I have two questions, I've completed. I am partially sure that the answers I've obtained are correct, and all I really want is confirmation on whether they are correct or not. If not, what am I doing wrong?(adsbygoogle = window.adsbygoogle || []).push({});

Question 1:

If the joint probability distribution of X1 and X2 is given by:

F(X1,X2) = (X1X2)/36

Where X1 = 1,2,3, and X2 = 1,2,3

a) Find the probability distribution of X1X2.

Solution:

Y = X1X2

X2 = Y/X1

dX2/dY = 1/X1

g(X1,Y) = [X1(Y/X1)/36][|1/X1|] = Y/36X1, For X1=1,2,3, and Y>0

Therefore, h(y) = Integral from 1 to 3: (Y/36X1)dX1 = Yln3/36, for Y>0

h(y) = 0, elsewhere.

b) Find the probability distribution of X1/X2

Solution:

Y = X1/X2

X2 = X1/Y

dX2/dY = -X1/Y^2

g(X1,Y) = [[X1(X1/Y)]/36][|-X1/Y^2|] = X1^3/36Y^3, For X1 = 1,2,3, and Y>0

Therefore, h(Y) = Integral from 1 to 3: (X1^3/36Y^3)dX1 = 5/9Y^3, for Y>0

h(y) = 0, elsewhere

Question 2:

Consider two random variables X and Y with the joint probability density:

f(X,Y) = {12XY(1-Y), for 0<X<1, 0<y<1.

0, elsewhere

Find the probability density of Z=XY^2 to determine the joint probability density of Y and Z and then integrating out Y.

Solution:

Z = XY^2

X = Z/Y^2

dX/dZ = 1/Y^2

g(Y,Z) = 12(Z/Y^2)(Y)(|1/Y^2|) = 12Z/Y^3, for 0<Y<1, and 0<Z<1

h(y) = Integral from 0 to 1: (12Z/Y^3)dy = Infinity, does not exist.

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References: Functions of Random Variables - Transformation Technique of Several Variables.

Thanks in advance. I appreciate it.

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# Homework Help: Statistical Probability Distributions

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