The random variables X and Y have the joint PMF
pX,Y(x,y)={c⋅(x+y)2,0,if x∈{1,2,4} and y∈{1,3},otherwise.
All answers in this problem should be numerical.
Find the value of the constant c.
c=
c=1/128
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= 1/4
To find the value of the constant c, we need to normalize the joint probability mass function (PMF) such that the sum of all probabilities equals 1.
We can start by calculating the sum of all probabilities. Considering the given condition for x and y, the sum can be written as:
c * [(1+1)^2 + (1+3)^2 + (1+1)^2 + (1+3)^2 + (2+1)^2 + (2+3)^2 + (4+1)^2 + (4+3)^2]
= c * [2^2 + 4^2 + 2^2 + 4^2 + 3^2 + 5^2 + 5^2 + 7^2]
= c * [4 + 16 + 4 + 16 + 9 + 25 + 25 + 49]
= c * 148
Since the sum of all probabilities should be equal to 1, we set the above expression equal to 1 and solve for c:
c * 148 = 1
Dividing both sides of the equation by 148:
c = 1/148
Therefore, the value of the constant c is 1/148.
To find the value of the constant c, we need to apply the properties of a joint probability mass function (PMF). The sum of the probabilities over all possible values of the random variables must be equal to 1.
In this case, we need to sum the probabilities over all possible values of x and y, and set it equal to 1.
pX,Y(x,y) = c*(x+y)^2
We can start by listing all possible values of x and y according to the given conditions:
x = 1,2,4
y = 1,3
Now, we can substitute these values into the joint PMF and sum up the probabilities:
P(X,Y) = ΣΣ pX,Y(x,y)
P(X,Y) = c*(1+1)^2 + c*(1+3)^2 + c*(2+1)^2 + c*(2+3)^2 + c*(4+1)^2 + c*(4+3)^2
P(X,Y) = c*4 + c*16 + c*9 + c*25 + c*25 + c*49
P(X,Y) = c*(4 + 16 + 9 + 25 + 25 + 49)
Now, we set the total probability equal to 1:
1 = c*(4 + 16 + 9 + 25 + 25 + 49)
Simplifying the equation:
1 = c*(128)
Now, we solve for c:
c = 1/128
Therefore, the value of the constant c is 1/128.