For every question related to the concept of conditional expectation of a random variable with respect to a $\sigma$-algebra. It should be used with the tag (probability-theory) or (probability), and other ones if needed.
Questions tagged [conditional-expectation]
4197 questions
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votes
1 answer
Conditional Expectation $E(X\mid X+Y=k)$ when Joint pdf is given.
$$f(x,y)=4 e^{-2(x+y)}$$ with $x,y > 0$.
What would be $$E(X\mid X+Y= 4), ?$$
I tried by putting Y=$4$-X and then using usual formula of conditional expectation but that doesn't seem to be correct. Do we have to use transformation? I know here $X$…
Ankita Goyal
- 103
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votes
2 answers
Given joint pdf $f(x,y,z) = \frac{2}{3}(x+y+z)$, how to find $E(X+Y|Z)$?
As the title says, how do you find $E(X+Y|Z)$ given $f(x,y,z) = \frac{2}{3}(x+y+z)$?
Not sure how to go about approaching this problem. The furthest I got was:
$$
E(X+Y|Z) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x+y)f(x+y|z)dxdy
$$
And now…
doctopus
- 491
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votes
1 answer
How to prove below?
Given
\begin{split}
A =~& \theta(X) \cdot D + g(X) + \epsilon ~~~&~~~ \mathbf E[\epsilon | X] = 0 \\
D =~& f(X) + \eta & \mathbf E[\eta \mid X] = 0 \\
~& \mathbf E[\eta \cdot \epsilon | X] = 0
\end{split}
How to prove below?
\begin{split}
A -…
Chandra
- 101
-1
votes
1 answer
Proof involving conditional expectation
I am trying to prove or disprove the following: $\mathbb{E}[\mathbb{E}(X\mid Y)^2]-\mathbb{E}[X \mathbb{E}(X\mid Y)]=0$. Using formulas for covariance or something like that just gives back the same problem.
AsMek
- 185
-1
votes
1 answer
Show conditional independance
As part of my excercise i have to show that X and Y are conditionally independent under the condition $(\lfloor X\rfloor,\lfloor Y\rfloor)$
where $X,Y \sim exp(\lambda)$
I previously computed $E[X|\lfloor X\rfloor]$ and I know that I have to show…
Max
- 923
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- 18
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votes
1 answer
$\mathbb E[X_i\mid X_1,...,X_n]=\mathbb E[X_i]$
Do you agree that if $(X_i)$ is a sequence of i.d.d. random variable, then for all $i$
$$\mathbb E[X_i\mid X_1,...,X_n]=\mathbb E[X_i]\ \ \ ?$$
idm
- 11,824