# Stuck again on disributions

• Nov 27th 2006, 02:57 PM
SwedishMan
Stuck again on disributions
I have another question. Again I'm in a college probability class studying multivarate distributions.

Here goes:
In a clinical study of a new drug formulated to reduce the effects of arthritis, researchers found that the proportion p of patients who respond favourably to the drug is a random variable that varies from batch to batch of the drug. Assume that p has a probability density function given by
f(p)={12*(p^(2))*(1-p), whenever p is between 0 and 1 inclusive
{0, whenever p is elsewhere.
Suppose that n patients are injected with portions of the drug taken from the same batch. Let Y denote the number showing a favourable response.
(A) Find the unconditional probability distribution of Y for general n.
(B) Find E(Y) for n = 2.

I'm confused because there's no Y1 and Y2. I would love any help.
• Nov 28th 2006, 08:53 AM
CaptainBlack
Quote:

Originally Posted by SwedishMan
I have another question. Again I'm in a college probability class studying multivarate distributions.

Here goes:
In a clinical study of a new drug formulated to reduce the effects of arthritis, researchers found that the proportion p of patients who respond favourably to the drug is a random variable that varies from batch to batch of the drug. Assume that p has a probability density function given by
f(p)={12*(p^(2))*(1-p), whenever p is between 0 and 1 inclusive
{0, whenever p is elsewhere.
Suppose that n patients are injected with portions of the drug taken from the same batch. Let Y denote the number showing a favourable response.
(A) Find the unconditional probability distribution of Y for general n.

The joint distribution is:

P(y,n,p)=p(y,n|p)f(p)

the probability that p takes value p and the number of favourable responses
is y. Now Y is a binomially distributed RV with probability of a fav. response for
an individual patient p, and number of patients n, so:

p(y,n|p)=n!/[(n-y)!y!)] p^y (1-p)^(n-y)

The unconditional (marginal?) distribution of Y is:

g(y,n)=int(0,1) P(y,n,p) dp

Quote:

(B) Find E(Y) for n = 2.
E(Y|n=2)= sum(y=0,infty) g(y,2)

RonL
• Nov 28th 2006, 09:09 AM
SwedishMan
Thank you so much.