5 votes
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Expectation of $X$ conditional on $aX+(1-a)Y>c$.

I don't know if you are looking for a closed and simplified solution based on the values of $a$ and $c$, but one approach for any problems in the conditional expectation is as follows: As you have ...
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  • 620
3 votes
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Calculating $E[X^2|X+Y]$ for $X,Y\sim N(0,1)$ independent

Actually $E(X^{2}|X+Y)=\frac 1 4(X+Y)^{2}+\frac 1 2$ if $X,Y$ is i.i.d. $N(0,1)$. To see this note that $E(X^{2}|X+Y)=E(Y^{2}|X+Y)$ since the joint distribution of $(X,Y)$ does not change when you ...
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  • 6,029
2 votes
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Conditional expectation with multiple random variables

It's the total law of probability: take the partition $\{C_1,C_2\}$ then $$ \mathbb{E}[N_7|T] = \sum_{i=1}^{2} \mathbb{E}[N_7|T \cap C_i] \mathbb{P}[C_1|T]$$
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  • 724
2 votes
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Conditional expectation and equality in law

Yes, they have the same distribution. More generally, a similar result holds for any measurable function $h:\mathbb{R}\rightarrow\mathbb{R}$ (you can use $h(x)=1_A(x)$ if you like). Setup: Let $X:\...
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  • 20.7k
2 votes

Calculating $E[X^2|X+Y]$ for $X,Y\sim N(0,1)$ independent

As a consequence of the formula $$\operatorname{Var}(X \: | \: X+Y) = \operatorname{E}[X^2 \: | \: X+Y] - \operatorname{E}[X \: | \: X+Y]^2,$$ we get that $$\operatorname{E}[X^2 \: | \: X+Y] = \...
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2 votes
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Two questions about conditional expectations

(1). The most general formulation of the property I am aware of states that over a $\sigma$-finite measure space $(X,\mathscr{A},\mu)$, if $\mathscr{G}\subseteq \mathscr{A}$ is a sub-$\sigma$-algebra ...
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  • 9,216
2 votes
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Why is the conditional probability distribution in terms of $\omega$?

By definition of $P(\cdot | \cdot)$, the object $\mu_{X|\mathcal{G}}(B | \mathcal{G}) \stackrel{\text{def}}{=} P(X^{-1}(B)|\mathcal{G})$ is an $\mathbb{R}$-valued $\mathcal{G}$-measurable random ...
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1 vote
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Markov Chain wrt filtration Definition?

(i) The LHS is the probability that starting at $X_0 = x$ and given $\mathcal{F}_t$, the information of the filtration up to time $t$, the chain ends up at $y$ at time $t + 1$. The RHS is the ...
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  • 1,279
1 vote

Does every $\sigma$-algebra contain a countable `disjoint' basis?

The power set of $\mathbb R$ is an easy counter-example: Every singleton set would be a union of some of the $A_i$s which means that every singleton set is one of the $A_i$'s making the collection $(...
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  • 6,029
1 vote
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Does every $\sigma$-algebra contain a countable `disjoint' basis?

The Borel sigma algebra $\mathcal B$ on $\Bbb R$ is a counterexample. If there were such a partition and $x<y$ were distinct element of $A_i$ then every measurable set would contain either both $x$ ...
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1 vote
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Proof of monotonicity of conditional expectation

$\newcommand{\field}{\mathscr{M}}$ Let $G$ be defined as you described. You cannot derive $G \notin \field$ -- in fact, because $E[Y|\field] - E[X|\field]$ is $\field$-measurable, it is certain that $...
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1 vote

Where does the randomness come in for conditional expectations $\mathbb{E}[X | \mathcal{F}]$?

I think this a legitimate question as intuition as opposed to mere manipulation of symbols is very important in mathematics in general and especially in probability theory. That said, it may be ...
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  • 660
1 vote
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Linearity in the conditional for conditional expectation

Conditioning on random variables means you condition on their generated sigma algebra, so unless there is a specific structure you are asking to decompose $\sigma(X+Z)$ which could be "many ...
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1 vote

Conditional distribution function of one random variable given the sum of two

In the second equality I obtained the density of (X,T), where T=X+Y, in this way: $f_{X,T}(x,t)=f_{X,Y}(t−y,y)=f_X(t−y)f_Y(y)$ by the independence property. Is this correct? No. You have the right ...
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1 vote
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Showing expectation of a finite sum of a sequence of random variables, squared

For each $j\ge 0$, note the event $B_j\in\mathcal F_j = \sigma(X_0,\dots, X_j)$. Assuming the condition ($\ast$) $E[X_n\mid\mathcal F_{n-1}]=0$ holds for every $n$, we have \begin{align*} E[(\sum_{i=0}...
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  • 19.7k
1 vote

Conditional expectation of X given (X, Y, Z)

If $\sigma(X)$ is the $\sigma$-algebra generated by the random variable $X$, then $$E[X| X,Y,Z] = E[X| \sigma(X,Y,Z)]$$ Now, $X$ is $\sigma(X)$ measurable, and $\sigma(X) \subset \sigma(X,Y,Z)$, so in ...
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