Probability question using r.v. X and Y what is W=min(X,Y) ?

[LeeroyJenkins]

Homework Statement

Y
-----------1----2-----3------4
x---1----1/4----0-----0------0
----2----1/8----1/8---0------0
----3----1/12---1/12--1/12---0
----4----1/16---1/16--1/16--1/16

The Joint PMF of random varables X and Y is given in the table above
Let W=min(X,Y) and Z = max(X,Y)

Find the marginal PMF Pw(w) of w ?
(I don't understand what is meant by W=min(X,Y)

Find the marginal PMF Pz(Z) of Z?
Z = man(Z,Y)

Homework Equations

Pw(w) = SUM Pxy(x,y) ; where the sum starts at (x,y):g(x,y) = w

The Attempt at a Solution

Pw(1) = 0
Pw(2) = 0
Pw(3) = 0
Pw(4) = 0

or

Pw(1) = 1/4 as 1 is the min of x and y

Pz(1) = 1/16
Pz(2) = 1/16
Pz(3) = 1/16
Pz(4) = 1/16

Thanks in advance

 

[mighty2000]

for any help!

Hi there,

To answer your first question, W=min(X,Y) means that W is the minimum value between X and Y. So for each (x,y) pair, W will take on the smaller of the two values. For example, in the first row of the table, W would equal 1 because 1 is the smaller value between X=1 and Y=4.

To find the marginal PMF Pw(w), you would need to sum the probabilities of all (x,y) pairs where W=w. So for w=1, you would have Pw(1) = P(X=1,Y=1) = 1/4. For w=2, you would have Pw(2) = P(X=1,Y=2) + P(X=2,Y=1) = 1/8 + 1/8 = 1/4. And so on for w=3 and w=4.

For the second question, you are correct that Z=max(X,Y). So for each (x,y) pair, Z will take on the larger of the two values. For example, in the first row of the table, Z would equal 4 because 4 is the larger value between X=1 and Y=4.

To find the marginal PMF Pz(z), you would need to sum the probabilities of all (x,y) pairs where Z=z. So for z=1, you would have Pz(1) = P(X=1,Y=1) = 1/16. For z=2, you would have Pz(2) = P(X=1,Y=2) + P(X=2,Y=1) = 1/8 + 1/8 = 1/4. And so on for z=3 and z=4.

I hope this helps clarify things for you. Good luck with your studies!