Question

A diagnostic test for a certain disease is applied to n individuals known to not have the disease. Let X = the number among the n test results that are positive (indicating presence of the disease, so X is the number of false positives) and p = the probability that a disease-free individual's test result is positive (i.e., p is the true proportion of test results from disease-free individuals that are positive). Assume that only X is available rather than the actual sequence of test results.
(a) Derive the maximum likelihood estimator of p.


If n = 25 and x = 3, what is the estimate?

(b) Is the estimator of part (a) unbiased?

(c) If n = 25 and x = 3, what is the mle of the probability (1 - p)5 that none of the next five tests done on disease-free individuals are positive? (Round your answer to four decimal places.)

Answer 1

Answer:

Part a

The maximum likelihood estimator of p, is the sample proportion,  $\bar p=\frac{x}{n}$

And the proportion of the estimate at n=25 and x = 3  is 0.12.

Part b

The maximum likelihood estimate from part (a) is an unbiased estimator.

Part c

The maximum likelihood estimate of the probability is 0.5277.

Step-by-step explanation:

a)

The likelihood function is, $L(p;x) =\frac{n!}{x!(n-x)!}p*(1- p)^{n-x}$

Applying log on both sides:

$\log L(p;x) =\log\frac{n!}{x!(n-x)!}+x\log p+(n-x)log(1-p)$

Find the differentiation of log L with respect to P and equate to zero to find the MLE for .

$\frac{d}{dp}\log L(p,x)=\frac{x}{p}-\frac{n-x}{1-p}\\\\=\frac{x}{p}-\frac{(n-x)}{1-p}\\\\=x(1-p)-(n-x)p=0\\=x-xp-np+xp=0\\=x-np=0\\=x=np\\=\bar p=\frac{x}{n}$

Calculate the estimate of the function if n=25 and x = 3.

$\bar p=\frac{x}{n}\\\\=\frac{3}{25}\\\\= 0.12$

Hint for next step

The MLE for the population proportion is the sample proportion and there is a 0.12 proportion of estimate at n = 25 and x = 3  

b)

Consider  

$E(\bar p)\\\\E(\bar p)=E(\frac{X}{n})\\\\=\frac{1}{n}E(X)\\\\=\frac{1}{n}(np)(since\, E(X) =np)\\\\=p$

Since $E(\bar p)=p$, the maximum likelihood estimate from part (a) is an unbiased estimator.

Hint for next step

Based on the part (a) calculation, since it is proved that , the maximum likelihood estimate is unbiased estimator for population proportion

(c)

Calculate the maximum likelihood estimate of the probability $(1-p)^5$.

By the Invariance property of MLE, to find the MLE of $(1-P)^5$, the MLE of p can be used.

That is, the MLE of $(1-p)^5$ is $(1-\bar p)^5$

Therefore,

MLE of the probability,

$(1- p)^5 = (1-\bar P)^5\\\\=(1-\frac{3}{25})^5\\\\=(1-0.12)^5\\=(0.88)^5\\=0.5277$