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17 votes
2 answers
5k views

Conditional expectation equals random variable almost sure

Let $X$ be in $\mathfrak{L}^1(\Omega,\mathfrak{F},P)$ and $\mathfrak{G}\subset \mathfrak{F}$. Prove that if $X$ and $E(X|\mathfrak{G})$ have same distribution, then they are equal almost surely. I ...
Marc's user avatar
  • 2,094
14 votes
1 answer
12k views

Independence and conditional expectation

So, it's pretty clear that for independent $X,Y\in L_1(P)$ (with $E(X|Y)=E(X|\sigma(Y))$), we have $E(X|Y)=E(X)$. It is also quite easy to construct an example (for instance, $X=Y=1$) which shows that ...
user73048's user avatar
  • 299
2 votes
0 answers
420 views

If $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable, then $\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$

I'm reading a proposition given without proof in this note. Proposition 12.4. Let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}, X, Y$ be two random variables such that $X$ is independent of $...
Analyst's user avatar
  • 5,817
11 votes
1 answer
1k views

Does almost sure convergence and $L^1$-convergence imply almost sure convergence of the conditional expectation?

Question. Let $ X_{n}, X $ be random variables on some probability space $ ( \Omega, \mathcal{F},\mathbb{P} ) $ and let $ \mathcal{G} \subset \mathcal{F} $ be a sub-$\sigma$-algebra. Moreover ...
Pass Stoneke's user avatar
5 votes
0 answers
271 views

Sigma algebra generated by a homeomorphic random variable

Let $\Omega = [0,1]$ be our probability space with sigma algebra of borel sets on $[0,1]$ and Lebesgue measure on $[0,1]$. Let Y be a random variable such that $Y(\omega) = Y(1-\omega)$ for every $\...
Hagrid's user avatar
  • 2,601
5 votes
1 answer
973 views

Compute the conditional expectation $E(Y|X)$ for a measurable function $Y$ and a random variable $X$ taking values on $[0,1)$

Good day, Currently I am working with "Probability: Theory and Examples" by Durrett and while getting familiar with conditional expectations I got to this problem: Consider the Lebesgue ...
Cahn's user avatar
  • 4,483
3 votes
1 answer
425 views

$X = E(Y | \sigma(X)) $ and $Y = E(X | \sigma(Y))$

Suppose $X, Y$ are random variables in $L^2$ such that $$X = E(Y | \sigma(X)) $$ $$Y = E(X | \sigma(Y))$$ Then I want to show that $X=Y$ almost everywhere. What I've done: By conditional Jensen $$...
WLOG's user avatar
  • 11.5k
2 votes
1 answer
60 views

"co-relatedness" of conditional expectation of two independent random variables

Let $\mathcal{F}$ be a $\sigma$-algebra and $X,Y$ be two independent (not just uncorrelated) random variables, I wonder if the following statement true $$\mathbb E(XY|\mathcal{F})=\mathbb E(X|\...
No One's user avatar
  • 8,039
1 vote
1 answer
117 views

Elementary explanation of getting two consecutive $6$'s in a die roll experiment

I know that there are already numerous questions that adress this problem. However, I am not interested in a soltuion at all but in an explanation of a particular solution (see https://math....
Philipp's user avatar
  • 4,564
1 vote
1 answer
2k views

Conditional expectation in Poisson point process

Considering a Poisson process with parameter $\lambda$, let $N(t_2)$ denote the number of events in $(0,t_2]$ and $N(t_1, t_3)$ denote the number of events in $(t_1,t_3]$, under the assumption that $0&...
Orest Xherija's user avatar
0 votes
1 answer
134 views

For $X,Y $ random variables, $h $ a function, show that $E (Xh(Y)|Y)=h (Y)E (X|Y) $ almost surely

Question in the title: For $X,Y $ random variables, $h $ a function, show that $E (Xh(Y)|Y)=h (Y)E (X|Y) $ almost surely My main problem is that I don't even understand what $E (Xh(Y)|Y)$ means... ...
user2520938's user avatar
  • 7,305
0 votes
0 answers
661 views

A question on conditional expectation leading to zero covariance and vice versa

In my probability class I was tackled with this seemingly weird question involving conditional expectation: Let X,Y be two random variables (it is not mentioned whether or not they are discrete or ...
Croc2Alpha's user avatar
  • 3,847