Thread: Conditional Expectation of a joint PMF

1. Conditional Expectation of a joint PMF

(please refer to attached image)
The question appears to be simple enough, but i have two queries

A) does E[X1 X2] mean the same as E[X1 | X2]

B) If not/so, how exactly do I go about computing this. I've seen a few formulas in my lectures notes for computing conditional expectations for discrete random variables,
however I find it difficult to understand and apply the notation/procedure.

Any help is appreciated!

edit: ok, after some more research, i've found that
E(X1 X2] simply means The expectations of X1 and X2 multiplied by each other.

so, what I want to ask now is this.
is the PMF of X1, given that table:

X1 | -1 | 0 | 1 |
px(X1)| 1/3 | 0 | 1/3 |

And finally, how do i find out if X1 and X2 are independent?

2. Re: Conditional Expectation of a joint PMF

Hey 99.95.

For a) the answer is no. E[X1*X2] is not the same as E[X1|X2]. E[X1X2] = Integral over region for (X1,X2) x1x2*f(x1,x2)dx1dx2 whereas E[X1|X2=x2] = Integral x1*f(x1|X2=x2)dx1.

X1 and X2 are independent if for all values of X1 and X2 you have the joint distribution to be separable. In other words:

P(X1 = x1, X2 = x2) = P(X1 = x2)P(X2 = x2) for all x1 and x2.

To compute conditional expectations you need to find the conditional distribution. To do this use the relationship P(A|B) = P(A and B)/P(B) and then integrate with respect to the mean of A.

To get you started on the table exercise, note that

E[X1X2] = -1*-1*P(X1=-1,X2=-1) + 0*-1*P(X1=0,X2=-1) + ... + 1*1*P(X1=1,X2=1).

3. Re: Conditional Expectation of a joint PMF

thanks chiro,
i forgot to update but yes i figured that out. Although I don't understand what I am supposed to integrate over? or is that when i'm given a PDF instead of in tabular form.

i found E[X1X2] = 0, E[Y1Y2]=0 and E[Z1Z2] = -2/3

I also just wanted to ask, instead of computing the E[X1X2] and E[Y1Y2] by -1*-1*P(X1=-1,X2=-1)+...+
is there some way to say, 'it is symmetrical around 0, so E[X1X2] = 0?