# Conditional expectation of joint PMF

#### nacho

##### Active member
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?

EDIT 2: okay, is this correct

for E[X1 X2]
i do:

(-1)(-1)*(1/6) + ....

That is multiply each (X1,X2) and then multiply that by the probability of its occurrence, and add them all up?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
okay, is this correct

for E[X1 X2]
i do:

(-1)(-1)*(1/6) + ....

That is multiply each (X1,X2) and then multiply that by the probability of its occurrence, and add them all up?
Yes.

If not, then let me add that to find E[X1|X2=1] you find each pair (x1,x2) such that x2=1 and do:
((-1)*(1/6) + ...) / ((1/6) + ...)

#### chisigma

##### Well-known member
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!...
In formal notation $\displaystyle E [X_{1}\ X_{2}]$ means the expected value of the r.v. $\displaystyle X = X_{1}\ X_{2}$ and $\displaystyle E [X_{1} | X_{2}]$ means the expected value of the r.v. $\displaystyle X_{1}$ conditioned by the r.v. $\displaystyle X_{2}$, i.e...

$\displaystyle E [X_{1}| X_{2}] = \sum x_{1} P \{X_{1}=x_{1} | X_{2} = x_{2}\}$

Kind regards

$\chi$ $\sigma$