Consider the following joint PMF of the random variables X and Y:
pX,Y(x,y)={1/72⋅(x2+y2),if x∈{1,2,4} and y∈{1,3}, 0, otherwise}.
1. P(Y<X)=
2. P(Y=X)=
3. Find the marginal PMF pX(x).
pX(1)=
pX(2)=
pX(3)=
pX(4)=
4. Find E[X] and E[XY].
E[X]=
E[XY]=
5. var(X)=
6. Let A denote the event X≥Y. Find E[X∣A].
E[X∣A]=
1.47/22
2.1/36
3a.12/72
3b.1/4
3c.0
3d.42/72
4.3
5.???
6.1.5
7.??
1???
2 2/72
3a 12/72
3b 1/4
3c 0
3d 42/72
4a 3
4b 61/9
5 3/2
6 173/47
1. 47/72
c= 5/64
P(Y<X)= 83/128
P(Y=X)= 1/32
P(X=1)= 10/64
P(X=2)= 17/64
P(X=3)= 0
P(X=4)= 37/64
E[X]= 3
E[XY]= 227/32
var(X)= 3/2
c= 1/128
P(Y<X)= 83/128
P(Y=X)= 1/32
P(X=1)= 10/64
P(X=2)= 17/64
P(X=3)= 0
P(X=4)= 37/64
E[X]= 3
E[XY]= 227/32
VAR(X)= 3/2
To find the answers to the given questions, let's break it down step by step:
1. P(Y<X):
To find P(Y<X), we need to sum up the probabilities of all the cases where Y is less than X. In this joint PMF, there are three possible cases where Y<X: (1,1), (2,1), and (2,3). So, we need to calculate:
P(Y<X) = pX,Y(1,1) + pX,Y(2,1) + pX,Y(2,3)
2. P(Y=X):
To find P(Y=X), we need to sum up the probabilities of all the cases where Y is equal to X. In this joint PMF, there is only one possible case where Y=X: (1,1). So, we just need to calculate:
P(Y=X) = pX,Y(1,1)
3. Find the marginal PMF pX(x):
To find the marginal PMF pX(x), we need to sum up the joint probabilities over all the possible values of Y for a given value of X. In this case, we have four possible values of X: 1, 2, 3, and 4.
To find pX(1), we sum up the joint probabilities pX,Y(1,1) and pX,Y(1,3).
To find pX(2), we sum up the joint probabilities pX,Y(2,1) and pX,Y(2,3).
To find pX(3), there is no joint probability with X=3.
To find pX(4), we sum up the joint probability pX,Y(4,1).
4. Find E[X] and E[XY]:
To find the expected value E[X], we need to multiply each possible value of X by its corresponding probability pX(x) and sum them up.
To find the expected value E[XY], we need to multiply each possible value of X and Y by their joint probability pX,Y(x,y) and sum them up.
5. var(X):
To find the variance var(X), we need to find E[X^2] - (E[X])^2. This involves calculating the expected value of X squared.
6. Let A denote the event X≥Y. Find E[X∣A]:
To find E[X∣A], we need to find the expected value of X, given that X≥Y. This involves finding the conditional probability p(A) and calculating the conditional expected value.
Using these steps, you can find the answers to the given questions by plugging in the appropriate values from the joint PMF provided.