# Finding conditional variance?

1. Sep 28, 2005

### grimster

the discrete prob distribution

X/Y - G - D
0 - 0,1 - 0,15
1 - 0,1 - 0,3
2 - 0,05 - 0,3

this is what i have so far:
E[X|Y=D]=0,2
E[X|Y=g]=0,9
E[X]=0,725
E[X^2|Y=D]=0,3
E[X^2|Y=G]=1,5
Var(X|Y=G)=0,69
Var(X|Y=D)=0,26

i.e. [X]=0,2*0,25 + 0,9*0,75=0,725

is the previous correct and how do i find Var(X)?
the conditional variance forumal is:
Var(X)=E[Var(X|Y)] + Var(E[X|Y])

2. Sep 28, 2005

### Millie

I am not sure but it seems easier to calculate Var[X]= E[X^2]- (E[X])^2
you already have E[X]. you can try to calculate E[X^2] by conditioning on Y.

3. Sep 28, 2005

### grimster

how do i do that?

secondly, i think i've might have made a mistake. i think i forgot to calculate the conditional probability mass function.

that would mean distribution, given Y=G) is i.e.
0 - 0,2
1 - 0,4
2 - 04

so then E[X|Y=G]=0,4+0,8=1,2

what is it? 1,2 or 0,9? i think the former 'result' is correct. so forget the other results, these are the corrected(?) ones

E[X|Y=G]=1,2
E[X|Y=D]=0,8
E[X^2|Y=G]=2
E[X^2|Y=D]=1,2
Var(X|Y=G)=0,56
Var(X|Y=D)=0,56
E[X]=1,1

i want to find Var(X)...

4. Oct 1, 2005

### EnumaElish

Your notation is confusing -- at least for me. What does the table
stand for?