# Statistical Probability Distributions

1. Nov 30, 2004

### Fenix

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?

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.

Last edited: Nov 30, 2004
2. Dec 1, 2004

### ehild

These are discrete variables, why do you integrate?

Make tables for probabilities P(X1=i, X2=j) for i, j = 1, 2, 3. For example, P(X1=2, X2=3)=(2*3)/36 = 1/6. You get the same probability when X1=3 and X2=2: P(X1=3, X2=2)=1/6.

Now find the possible values for X1*X2, and the possible values of both X1 and X2 which yield that product.

X1*X2 can have the value of 6 either with X1=2 and X2=3 or with X1=3 and X2=2.

So P(X1X2=6)=P(2,3)+P(3,2)=2*1/6=1/3.

Try to proceed this way.

ehild

3. Dec 1, 2004

### Fenix

You're right, that was a careless mistake.

I've corrected it now.

Instead of integration for Part a)

a)

h(y) = Sigma from X1=1 to 3: (Y/36X1) = Y/36+Y/72+Y/108 = 11Y/216, Y>0
h(y) = 0, elsewhere.

b)

h(y) = Sigma from X1=1 to 3: (X1^3/36Y^3) = (1^3+2^3+3^3)/36Y^3
= 36/36Y^3 = 1/Y^3, Y>0
h(y) = 0, elsewhere.

So how about now? Any other errors?

Also, what about the second question? Doesn't it strike you as odd that the answer is infinity?...

4. Dec 2, 2004

### ehild

I am afraid you did it entirely wrong. The sum of h(Y)-s for all possible values of x1*x2 which are 1, 2, 3, 4, 6, 9, must be 1. You can check yourself if your values meet this requirement. They do not.

You got h(Y)=11Y/216. The sum for all possible Y-s is 25*11/216 instead of 1.

Now again.

Y=1 can be only if both X1 and X2 is 1. f(X1,X2)=(X1X2)/36=1/36, so h(1)=1/36.

Y=2 is possible if either X1=1 and X2=2 or X1=2 and X2=1. f(1,2)=f(2,1)=2/36.
h(2)=f(1,2)+f(2,1)= 4/36.

Y=3 is obtained by x1=1 and x2=3 or vice versa. f(1,3)=f(3,1)=3/36, h(3)=6/36.

Yo get in the same way that h(4)= f(2*2) = 4/36;h(6)=f(2.3)+f(3,2)=12/36; h(9)=f(3,3)=9/36.

The sum of all h-s is (1+4+6+12+9)/36 =1.

Continue with question b....

ehild