PDA

View Full Version : Bayes Problem



jamesmartinn
02-04-2010, 10:19 PM
A diagnostic test is designed to detect whether subjects have a certain disease. A positive test outcome predicts that a subject has the disease. Given that a subject has the disease, the probability the diagnostic test is positive is called the sensitivity. Given that the subjet doesn't have the disease the probability that the test is negative is called specificity.

Consider a 2x2 table having the true status as the ROW variable and the diagnosis as the COLUMN variable. If "Positive" is the first level of each classification,then sensitivity is PI1 and specificity is 1-PI2. Let GAMMA denote the probability that the subject has the disease.

a) Given that the diagnosis is positive, use Bayes Theorem to show that the probability a subject has the disease is

PI1 * GAMMA / PI1 * GAMMA + PI2 * (1-GAMMA)

b) Suppose a diagnostic test for HIV+ has sensitivity and specificity equal to 0.95 if GAMMA = 0.005, find the probability that a subject is HIV+, given that the diagnostic test is positive

c) To better understand the answer in part (b), find the four joint probabilities for the 2x2 table and discuss their relative sizes in the two cells that refer to a positive result.



A) By Bayes,

P(true status = pos | diagnosis = pos) = P (diagnosis = pos | true status = pos) * P (true status = pos) / P (diagnosis = pos | true status = pos) * P (true status = pos) + P (diagnosis = pos | true status = neg) * P (true status = neg)

Substituting in variables gives: PI1 * GAMMA / PI1 * GAMMA + PI2 * (1-GAMMA)


B) 0.95 * 0.005 / 0.95 * 0.005 + 0.05 * 0.005 = 0.95

C)

The 2x2 Table is:
-------------------Diagnosis---------------
-------------------------Positive----Negative--
True Status:-----Positive .95----------.05
True Status:-----Negative.05----------.95

I'm not sure how I should discuss this. This assignment question is from a chapter on Odds Ratio, Relative Risk, and Difference of Probabilities; any thoughts?


Thanks!

GhostofPythagoras
02-05-2010, 03:22 PM
For C) the joint probabilities are simply the values found in the 2x2 table.

For instance the Prob that one has a positive diagnosis AND has the disease is 0.95 (upper left value in the table)

The Probability that one has a positive diagnosis AND does not have the disease is 0.05 (lower left value in table).

These are your two joint probabilities associated with a positive diagnosis.

jamesmartinn
02-06-2010, 11:08 AM
Thanks for the reply!

So, I'm taking part's A and B are fine?

I'm still stuck on part C. I know which two cells to look at but I'm not sure how to "comment" on them. I was thinking relative risk or odds ratio? Any ideas on how to further tackle this?


Thanks :D