Given X = {0, 1} with X = 1 indicating disease. Given their status, a diagnostic test returns a classification of Y E {0, 1} with Y = 1 indicating a positive result: X~ Ber(p) | YX ~ Ber(px)

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Kindly solve the following

Given X = {0, 1} with X = 1 indicating disease. Given their status, a diagnostic test returns a classification of
YE {0, 1} with Y = 1 indicating a positive result:
X~ Ber(p) |
Y | X~ Ber(px)
Recall that the pmf of a generic Ber(π) random variable, W, is
fw (w) = pw(1 p)¹-w 0 ≤ ≤1
for w = {0, 1} and fw (w) = 0 elsewhere.
a. solve for the expression for P(X = 1 | Y = y) only in terms of y = 2², and p.
b. Prove that the conditional variance of X is
V[X | Y = y] = P(X = 1 | Y = y)(1 − P(X = 1 | Y = y))
Transcribed Image Text:Given X = {0, 1} with X = 1 indicating disease. Given their status, a diagnostic test returns a classification of YE {0, 1} with Y = 1 indicating a positive result: X~ Ber(p) | Y | X~ Ber(px) Recall that the pmf of a generic Ber(π) random variable, W, is fw (w) = pw(1 p)¹-w 0 ≤ ≤1 for w = {0, 1} and fw (w) = 0 elsewhere. a. solve for the expression for P(X = 1 | Y = y) only in terms of y = 2², and p. b. Prove that the conditional variance of X is V[X | Y = y] = P(X = 1 | Y = y)(1 − P(X = 1 | Y = y))
Expert Solution
Step 1

Given information:

Statistics homework question answer, step 1, image 1

Statistics homework question answer, step 1, image 2

Consider, the joint distribution for X and Y is,

Statistics homework question answer, step 1, image 3

 

 

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