3 votes
Accepted

Intuitively, why does the conditional expectation of X given the trivial sigma algebra equals the expected value of X itself?

Let $F_0=\text{{$\emptyset, \Omega$}}$ be the trivial sigma algebra for a probability space $(\Omega, F, P)$. Let $X$ be some integrable random variable defined on $(\Omega, F, P)$. Intuition that can ...
ShaftSinker's user avatar
3 votes

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

Your idea can work as follows. For $i\in \{2,3,\dots \}$, we can show that for $\color{blue}{t\in (0,1)}$: $$\color{blue}{F_{X_{N}|N=i}(t)=1-\left ( 1+ \frac{1}{i-1} \right ) (1-t) +\frac{1}{i-1} (1-t)...
Amir's user avatar
  • 4,784
3 votes
Accepted

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

This step is wrong: $$\int_0^1\mathbb P[X\leq t\cap X> 1-Y|Y<1, Y=y](i-1)y^{i-2}\ \text dy= \int_0^1(t-1+y)(i-1)y^{i-2}\ \text dy$$ You can't say $\mathbb P[X\leq t\cap X> 1-Y|Y<1, Y=y]=t-...
Noble Mushtak's user avatar
3 votes

Is it true that $E[X|Y] = \rho \frac{\sigma_X}{\sigma_Y} Y$ if $X$ and $Y$ are not jointly gaussian?

Here is a counterexample when $X$ and $Y$ are centred Gaussian random variables: Let $f$ be the inverse of the cdf of the standard Gaussian distribution and $U$ be uniformly distributed on $(0,1)$, so ...
Will's user avatar
  • 6,972
2 votes
Accepted

Definition of Left-Closable Martingale

I haven't run into this terminology before, but here is my guess. The term closed probably refers to the set of indices $n$ where the martingale is defined. That is, consider the domain $$ \{ n : X_{n}...
Jacob Maibach's user avatar
2 votes

Is it true that $E[X|Y] = \rho \frac{\sigma_X}{\sigma_Y} Y$ if $X$ and $Y$ are not jointly gaussian?

Let $X,Z$ be iid standard Gaussian. Let $Y= |Z| \, sgn(X)$. This forces $X$ and $Y$ to have same sign; in terms of the joint density function, it amounts to erasing two quadrants and multiply by two ...
leonbloy's user avatar
  • 63.6k
2 votes

Inequality regarding expectations

It follows immediately by linearity. Specifically, we know that $$E(a X + b Y | \mathcal{H}) = a E(X| \mathcal{H}) + b E(X|\mathcal{H}).$$ Now, observe that $$|E(X|\mathcal{H})| = |E(X^+|\mathcal{H}) -...
Oscar's user avatar
  • 866
2 votes
Accepted

Alternative Proof of Hoeffding's Lemma / solve the equation $E[1_A | X] = (1+X)/2$ given $X$

You totally have the right idea with your ansatz for $Y$ ! Indeed, when looking at the equations you obtained for $P(A\mid X)$ and $P(B\mid X)$, one can see that both expressions look suspiciously ...
Stratos supports the strike's user avatar
2 votes

$E[X|X^2+Y^2] = 0$ when $X$ and $Y$ are independent standard normals.

Let $A$ be a Borel set in $\mathbb R$. Then, $\int_ {(X^{2}+Y^{2}\in A)}XdP=\int_ {(x^{2}+y^{2}\in A)}xdF_{X,Y} (x,y)$ and $\int_ {(X^{2}+Y^{2}\in A)}(-X)dP=\int_ {(x^{2}+y^{2}\in A)}xdF_{-X,Y} (x,y)$....
geetha290krm's user avatar
  • 37.1k
1 vote

Exercise 7.14 First course in probability

Here is the outline for my intuitive way to get the answer : Number of stages to get $m$ White Balls is $S(m)$ Then we get $S(m)=S(1)+S(m-1)$ That will then give $S(m)=S(1)+S(1)+S(m-2)$ Eventually we ...
Prem's user avatar
  • 9,859
1 vote

Finding the expectation of Gaussian a RV conditioned on affine transformation of jointly Gaussian Vector

The doctrine is the following: if $(X,Y)$ is Gaussian with covariance $$\left[\begin{array}{cc}A&B^T\\B&C\end{array}\right]$$ such that $C^{-1}$ exists, then $X|Y$ is Gaussian with expectation ...
Letac Gérard's user avatar
1 vote

Definition of Left-Closable Martingale

To build on Jacob Maibach's answer on the positive integers being open on the right, later on in Resnick's book he discusses reversed/backwards martingales: given a DECREASING family of $\sigma$-...
picklechu's user avatar
1 vote

Expectation of the indicator function

$\{k\}$ is not independent of $\mathcal F_n$. Since $\mathcal F_n$ is generated by a partition, $$ \mathbb E\left[X_{n+1}\mid\mathcal F_n\right]=\sum_{k=1}^n \mathbb E\left[X_{n+1}\mathbf{1}_{\{k\}}\...
Davide Giraudo's user avatar
1 vote

Expectation of the squared conditional expectation

You are going wrong on the fourth line. You cannot distribute the second conditional expectation outside the scope of the outer integral. Both are conditioned on the same variable $(x)$. $\quad\...
Graham Kemp's user avatar
1 vote
Accepted

joint distribution and conditional expectation

Suppose $X$ and $Y$ are jointly normal; then $$Z = E(X | \sigma (Y)) = \mu_X + \frac{\sigma_X}{\sigma_Y}\rho (Y - \mu_Y)$$ which is readily seen to be normally distributed as well, and the vector $(Z,...
Jose Avilez's user avatar
  • 12.9k
1 vote

Joint distribution of two conditional distributions

If $Y$ and $Z$ are conditionally independent when given $X$, then $$\begin{align}\mathsf P_{X,Y\mid Z=z}(x,y\mid z)=\mathsf P_{Y\mid X=x}(y)\,\mathsf P_{X\mid Z=z}(x)\\\mathsf P_{Y\mid Z=z_0}\!(E)=\...
Graham Kemp's user avatar
1 vote

Calculating $\mathbb{E}[2\sin (\pi Z)|\cos (\pi Z)]$ when $Z$ is uniform on $[0,2]$

I have an upcoming exam, so I wanted to tackle this question as an exercise. We use the second approach suggested by the accepted answer. Let $X=\cos(\pi Z)$ and $Y=\sin(\pi Z)$ where $Z \sim \text{...
Ata Keskin's user avatar
1 vote

Expected number of die rolls to get 6 given that all rolls are even.

The probability of rolling a 6 is $\frac16$ and of a 2 or 4 is $\frac26$. The conditional expectation is therefore $$\dfrac{1\times \frac16+2 \times \frac16\left(\frac26\right)^1 +3 \times \frac16\...
Henry's user avatar
  • 157k

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