3. Let X, Y be independent Bernoulli(1/2) random variables. Let Z be a random variable that takes the value 1 if X+Y= 1, and 0 otherwise. Show that X, Y, Z are pairwise, but not mutually, independent. Note: To show that they are pairwise independent, show that each of the pairwise joint PMFs factorize into the product of the individual PMFs: px,y = PxPY, Px,z = PxPz and Pyz = PyPz. That is, show that px,y (x,y) = px (x)py (y) Px,z(x,z) = Px(x)pz(z) py,z(y, z) = py(y)pz(z) for all x, y, z. Then, to show that they are not mutually independent, show that the joint PMF px,y,z ‡ pxpypz. That is, show that Px,y,z(x, y, z) ‡ px(x)py (y)Pz(z), for some x, y, z.
3. Let X, Y be independent Bernoulli(1/2) random variables. Let Z be a random variable that takes the value 1 if X+Y= 1, and 0 otherwise. Show that X, Y, Z are pairwise, but not mutually, independent. Note: To show that they are pairwise independent, show that each of the pairwise joint PMFs factorize into the product of the individual PMFs: px,y = PxPY, Px,z = PxPz and Pyz = PyPz. That is, show that px,y (x,y) = px (x)py (y) Px,z(x,z) = Px(x)pz(z) py,z(y, z) = py(y)pz(z) for all x, y, z. Then, to show that they are not mutually independent, show that the joint PMF px,y,z ‡ pxpypz. That is, show that Px,y,z(x, y, z) ‡ px(x)py (y)Pz(z), for some x, y, z.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:3. Let X, Y be independent Bernoulli(1/2) random variables. Let Z be a random
variable that takes the value 1 if X+Y= 1, and 0 otherwise. Show that X, Y, Z
are pairwise, but not mutually, independent.
Note: To show that they are pairwise independent, show that each of the
pairwise joint PMFs factorize into the product of the individual PMFs: px,y =
PxPY, Px,z = PxPz and Pyz = PyPz. That is, show that
px,y (x,y) = px (x)py (y)
Px,z(x,z) = Px(x)pz(z)
py,z(y, z) = py(y)pz(z)
for all x, y, z. Then, to show that they are not mutually independent, show that
the joint PMF px,y,z ‡ pxpypz. That is, show that
Px,y,z(x, y, z) ‡ px(x)py (y)Pz(z),
for some x, y, z.
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