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:A. c=1/100B. P(x==y)=3/10C. [tex]f_{x}(x)=\frac{x+1}{10}\ y\ f_{y}(y)=\frac{y+1}{10}[/tex]D. P(x≤y | y=2)=9/50E. P(x≤y)= 7/20Step-by-step explanation:A. To answer the question we use a really important property about mass functions of probability: If [tex]f_{xy}\[/tex] is a probability mass function, then (for a discrete variables because that´s our case)[tex]\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{f_{xy}}}=1\\[/tex]So, for our problem we have:[tex]\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{c(x_{i}+1)(y_{j}+1)}}=1\\[/tex]And we solve for c:[tex]c\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{(x_{i}+1)(y_{j}+1)}}=1\\[/tex][tex]c\sum_{i=1}^{\infty}{(x_{i}+1)\sum_{j=1}^{\infty}{(y_{j}+1)}}=1\\[/tex]Because x and y only takes values in [0,1,2,3] we know:[tex]\sum_{i=1}^{\infty}{(x_{i}+1)=(0+1)+(1+1)+(2+1)+(3+1)=10[/tex]And we use that to solve for c:[tex]c\sum_{i=1}^{\infty}{(x_{i}+1)(10)=1\\[/tex][tex]c(10)(10)=1\\[/tex]Finally we have c=1/100B. 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=1/100P(x=1,y=1)=4/100 (putting x=1 and y=1 in the function)P(x=2,y=2)=9/100P(x=3,y=3)=16/100And for the last part we add them:P(X==Y)=1/100+4/100+9/100+16/100=3/10This is the answerC. We find (in discrete variables) the marginal probability mass function just solving the function for one of the variables as follows:[tex]f_{x}(x)=\sum_{j=1}^{\infty}{\frac{(x+1)(y_{j}+1)}{100}[/tex][tex]f_{x}(x)=\frac{10(x+1)}{100}[/tex][tex]f_{x}(x)=\frac{1}{10}(x_{i}+1)\\[/tex]And we do the same for y:[tex]f_{y}(y)=\frac{1}{10}(y_{i}+1)\\[/tex]These functions are the marginal probability mass functions for x and yD. 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=3/100P(x=1,y=2)=6/100P(x=2,y=2)=9/100And we add them to find our answerP(x≤y | y=2)=3/100+6/100+9/100=9/50E. 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)=2/100P(x=0,y=2)=3/100P(x=1,y=2)=6/100P(x=0,y=3)=4/100P(x=1,y=3)=8/100P(x=2,y=3)=12/100We add them to get our final answer:P(x≤y)=7/20