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)
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)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 31E
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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))](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa77d52c3-5480-47b7-8f2c-45b52855bf59%2Fc560b7e2-b877-4a77-ba74-955008821253%2F5x7lcn6_processed.png&w=3840&q=75)
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))
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