Question

1. On the basis of a physical examination, a doctor determines the probability of no tumour   (event labelled C for ‘clear’), a benign tumour (B) or a malignant tumour (M) as 0.7, 0.2 and 0.1 respectively.

A further, in depth, test is conducted on the patient which can yield either a negative (N) result or positive (P). The test gives a negative result with probability 0.9 if no tumour is present (i.e. P(N|C) = 0.9). The test gives a negative result with probability 0.8 if there is a benign tumour and 0.2 if there is a malignant tumour.

(i) Given this information calculate the joint and marginal probabilities and display in the table below.

	Positive (P)	Negative (N)
Clear          (C)
Benign        (B)
Malignant   (M)
			1

2. What is the marginal probability the test result will be negative?
3. Obtain the posterior probability distribution for the patient when the test result is

            a) positive,   b) negative

4. Comment on how the test results change the doctor’s view of the presence of a tumour.

2. (i) A continuous variable X defined on the interval (1, ∞) has p.d.f given by f(x) = 1/x²

Derive the corresponding cumulative density function and graph it

(ii) Can the function f(x) = 1/x² define a p.d.f over the interval (2, ∞)? Explain your answer.

3. In the context of a discrete variable, X, show that the formula for variance of the variable can be written as Var(X) = E(X²) – [E(X)]²

4. Suppose that f(x) = 1/5 , x = 1, 2, 3, 4, 5 and zero elsewhere is the p.d.f. of the discrete random variable X. Compute E(X) and E(X²).

Use these results to find E[(X+2)²]

5. In question 4, calculate Var(X).

Answer 1

(i)

	Positive (P)	Negative (N)	Total
Clear          (C)	0.07	0.63	0.7
Benign        (B)	0.04	0.16	0.2
Malignant   (M)	0.08	0.02	0.1
Total	0.19	0.81	1

Given,

P(C) = 0.7, P(B) = 0.2, P(M) = 0.1

P(N | C) = 0.9

=> P(N and C) / P(C) = 0.9

=> P(N and C) = 0.9 * P(C) = 0.9 * 0.7 = 0.63

P(P and C) = P(C) -  P(N and C) = 0.7 - 0.63 = 0.07

P(N | B) = 0.8

=> P(N and B) / P(B) = 0.8

=> P(N and B) = 0.8 * P(B) = 0.8 * 0.2 = 0.16

P(P and B) = P(B) -  P(N and B) = 0.2 - 0.16 = 0.04

P(N | M) = 0.2

=> P(N and M) / P(M) = 0.2

=> P(N and M) = 0.2 * P(M) = 0.2 * 0.1 = 0.02

P(P and M) = P(M) -  P(N and M) = 0.1 - 0.02 = 0.08

P(P) = 0.07 + 0.04 + 0.08 = 0.19

P(N) = 0.63 + 0.16 + 0.02 = 0.81

(ii)

From the table, the marginal probability the test result will be negative = P(N) = 0.81

(iii)

a)

P(P | C) = 1 - P(N | C) = 1 - 0.9 = 0.1

P(P | B) = 1 - P(N | C) = 1 - 0.8 = 0.2

P(P | M) = 1 - P(N | C) = 1 - 0.2 = 0.8

Posterior probability distribution for the patient when the test result is positive are,

P(C | P) = P(P | C) P(C) / P(P) {By Bayes theorem}

= 0.1 * 0.7 / 0.19

= 0.3684211

P(B | P) = P(P | B) P(B) / P(P) {By Bayes theorem}

= 0.2 * 0.2 / 0.19

= 0.2105263

P(M | P) = P(P | M) P(M) / P(P) {By Bayes theorem}

= 0.8 * 0.1 / 0.19

= 0.4210526

(ii)

Posterior probability distribution for the patient when the test result is negative are,

P(C | N) = P(N | C) P(C) / P(N) {By Bayes theorem}

= 0.9 * 0.7 / 0.81

= 0.7777778

P(B | N) = P(N | B) P(B) / P(N) {By Bayes theorem}

= 0.8 * 0.2 / 0.81

= 0.1975309

P(M | N) = P(P | N) P(M) / P(N) {By Bayes theorem}

= 0.2 * 0.1 / 0.81

= 0.02469136

(iv)

The chance of malignant tumor is high (0.4210526) when the test is positive,

The chance of malignant tumor is low (0.02469136) when the test is negative,

Thus, for negative result,  doctor’s view will be of absence of a tumour.