar{X} > z_{\alpha_0})$$

Proof: For either choice of $\Theta_0$ this test has level $\alpha_0$ because for $\mu\le 0$ we have

$$P_\mu(n^{1/2}\bar{X} > z_{\alpha_0})$$

$$= P_\mu(n^{1/2}(\bar{X}-\mu) > z_{\alpha_0}-n^{1/2}\mu)$$

$$= P(N(0,1) > z_{\alpha_0}-n^{1/2}\mu)$$

$$\le P(N(0,1) > z_{\alpha_0})$$

$$= \alpha_0$$

(Notice the use of $\mu\le 0$. The central point is that the critical point is determined by the behaviour on the edge of the null hypothesis.) Now if $\phi$ is any other level $\alpha_0$ test then we have

$$E_0(\phi(X_1,\ldots,X_n)) \le \alpha_0$$

Fix a $\mu > 0$. According to the NP lemma

$$E_\mu(\phi(X_1,\ldots,X_n)) \le E_\mu(\phi_\mu(X_1,\ldots,X_n))$$

where $\phi_\mu$ rejects if

$$f_\mu(X_1,\ldots,X_n)/f_0(X_1,\ldots,X_n) > \lambda$$

for a suitable $\lambda$. But we just checked that this test had a rejection region of the form

$$n^{1/2}\bar{X} > z_{\alpha_0}$$

which is the rejection region of $\phi^*$. The NP lemma produces the same test for every $\mu > 0$ chosen as an alternative. So we have shown that $\phi_\mu=\phi^*$ for any $\mu > 0$.

This phenomenon is somewhat general. What happened was this. For any $\mu > \mu_0$ the likelihood ratio $f_\mu/f_0$ is an increasing function of $\sum X_i$. The rejection region of the NP test is thus always a region of the form $\sum X_i > k$. The value of the constant $k$ is determined by the requirement that the test have level $\alpha_0$ and this depends on $\mu_0$ not on $\mu_1$.

Definition: The family $f_\theta;\theta\in \Theta\subset R$ has monotone likelihood ratio with respect to a statistic $T(X)$ if for each $\theta_1>\theta_0$ the likelihood ratio $f_{\theta_1}(X) / f_{\theta_0}(X)$ is a monotone increasing function of $T(X)$.

Theorem: For a monotone likelihood ratio family the Uniformly Most Powerful level $\alpha$ test of $\theta \le \theta_0$ (or of $\theta=\theta_0$) against the alternative $\theta>\theta_0$ is

$$\phi(x) =\left\{\begin{array}{ll} 1 & T(x) > t_\alpha \\ \gamma & T(X)=t_\alpha \\ 0 & T(x) < t_\alpha \end{array}\right.$$

where $P_{\theta_0}(T(X) > t_\alpha)+\gamma P_{\theta_0}(T(X) = t_\alpha) = \alpha_0$.

Typical family where this works: one parameter exponential family. Usually there is no UMP test.

Example: test $\mu=\mu_0$ against two sided alternative $\mu\neq\mu_0$. There is no UMP level $\alpha$ test.

If there were its power at $\mu > \mu_0$ would have to be as high as that of the one sided level $\alpha$ test and so its rejection region would have to be the same as that test, rejecting for large positive values of $\bar{X} -\mu_0$. But it also has to have power as good as the one sided test for the alternative $\mu < \mu_0$ and so would have to reject for large negative values of $\bar{X} -\mu_0$. This would make its level too large.

Favourite test: usual 2 sided test rejects for large values of $\vert\bar{X} -\mu_0\vert$. Test maximizes power subject to two constraints: first, level $\alpha$; second power is minimized at $\mu=\mu_0$. Second condition means power on alternative is larger than on the null.

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