Jane chooses a number X at random from the set of numbers{1, 2, 3, 4}, so thatP(X = k) =14for k = 1, 2, 3, 4.She then chooses a number Y at random from the subset ofnumbers {X, ...,4}; for example, ifX = 3, then Y is chosen atrandom from {3,4}.(i) Find the joint probability distribution of X and Y and displayit in the form of a two-way table.[5 marks](ii) Find the marginal probability distribution of Y , and hencefind E(Y ) and V ar(Y ).[4 marks](iii) Show that Cov(X, Y ) = 5/8.[4 marks](iv) Find the probability distribution of U = X + Y . [7

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Jazz Frenia · Accepted Answer

We are given that Jane chooses a number X at random from the set of numbers {1,2,3,4}, and P(X=k)=14 for k=1,2,3,4. We are also given that she chooses a number Y at random from the subset of numbers {X,...,4}.To find the joint probability distribution of X and Y, we need to calculate P(X=i,Y=j) for all i and j. We can do this by considering all possible values of X and Y.If X=1, then Y can be either 1, 2, 3, or 4. Since Y is chosen uniformly at random from {1,2,3,4}, we haveP(X=1,Y=j)=P(Y=j|X=1)P(X=1)=14·14=116.Similarly, if X=2, then Y can be either 2, 3, or 4, soP(X=2,Y=j)=P(Y=j|X=2)P(X=2)=13·14=112.If X=3, then Y can be either 3 or 4, soP(X=3,Y=j)=P(Y=j|X=3)P(X=3)=12·14=18.Finally, if X=4, then Y can only be 4, soP(X=4,Y=4)=P(Y=4|X=4)P(X=4)=1·14=14.Therefore, the joint probability distribution of X and Y is given by the following two-way table:Y=1Y=2Y=3Y=4X=1116116116116X=20112112112X=3001818X=400014And we can verify that the sum of all the probabilities in the table is 1.

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