This is a joint PDF question. (c) Draw up a table showing the joint probability function Px,y(x,y) of X and Y.

a random variable Y has probability function P(Y=2)= 0.6, P(Y=3)=0.2 and P(Y=4)=0.2.


once Y has been observed, Y fair coins are tossed; let X denote the number of heads which are observed.





(a) Write downt the conditional expectation of X given Y= y, for each y=2,3,4.


hence express E[X l Y] as a function of Y.


(b) Find E(Y) and use this to calculate E(X).


(c) Draw up a table showing the joint probability function Px,y(x,y) of X and Y.


(d) find the marginal probability function of X and verify that the expectation of this distribution agrees with the value calculated in (b)

This is a joint PDF question. (c) Draw up a table showing the joint probability function Px,y(x,y) of X and Y.
.. X ....... 0 ................. 1 ................ 2 ................ 3 ............... 4


Y=2 . (.25)(.6) ...... (.5)(.6) ........ (.25)(.6) ............ 0 .............. 0


3 .... (.125)(.2) .... (.375)(.2) ...... (.375)(.2) ..... (.125)(.2) ...... 0


4 ... (.0625)(.2) ..... (.25)(.2) ... (.375)(.2) . (.25)(.2) . (.0625)(.2)








If y = 2, 0 head: 1/4 , 1 head: 2/4, 2 heads: 1/4 (then multiply 0.6)





If y = 3, 0 H: 1/8 , 1H: 3/8 , 2H: 3/8 , 3H: 1/8





If y = 4: 0H: 1/16, 1H: 4/16 , 2H: 6/16 3H: 4/16 4H: 1/16





Also, E[Y] = 2*.6 + 3*.2 + 4*.2 = 2.6


I had now given you the table. Multiply to get the values and use that to determine the rest of the answers. ©
Reply:a)





E[X|Y=2] = 0.5*2 = 1


E[X|Y=3] = 0.5*3 = 1.5


E[X|Y=4] = 0.5*4 = 2


E[X|Y=y] = 0.5*y





b)





E[Y] = 2*0.6 + 3*0.2 + 4*0.2 = 2.6





E[X] = E[ E[ X|Y ] ] = Sum over all y of E[X|Y=y]P[Y=y]


= 1 * 0.6 + 1.5 * 0.2 + 2 * 0.2 = 1.3





c)


These are all found using the binomial distribution and mulitplied by P[Y=y]





Y = y | 2_________3___________4_______total P[X= x]





X = 0 | 0.25*0.6____0.125*0.2____0.0625*0.2


X = 1 | 0.5*0.6_____0.375*0.2____0.25*0.2


X = 2 | 0.25*0.6____0.375*0.2_____0.375*0.2


X = 3 | 0_________0.125*0.2_____0.25*0.2


X = 4 | 0_________0___________0.0625*0.2





total P[Y=y]





to finish the table sum the columns to get P[Y=y] and sum the rows for P[X=x]