Suppose we have a box with two pennies and one dime. We randomly sample one coin

Question

Suppose we have a box with two pennies and one dime. We randomly sample one coin from the box and flip. The penny has probability 1/8 of coming up heads and the dime has probability 1/4 of coming up heads. Define the random variables X and Y as follows. X = {1 if coin chosen is a penny 2 if coin chosen is a dime Y ={1 if get heads 2 if get tails Then we have Pr(X = 1) = 2/3, Pr(Y = 1| X = 1) = 1/8, and Pr(F = 1|X = 2) = 1/4 Find p(X, Y). That is find the joint distribution of X and Y. Give your answer as a table in the same format as those in questions 4 and 8. Are X and Y independent? Provide support for your answer, by showing certain probabilities are either equal (if showing independence) or not equal (if showing dependence). Find E(X) and E(Y). Find cov(X, Y).

Explanation / Answer

given

P(X=1) =2/3 P(Y=1|X=1)=1/8 so P(Y=2|X=1) =7/8

P(X=2) =1/3 P(Y=1|X=2) =1/4 so P(Y=2 |X=2) =3/4

a)

since P(Y|X)=P(Y and X)/P(X)

then P(Y and X) =P(X) P(Y|X)

b)

P(X=1,Y=1) =1/12

P(X=1)P(Y=1) =(2/3)*(1/6) =1/9

since P(X=1,Y=1) not equal to P(X=1)*P(Y=1)

hence X and Y are Not independent

c)

E(X) =1*(2/3) +2*(1/3) =4/3

E(Y)=1*(1/6)+2*(5/6) =11/6

d)

E(XY) =1*(1/12) +2*(1/12) +2*(7/12) +4*(3/12) = (1/12)+(2/12)+(14/12)+(12/12) =29/12

Cov(X,Y)=E(XY)-E(X)E(Y)

=(29/12) -(4/3)*(11/6)

=(29/12) -22/9

=-3/108

=-1/36

YX 1 2 1 1/12 1/12 2 7/12 3/12

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.