2.3 The neyman-pearson criterion (Page 2/3)

Signal and information processing Page 3 / 3

The neyman-pearson lemma: general case

In our initial statement of the Neyman-Pearson Lemma, we assumed that for all , the set $x x x$ had probability zero under $_{0}$ . This eliminated many important problems from consideration, including tests of discrete data.In this section we remove this restriction.

It is helpful to introduce a more general way of writing decision rules. Let be a function of the data $x$ with $x 01$ . defines the decision rule "declare $_{1}$ with probability $x$ ." In other words, upon observing $x$ , we flip a " $x$ coin." If it turns up heads, we declare $_{1}$ ; otherwise we declare $_{0}$ . Thus far, we have only considered rules with $x 01$

Neyman-pearson lemma

Consider the hypothesis testing problem $_{0} : x f_{0} x$ $_{1} : x f_{1} x$ where $f_{0}$ and $f_{1}$ are both pdfs or both pmfs. Let $01$ be the size (false-alarm probability) constraint. The decision rule $x 1 x x 0 x$ is the most powerful test of size , where and are uniquely determined by requiring $P_{F}$ . If $0$ , we take ∞ , $0$ . This test is unique up to sets of probability zero under $_{0}$ and $_{1}$ .

When

x 0

for certain

, we choose

and

as follows: Write

P_{F} x x

Choose

such that

x x

Then choose

such that

x x

Repetition code

Suppose we have a friend who is trying to transmit a bit (0 or 1) to us over a noisy channel. Thechannel causes an error in the transmission (that is, the bit is flipped) with probability $p$ , where $0 p 12$ , and $p$ is known. In order to increase the chance of a successful transmission,our friend sends the same bit $N$ times. Assume the $N$ transmissions are statistically independent. Under these assumptions, the bitsyou receive are Bernoulli random variables: $x_{n} Bernoulli$ . We are faced with the following hypothesis test: $_{0} : p (0 sent)$ $_{1} : 1 p (1 sent)$ We decide to decode the received sequence $x x_{1} x_{N}$ by designing a Neyman-Pearson rule. The likelihood ratio is

x n 1 N 1 p x_{n} p 1 x_{n} n 1 N p x_{n} 1 p 1 x_{n} 1 p k p N k p k 1 p N k 1 p p 2 k N

where

k n 1 N x_{n}

is the number of 1s received.

k

is a sufficient statistic for

The LRT is

1 p p 2 k N_{_{0}}^{_{1}}

Taking the natural logarithm of both sides and rearranging, we have

k_{_{0}}^{_{1}} N 2 12 1 p p

The false alarm probability is

P_{F} k k k 1 N N k p k 1 p N k N p 1 p N

and

are chosen so that

P_{F}

, as described above.

The corresponding detection probability is

P_{D} k k k 1 N N k 1 p k p N k N 1 p p N

Problems

Design a hypothesis testing problem involving continous random variables such that $x 0$ for certain values of . Write down the false-alarm probability as a function of thethreshold. Make as general a statement as possible about when the technical condition is satisfied.

Consider the scalar hypothesis testing problem $_{0} : x f_{0} x$ $_{1} : x f_{1} x$ where $f_{i} x 1 1 x i 2, i 01$

Write down the likelihood ratio test.

Determine the decision regions as a function of $_{1}$ for all $0$ . Draw a representative of each. What are the "critical" values of ?

There are five distinct cases.

Compute the size and power ( $P_{F}$ and $P_{D}$ ) in terms of the threshold $_{1}$ and plot the ROC.

x 1 1 x 2 x

Suppose we decide to use a simple threshold test $x_{_{0}}^{_{1}}$ instead of the Neyman-Pearson rule. Does our performance $_{0}$ suffer much? Plot the ROC for this decision rule on the same graph as for the previous ROC.

Suppose we observe $N$ independent realizations of a Poisson random variable $k$ with intensity parameter : $f k k k$ We must decide which of two intensities is in effect: $_{0} :_{0}$ $_{1} :_{1}$ where $_{0}_{1}$ .

Write down the likelihood ratio test.

Simplify the LRT to a test statistic involving only a sufficient statistic. Apply a monotonicallyincreasing transformation to simplify further.

Determine the distribution of the sufficient statistic under both hypotheses.

Use the characteristic function to show that a sum of IID Poissonvariates is again Poisson distributed.

Derive an expression for the probability of error.

Assuming the two hypotheses are equally likely, and $_{0} 5$ and $_{1} 6$ , what is the minimum number $N$ of observations needed to attain a false-alarm probability no greater than0.01?

If you have numerical trouble, try rewriting the log-factorial so as to avoidevaluating the factorial of large integers.

In , suppose $p 0.1$ . What is the smallest value of $N$ needed to ensure $P_{F} 0.01$ ? What is $P_{D}$ in this case?

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Source: OpenStax, Signal and information processing for sonar. OpenStax CNX. Dec 04, 2007 Download for free at http://cnx.org/content/col10422/1.5

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