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1. ## Conditional Expectation

Let X be the number of rolls of a fair die before getting a 6, and let Y be the number of rolls before the 1st even number. Find E[X / Y=5]
Can you calculate this using the memoryless property?

2. These might jog your thought process http://www.actuarialoutpost.com/actu...d.php?t=181444

3. Originally Posted by in_d_ky
Let X be the number of rolls of a fair die before getting a 6, and let Y be the number of rolls before the 1st even number. Find E[X / Y=5]
Can you calculate this using the memoryless property?
I wrote up two different solutions for this similar problem myself. Just keep in mind that 6 is even. You have to take that into consideration when you compute probabilities. If you are still having trouble,
contact me. R_T

A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y the number of rolls needed to obtain a 6. Calculate E(X given Y=2).

So I thought since the number of rolls needed to obtain a 5 is independent from the number of rolls needed to obtain a 6. And since p=1/6 I thought maybe the answer is just 1/(1/6) which the expected value of a geometric distribution. But that's not correct.The actual answer is 6.6. Can someone please tell me what I am doing wrong?

You are missing the fact that once you are told that the second toss resulted
in a 6, the expected number of tosses required to obtain your first 5 changes.

What makes you think X and Y are independent? If they were,
Pr[X=2 and Y=2] would = P[X=2]P[Y =2].
The former probability is zero; the latter is ((5/36)^2);

What follows is a solution derived from first principles; an alternate solution appears below this one.

Let Px(X) be the conditional distribution of X given Y=2
Px(1) = 5 on first roll; 6 on second =((1/6)^2)
Px(2) = 5 on second roll(impossible since Y=2) =0
Px(3) = no 5 or 6 on the first roll; 6 on the second roll; 5 on third roll
=(2/3)((1/6)^2)
Px(4) = no 5 or 6 on first roll; 6 on second roll; no 5 on third roll; 5 on the fourth roll
=(2/3)((1/6)^2)(5/6)

E[X given Y=2]
=1((1/6)^2) + 2(0) +3[(2/3)(1/6)^2] + 4[(2/3)((1/6)^2)(5/6)] +5[(2/3)((1/6)^2)((5/6)^2)] + ........../P(Y=2)

=((1/6)^2)[1 + 3(2/3) +4(2/3)(5/6) + 5(2/3)((5/6)^2) + ........../P(Y=2)
=((1/6)^2)[1+ (2/3)[3+4(5/6) +5((5/6)^2) +............../P(Y=2)
=((1/6)^2[1+ (2/3)[48]] / (5/36)
=(33/36)/(5/36) = 33/5 = 6.6

For the alternate solution, we partition the conditional expectations into two mutualy exclusive parts. The first one imposes the restriction that X<2 and Y
=2. Of course,the only way this can happen is if X=1, in which case, our conditional expected value is one, since X represents the number of rolls needed fo the first 5: P(X=1 and Y=2)/P(Y=2) = 1/5 = .2. Now the second partition considers the case where E[(X) given X > 2and Y =2)]. Because of the memoryless property of the geometric distribution, E[X] given X > 2 has the same ditribution as E[X + 2]=
E[X] +2 = 6 + 2 = 8. So this is our conditional Expectatin of X given X>2 and Y=1. The probobility for this conditional expectation is just the complement of the first probability we calculated . Now remember that the first conditional expectation is 1 and we multiply that by the associated probability. We multiply 1(.2)
to obtain our second part, we multiply 8(.8) and then add the two products: 1(.2) + 8(.8) = 6.6.

Good Luck To You

4. Since 6 is even it must occur on or after the fifth toss. In the former case we have E[X given Y= 5]Pr[X=5 and Y=5]/Pr[Y=5] =5(1/3) =(16/3)
In the latter case, we have E[X given X>5 and Y=5]Pr[X>5 and Y=5]/Pr[Y=5]. Using the memoryless property of the Geometric Distribution,
E[X given X>5] = E[X+5] = E[X] + 5 = 6 + 5 =11 and Pr[X>5 aand Y=5]/Pr[Y=5] = (2/3). So E[X given Y=5] = 5 (1/3) + 11(2/3) = 9