Exercise 2.68

Answers

(a)

OR = \frac{P (D | C)}{P (D^{c} | C)} * \frac{P (D^{c} | C^{c})}{P (D | C^{c})}

Since the disease is rare among both exposed and not exposed groups, $P (D^{c} | C) \approx 1$ and $P (D^{c} | C^{c}) \approx 1$ . Thus,

OR \approx \frac{P (D | C)}{P (D | C^{c})} = RR

(b)

\frac{P (C, D) P (C^{c}, D^{c})}{P (C, D^{c}) P (C^{c}, D)} = \frac{P (C) P (D | C) P (C^{c}) P (D | C^{c})}{P (C) P (D^{c} | C) P (C^{c}) P (D | C^{c})} = OR

(c)

Since

P (C, D)

also equals

P (D) P (C | D)

, reversing the roles of

C

and

D

in part

b

gives the result.

fifthist

2021-12-05 00:00