I Expected value of X~Geom(p) given X+Y=z

  • Thread starter Thread starter BerriesAndCream
  • Start date Start date
  • Tags Tags
    Geometric Value
Click For Summary
The discussion centers on calculating the expected value E[X|Z=z] for independent geometric random variables X and Y, where Z=X+Y. The user derives that P(X=x|Z=z) equals 1/(z-1) and attempts to find E[X|Z=z] using this distribution. They conclude that E[X|Z=z] equals z/2, confirming the validity of their calculations. The conversation also touches on whether E[X|Z=z] or E[Y|Z=z] is larger, with the implication that both expected values are equal due to the uniform distribution of P(X=x|Z=z). The thread emphasizes the use of formulas for expected values in discrete uniform distributions for such calculations.
BerriesAndCream
Messages
7
Reaction score
2
TL;DR
finding a conditioned expected value
Hello everyone.

If X, Y are two independent geometric random variables of parameter p, and Z=X+Y, what's E[X|Z=z]?

I have calculated the distribution of P(Z=z) and I have then found that the conditional probability P(X=x|Z=z) equals 1/(z-1).
How can I now find the conditioned expected value?
 
Physics news on Phys.org
What is the value of ##E[X|Z=z]+E[Y|Z=z]##?

Which is the bigger of ##E[X|Z=z]## and ##E[Y|Z=z]## (trick question).
 
andrewkirk said:
What is the value of ##E[X|Z=z]+E[Y|Z=z]##?

Which is the bigger of ##E[X|Z=z]## and ##E[Y|Z=z]## (trick question).
1. E[X|Z=z] + E[Y|Z=z] = E[X+Y|Z=z] = E[Z|Z=z] = z?
2. Is there a way of knowing?
 
Could it be that since P(X=x|Z=z) = 1/(z-1), which is an uniform distribution, the expected value I'm looking for can simply be found by using the formula for the expected value of the uniform distribution?

edit: I tried to do it that way, but I couldn't find the right answer. So i tried by using the formula for the expected value. What I did not realise at first is that I had to sum from x=1 to x=z-1, right? By doing so I get that E[X|Z=z]=
Σx=1z-1 x·P(X=x|Z=z) =
Σx=1z-1 x/(z-1) = 1/(z-1)·Σx=1z-1 x =
1/(z-1) · (z-1)(z-1+1)/2 = z/2.

Is this valid?
 
Last edited:
BerriesAndCream said:
By doing so I get that E[X|Z=z]=
Σx=1z-1 x·P(X=x|Z=z) =
Σx=1z-1 x/(z-1) = 1/(z-1)·Σx=1z-1 x =
1/(z-1) · (z-1)(z-1+1)/2 = z/2.

Is this valid?
Yes it is.
There is another approach as described by @andrewkirk in the post #2. The first question is answered in the post #3. What about the trick question? Is ## E [X|Z=z] ## bigger than ## E [Y|Z=z] ##? or Is ## E [Y|Z=z] ## bigger than ## E [X|Z=z] ##? or What is the third option?
 
BerriesAndCream said:
Could it be that since P(X=x|Z=z) = 1/(z-1), which is an uniform distribution, the expected value I'm looking for can simply be found by using the formula for the expected value of the uniform distribution?
Yes, it can be found by using a formula for calculating an expected value of a discrete uniform distribution.
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K