Question

Consider two discrete random variables X, Y , taking values in {0, 1, 2, 3} each (for a total of 16 possible points). Their joint probability mass function is given by fXY (x, y) = c · x+1 y+1 . Answer the following questions. (a) What is c? (b) What is the probability that X = Y ? (c) Derive the marginal probability mass functions for both X and Y . (d) What is the probability that X ≤ Y given that Y = 2? (e) What is the probability that X ≤ Y ?

Answer 1

A. c=1/100

B. P(x==y)=3/10

C.
`f_(x)(x)=(x+1)/(10)\ y\ f_(y)(y)=(y+1)/(10)`

D. P(x≤y | y=2)=9/50

E. P(x≤y)= 7/20

Explanation:

A. To answer the question we use a really important property about mass functions of probability: If
`f_(xy)\` is a probability mass function, then (for a discrete variables because that´s our case)

`\sum_(i=1)^(\infty){\sum_(j=1)^(\infty){f_(xy)}}=1\\`

So, for our problem we have:

`\sum_(i=1)^(\infty){\sum_(j=1)^(\infty){c(x_(i)+1)(y_(j)+1)}}=1\\`

And we solve for c:

`c\sum_(i=1)^(\infty){\sum_(j=1)^(\infty){(x_(i)+1)(y_(j)+1)}}=1\\`

`c\sum_(i=1)^(\infty){(x_(i)+1)\sum_(j=1)^(\infty){(y_(j)+1)}}=1\\`

Because x and y only takes values in [0,1,2,3] we know:

`\sum_(i=1)^(\infty){(x_(i)+1)=(0+1)+(1+1)+(2+1)+(3+1)=10`

And we use that to solve for c:

`c\sum_(i=1)^(\infty){(x_(i)+1)(10)=1\\`

`c(10)(10)=1\\`

Finally we have c=1/100

B. For part B, we just choose the probabilities that we want because there are few values to choose and use the mass function to find them:

P(x=0,y=0)=(0+1)(0+1)/100=1/100

P(x=1,y=1)=4/100 (putting x=1 and y=1 in the function)

P(x=2,y=2)=9/100

P(x=3,y=3)=16/100

And for the last part we add them:

P(X==Y)=1/100+4/100+9/100+16/100=3/10

This is the answer

C. We find (in discrete variables) the marginal probability mass function just solving the function for one of the variables as follows:

`f_(x)(x)=\sum_(j=1)^(\infty){((x+1)(y_(j)+1))/(100)`

`f_(x)(x)=(10(x+1))/(100)`

`f_(x)(x)=(1)/(10)(x_(i)+1)\\`

And we do the same for y:

`f_(y)(y)=(1)/(10)(y_(i)+1)\\`

These functions are the marginal probability mass functions for x and y

D. Because there are few values of x and y that they can be. we choose the probabilities as follows:

P(x=0,y=2)=(0+1)(1+2)/100=3/100

P(x=1,y=2)=6/100

P(x=2,y=2)=9/100

And we add them to find our answer

P(x≤y | y=2)=3/100+6/100+9/100=9/50

E. We are working with a few amount of possible values, so we can use the part B (P(x==y)) and add them the probabilities that are left to complete P(x≤y)

P(x==y)=3/10 (B)

P(x=0,y=1)=2/100

P(x=0,y=2)=3/100

P(x=1,y=2)=6/100

P(x=0,y=3)=4/100

P(x=1,y=3)=8/100

P(x=2,y=3)=12/100

We add them to get our final answer:

P(x≤y)=7/20