All Questions
135
questions
3
votes
1
answer
99
views
Show that $E[(Y-E[Y|X])*(E[Y|X]-g(X))]=0$
$g$ is a measurable function and $X$ and $Y$ are continuous random variables, we need to show that:
$E[(Y-E[Y|X])*(E[Y|X]-g(X))]=0$
My attempt:
$E[(Y-E[Y|X])*(E[Y|X]-g(X))]=E[E[(Y-E[Y|X])*(E[Y|X]-...
- 2,585
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 $$...
- 11.5k
4
votes
0
answers
230
views
Calculation of Conditional Expectation $\Bbb E[f(X)\mid Y]$
$\newcommand{\Cov}{\operatorname{Cov}}$and thank you for taking the time to read this. :)
My question is about evaluating $\Bbb E[f(X) \mid Y]$ (a random variable in $Y$). There's plenty online (and ...
- 6,775
3
votes
1
answer
556
views
A necessary and sufficient condition for symmetry of a random variable
Prove that if $X$ is an integrable random variable, it has a symmetric distribution if and only if:
$$E(X|X^2)=0$$
Can anyone check my solution?
Firstly $$EX=0$$
Then we have:
$$E(E(X|X^2))=EX=0$$ ...
- 161
1
vote
1
answer
311
views
When the two conditional expectations are independent?
Consider $X,Y$ be two independent random variable
I want to know under what sigma-algebra $\mathcal{F}$, we can say the conditional expectation $E[X|\mathcal{F}]$ is independent of $E[Y|\mathcal{F}]$
...
- 1,248
3
votes
1
answer
208
views
Conditional expectations and random vectors.
Let $(\Omega, \sigma,P)$ a probability space and $Y$ a random variable on it, with $E|Y|<\infty$. Let $X_1,X_2$ random vectors with $\sigma(Y,X_1)$ independent of $\sigma(X_2)$. The problem is to ...
- 299
3
votes
1
answer
778
views
Prove that $S_n^2-s_n^2$ is martingale
Let $(X_i)$ be iid such that $EX_i = 0$ and $\operatorname{Var}X_i = \sigma_i^2$. Let $s_n^2 = \sum_{i=1}^n \sigma_i^2$ and $S_n = \sum_{i=1}^n X_i$. Prove that $S_n^2 - s_n^2 $ is martingale.
My ...
- 2,566
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 ...
- 2,094
1
vote
1
answer
262
views
Conditional expectation almost sure
If $X_1 = X_2$ on a measurable set $B \in \mathfrak F$ then $E(X_1\mid\mathfrak F)=E(X_2\mid\mathfrak F)$ almost sure on $B$.
- 2,094
5
votes
1
answer
2k
views
Conditional expectation constant on part of partition
I have a question about conditional expectation, while looking for the answer here on stackexchange I noticed that there are a few different definitions used, so I will first give the definitions I ...
- 51
3
votes
2
answers
122
views
What is the difference between $\mathbb E[Z|\mathcal G]=Y$ and $\mathbb E[Z|\mathcal G]\stackrel{\text{a.s.}}{=}Y$?
I'm somewhat confused by the definition of martingale:
Let $(\Omega, \mathcal F, \mathcal F_n, \mathbb P)$ be a filtered probability space. We call $(X_n)_{n\in\mathbb N}$ martingale if for all $n\in\...
- 7,720
2
votes
1
answer
1k
views
Conditional expectation of symmetric Sigma algebra
Another exercise with conditional expectation that I have problems with.
Let $\Omega=[-1,1]$, $\mathcal{F}=\mathcal{B}(\Omega)$,
$\mathbb{P}=\frac{1}{2}\lambda$. Let X be a $\mathcal{F}$-...
- 411
1
vote
1
answer
48
views
Bound on variance of random process when signal is known
I am reading this paper (link to a Nature paper, may not be accessible) and I encountered the following. I have very little experience in probability theory and I could not find much helpful in ...
- 6,175
4
votes
1
answer
894
views
Why $E[X|\mathcal{G}]=X$ if $X$ is $\mathcal{G}$-measurable?
If $X$ is a $\mathcal{G}$-measurable random variable, why $E[X|\mathcal{G}] = X$? I know the intuition (basicly we're conditioning on the same informations on which $X$ is defined, $\sigma(X)$, we can'...
- 41
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 ...
- 299