Question

Conditional expectation help please. Just checking my solution.

Answer 1

We roll 2 fair dice, find the joint pmf of X and Y where X is the value on first die, and Y is the largest roll of the two (joint pmf attached)

Answer 2

For part 2 I need to find \[E(Y|X=x)=\sum_{y}{y*P(y|X=x)}=\sum_{j=1}^{6}{j*P(j|X=x)}\]

Answer 3

I mean I need to find E(Y|X=x), and that's what I'm doing so far, is that correct?\[=1*1/36 + 2*3/36+ 3*5/36+4*7/36+5*9/36+6*11/36\]

Answer 4

And I just want to know if my joint pdf is correct and whether I'm taking the right approach for E(Y|X=x)

Answer 5

E(Y|X=x) is a function of x

you assuming x=1 for some reason

Answer 6

Hmm I thought it was:
P(Y|X=x) (meaning X can be any of xs):
if Y=1, X must be 1 (meaning both rolls are 1), so:
P(Y=1|X=x) = (1/36 + 0 + 0 + 0 + 0 + 0)/P(X=x) = (1/36)/(1) = 1/36

Answer 7

no

Answer 8

Ohhh is it:\[=1*x/36 + 2*3*x/36+ 3*5*x/36+4*7*x/36+5*9*x/36+6*11*x/36\]

Answer 9

I'm confused now, can you walk through this with me please, I just need one worked out example and I can't seem to find anything similar online, they all have X=2 or something. So:
\[P(X=x) = 1 \space because \space x \space can \space be \space whatever\]\[P(1|X=x)=P(1 \cap X=x)/P(X=x)\]\[\]

Answer 10

@ganeshie8 @Zarkon
Sorry guys I'm stuck, from the graph
\[P(1|X=x)=x*1/36\]?

Answer 11

@mathmale or @AravindG any ideas?

Answer 12

I just can't seem to grasp the intuition behind this.

Answer 13

dice

Answer 14

for 7

Answer 15

when u roll a 6 i mean lool

Answer 16

oops!

Answer 17

redoing it haha

Answer 18

haha xD

Answer 19

okay so 6 of 6s

Answer 20

5 of 6s i mean

Answer 21

Okay wait, what do you mean by 5 of 6? That a die lands on a 5?