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6 votes
2 answers
2k views

How does conditional expectation allow us to condition on zero-probability events?

Can someone explain and provide example(s) of where we need to appeal/use conditional expectation to determine a conditional probability given a zero-probability event has occurred? I've seen books ...
user avatar
6 votes
4 answers
312 views

$E(X \mid X > x)$ is increasing in $x$. Why?

For two points $x < x'$ and a random variable $X$, we must have $E(X\mid X > x )\leq E(X\mid X > x' )$. This is "obviously" true because the center of the truncated distribution ...
Galton's user avatar
  • 675
6 votes
1 answer
707 views

Application of Doob's inequality

Suppose that $X_n$ is a martingale with $X_0 = 0$ and $EX^2_n < \infty$. Show that $$P\left(\max_{1\leq m \leq n} {X_m} \geq \lambda\right) \leq \frac{EX^2_n}{EX^2_n+\lambda^2}$$ by using the ...
Sam Gates's user avatar
6 votes
1 answer
1k views

If $E(X\mid Y)=Z$, $E(Y\mid Z)=X$ and $E(Z\mid X)=Y$, then $X=Y=Z$ almost surely

Let $X$, $Y$, and $Z$ be three random variables in $L^{1}(\Omega, \mathcal{F}, P)$. Suppose that we have $E(X|Y)=Z$, $E(Y|Z)=X$, and $E(Z|X)=Y$. Show that $X=Y=Z$ a.s. I am able to show this in $L^{2}...
user avatar
6 votes
2 answers
143 views

A question on conditional probability in sde

Let $s \in[a, b]$ and $x \in \mathbb{R}$ be fixed and consider the following SIE: $$ X_{t}=x+\int_{s}^{t} \sigma\left(u, X_{u}\right) d B(u)+\int_{s}^{t} f\left(u, X_{u}\right) d u, \quad s \leq t \...
Talking Puppet's user avatar
6 votes
1 answer
245 views

If $M$ is a martingale and $σ_n↓$ and $τ_n↓τ$, then $\text E[M_{τ_n}\mid\mathcal F_{σ_n}]\xrightarrow{n→∞}\text E[M_τ\mid\mathcal F_σ]$

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$ $M$ be a $\operatorname P$-almost surely right-continuous $(\operatorname P,\...
0xbadf00d's user avatar
  • 13.9k
6 votes
2 answers
915 views

Is conditional entropy ever taken to be a random variable?

In probability theory, the conditional expectation $E(X|Y)$ and variance $V(X|Y)$ er usually taken to be random variables, st. the value of $E(X|Y)$ depends on what value $Y$ ends up taking. I've ...
Thomas Ahle's user avatar
  • 4,814
6 votes
1 answer
2k views

Change the order of conditional expectation of integration

I encountered this problem when learning SDE: $g(t,\omega)$ is a adapted process then $$\mathbb E\left(\int_a^b |g(t)|^2 \, dt \mid \mathcal F_a\right)=\int_a^b\mathbb E\left(|g(t)|^2\mid \mathcal ...
Lookout's user avatar
  • 2,161
6 votes
1 answer
89 views

Unusual behavior in a conditional expectation

Show an example of random variables $X$ and $Y$ such that $X$ and $Y$ are not independent but still $$\textbf{E}(X\mid Y) = \textbf{E}X$$ I tried looking at the simplest discrete probability ...
tosi3k's user avatar
  • 654
6 votes
1 answer
332 views

Limit and conditional expectation commute in a uniformly integrable sequence

I am thinking of the next proposition: Proposition. Let $(\Omega, \mathcal{F}, P)$ a probability space, and $\{X_n\}_{n=1,\cdots}$ a uniformly integrable r.v. sequence s.t. $X_n \rightarrow X ~ \text{...
mathmrk's user avatar
  • 183
6 votes
1 answer
243 views

If $X$ is independent of $Y$, is it true that $E(f(X)|Y) = E(f(X))$ for any function $f$?

If $X$ is independent of $Y$, is it true that $E(f(X)|Y) = E(f(X))$ for any function $f$? I cannot think of a specific example but am not able to prove it either. The best I have been able to do so ...
user123276's user avatar
  • 3,465
6 votes
2 answers
491 views

Understanding conditional expectation using measure-theoretical definition

Definition: For a random variables $X\in\mathbb R^{d_1}$ and $Y\in\mathbb R^{d_2}$, we define a conditional expectation of $X$ given $Y$ by any random variable $Z$ satisfying: there exists $g:\mathbb ...
WhyMeasureTheory's user avatar
6 votes
1 answer
61 views

Does independence of random variables imply independence of conditional expectations?

Does $X\perp\!\!\!\perp Y\implies\mathbb{E}[Z|X]\perp\!\!\!\perp\mathbb{E}[Z|Y]$ for any r.v. Z? I would think so, because $X\perp\!\!\!\perp Y$ implies $\sigma(X)\perp\!\!\!\perp\sigma(Y)$ and $\...
abeeisnotabug's user avatar
6 votes
1 answer
272 views

Proof that Expected Lifetime is longer than Remaining Lifetime if the Hazard Rate is increasing.

Let $X$ be a positive, continuous random variable. Denote the density of $X$ by $f(x)$ and its CDF by $F(x)$. Let $\bar{F}(x) = 1- F(x)$ be the survival function of $X$. Given that the Hazard Rate, \...
Mads Hulgaard's user avatar
6 votes
1 answer
1k views

Why are these two definitions of Markov property equivalent?

Question Suppose that $S$ is a finite or a countable subset of $\mathbb R$ and $(\xi_n)_{n\in\mathbb N}$ is an $S$-valued sequence of random variables. Then are these two definitions of Markov ...
zxcv's user avatar
  • 1,499

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