Problem 4: Consider the joint pmf given by the following table: 1 1 1/61/C0 S) X
ID: 2909974 • Letter: P
Question
Problem 4: Consider the joint pmf given by the following table: 1 1 1/61/C0 S) X000 11/61/0 S) (a) Compute the probability that X 0 and Y-1 so that the above table is a valid pmf. (b) Based on the joint pmf, compute the probability that X is nonzero and Y is nonzero. (c) Based on the joint pmf, compute the probability that X is nonzero or Y is nonzero. (d) Based on the joint pmf, compute the probability that X-Y. (e) Based on the joint pmf, compute and report the marginal pmf of X. (f) Based on the joint pmf, compute and report the marginal pmf of Y (g) Explain whether the two variables are independent based on probability.Explanation / Answer
a. Let P [X=0,Y=1] = P [0,1]
Since Total probability must be equal to 1.
therefore 1/6 +1/6 +0+0+0+ P[0,1] +1/6 +1/6 +0 = 1
or P [0,1] + 4/6 = 1
P [0,1] = 1 - 4/6 = 2/6
Thus P [X=0,Y=1] = 2/6.
b. P [X is nonzero and Y is nonzero] = P[-1,-1] + P[-1,1] + P[1,-1] + P[1,1]
= 1/6 + 0 + 1/6 + 0
= 2/6
c. P [ X is nonzero Or Y is nonzero] = 1 - P [ X is zero And Y is zero]
= 1 - P [X=0,Y=0]
=1 - P[0,0]
= 1 - 0 = 1
d. P [X=Y] = P[x=-1,Y=-1] + P[X=0,Y=0] +P[X=1,Y=1]
=1/6 +0 + 0 = 1/6
e. Now for the marginal distribution of X we have
P [ X=-1] = P[X=-1,Y=-1] +P[X=-1,Y=0] + P[X=-1,Y=1]
= 1/6 + 1/6 + 0
= 2/6
P [X=0] = P[X=0,Y=-1] +P[X=0,Y=0] + P[X=0,Y=1]
= 0 + 0 + 2/6
= 2/6
P [ X=1] = P[X=1,Y=-1] +P[X=1,Y=0] + P[X=1,Y=1]
= 1/6 +1/6 + 0
= 2/6
Thus the marginal pmf of X is
X : -1 0 1
P(x) : 2/6 2/6 2/6
f. Also for marginal pmf of Y we have
P[Y=-1] = P[X=-1,Y=-1] + P[X=0,Y=-1] +P[X=1,Y=-1]
= 1/6 +0 +1/6
= 2/6
P[Y=0] = P[X=-1,Y=0] + P[X=0,Y=0] +P[X=1,Y=0]
= 1/6 + 0 + 1/6
= 2/6
P[Y=1] = P[X=-1,Y=1] + P[X=0,Y=1] +P[X=1,Y=1]
= 0 +2/6 +0
=2/6
Thus the marginal pmf of Y is
Y : -1 0 1
P(y) : 2/6 2/6 2/6
Hope this helps.Thank You