Question

@satellite73, last one. Find E(X|Y=y)

Answer 1

Let (X,Y) have joint mass function as seen in the attachment. Find E(X|Y=y)

Answer 2

now you are making me think

Answer 3

Yes, this is the last remaining problem in my set :/

Answer 4

\[E(X|Y=y)=\sum_xxF_{x|y}(x|y)\] if i recall

Answer 5

unless you have a different formula to use

Answer 6

no that's correct. I've been doing that here as well..
http://openstudy.com/users/agent47#/updates/56c9602ee4b09397a170926f

Answer 7

link is not working

Answer 8

you got an example?

Answer 9

Does this help?

Answer 10

That was for this question

Answer 11

yeah but i have no idea how to apply it in your case here

Answer 12

I think we need to find the general formula for
P(X|Y=n) for n = 1, 2, .. k
What I've been doing so far was just plugging in numbers and trying to catch a pattern

Answer 13

the only thing is see in your example is that the final answer should be \[\frac{y+1}{2}\]

Answer 14

i mean the example you already did, not this one here

Answer 15

Hang on, sorry I may have missed something, and yes that's what i simplified my earlier example to.

Answer 16

That's the whole example, the question is asking me to compute E(X|Y=y) for that example.

Answer 17

yikes
can you show me what \[E(X|Y=1)\] is?

Answer 18

sec, let me try computing it..

Answer 19

i guess what i am asking is, do we replace n by 1 and sum over k , or what?

Answer 20

Isn't\[P(X=k|Y=n)\] just 0 when n > k and what they have when n <=k ?

Answer 21

my problem is that to be honest i don't know how to read this

Answer 22

To \(k,n\) correspond to \(X,Y\)?

Answer 23

So given a joint pmf, find E(X|Y=y).. to make this somewhat simpler for myself I can just say E(X=k|Y=n) maybe?\[P(X=k|Y=n)=C*2^{-k}/n, \space n \le k \space and \space 0 \space otherwise\]
@wio i think so

Answer 24

And is \(E\) expected value?

Answer 25

\[E(X=k|Y=n)=\sum_{k=1}^{\infty}k*P(X=k|Y=n)\]@wio yes

Answer 26

just dug out my copy of Ross and that is exactly what it says

Answer 27

Yea it's a screenshot from A Natural intro to probability

Answer 28

so now my only question is what is \[P(X=k|Y=n)\] in this case?

Answer 29

\[P(X=k \cap Y=n)/P(Y=n)\]

Answer 30

I think the top is already given, we just need to find P(Y=n)

Answer 31

In that case, my initial thought is: \[
E(X|Y=y)=\sum_{k=1}^{\infty}k\cdot p(k,y)
\]

Answer 32

\[P(X=k|Y=n)=\frac{C2^{-k}}{n} \space when \space n \le k \space and \space 0 \space otherwise\]right?

Answer 33

So do we split these into two sums?
k=1 to n and k=n to inf?

Answer 34

You can start \(k\) at \(y\):\[
E(X|Y=y)=\sum_{k=y}^{\infty}k\cdot p(k,y)
\]

Answer 35

This insures that \(n\) is never larger.

Answer 36

good luck, i gotta go
sorry i can't be more help here

Answer 37

no problem satellite, thanks for trying!

Answer 38

@wio so is that sum just:\[\sum_{k=y}^{\infty}{k*\frac{C*2^{-k}}{n}}=\frac{C}{n}\sum_{k=y}^{\infty}{k*2^{-k}}\]

Answer 39

Well, the \(n=y\).

Answer 40

Oh yea you're right. Also my professor mentioned that the constant C cancels out when you compute conditional expectation.

Answer 41

My reasoning was this: \[
\Pr(X=k) = p(k, y)
\]Then I used expected value definition on that.

Answer 42

Ah I see, So anyway we're left with\[\frac{C}{y}\sum_{k=y}^{\infty}\frac{k}{2^k}\].. Sums are not my strongest suit..

Answer 43

http://www.wolframalpha.com/input/?i=sum(k%2F2%5Ek,+k,+y,+infinity)

Answer 44

Is that it?

Answer 45

I think so, it's a rather obscure sum.

Answer 46

So what we have is:\[E(X|Y=y)=\sum_{k=y}^{\infty}k\cdot p(k,y)=\sum_{k=y}^{\infty}{k \cdot \frac{C \cdot 2^{-k}}{y}}=\]\[=\frac{C}{y}\sum_{k=y}^{\infty}{\frac{k}{2^k}}\] which equals to whatever wolframalpha came up with

Answer 47

Yeah, I think that looks good to me.

Answer 48

Cool, thanks wio. Also dividing that sum by y results in a rather nice-ish result on wolframalpha:\[\frac{2^{1-y}(y+1)}{y}\] http://www.wolframalpha.com/input/?i=(sum(k%2F2%5Ek,+k,+y,+infinity))%2Fy

Answer 49

So the end result ends up being C times that ^

Answer 50

It's be great if there were a way to test the solution though.

Answer 51

Yea, unfortunately I dont have the answer key :/