Suppose $X$, $Y$ are random variables and $E(X|Y)=Y, E(Y|X)=X$. Is this sufficient to show that $X=Y$? Or at least $X = Y$ almost surely?
52318
Assume that $X$ and $Y$ are two random variables such that $Y=E[X|Y]$ almost surely and $X= E[Y|X]$ almost surely. Prove that $X=Y$ almost surely.   The hint I was given is to evaluate: $$E[X-Y;X&gt;a,Y\leq a] + E[X-Y;X\leq a,Y\leq a]$$  which I can write as: $$\int_A(X-Y)dP +\int_B(X-Y)dP$$ where $A=\{X&gt;a, Y\leq a\}$ and $B=\{X\leq a,Y\leq a\}$.  But I need some more hints.
75564
How do I prove the following?     If $X$ and $Y$ are random variables on a probability space $(\Omega, F, \mathbb P)$, then so is $X+Y$.   The definition of a random variable is a function $X: \Omega \to \mathbb R$, with the property that $\{\omega\in\Omega: X(\omega)\leq x\}\in F$, for each $x\in\mathbb R$.    Furthermore, how to approach $X+Y$ and $\min\{X, Y\}$?
122536
Suppose $(\Omega,\mathcal F,P)$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $E[X\mid \mathcal G]$, how does one show that $X=E[X\mid \mathcal G]$ a.s.?
17237
Let $X$ be a random variable and let $g,f$ be real-valued, nondecreasing, and bounded.   Show that $\mathbb{E}[f(X)g(X)]\geq \mathbb{E}[f(X)]\mathbb{E}[g(X)]$  Having a hard time seeing where to start for some reasons, any hints?  I managed to write:  $\mathbb{E}[f(X)g(X)] = \frac{1}{4}\mathbb{E}[(f(X)+g(X))^2 - (f(X)-g(X))^2]$  But I'm not sure if this is in the right direction or not.
85891
Take any three random variables $X_1$, $X_2$, and $X_3$.   Is it possible for $X_1$ and $X_2$ to be dependent, $X_2$ and $X_3$ to be dependent, but $X_1$ and $X_3$ to be independent?  Is it possible for $X_1$ and $X_2$ to be independent, $X_2$ and $X_3$ to be independent, but $X_1$ and $X_3$ to be dependent?
67801
Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that  $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?  If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?),  I was  the  that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.  So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.  p.17  seems to suggest otherwise. idk  Help please?  Also this:    or    So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
70241
It's a really simple question. However I didn't see it in books and I tried to find the answer on the web but failed.  If I have two independent random variables, $X_1$ and $X_2$, then I define two other random variables $Y_1$ and $Y_2$, where $Y_1$ = $f_1(X_1)$ and $Y_2$ = $f_2(X_2)$.  Intuitively, $Y_1$ and $Y_2$ should be independent, and I can't find a counter example, but I am not sure. Could anyone tell me whether they are independent? Does it depend on some properties of $f_1$ and $f_2$?  Thank you.
24993
I don't really know how to start proving this question.  Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.  Show that $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$  Does anyone here have any idea for starting this question?
51873
The proof of linearity for expectation given random variables are independent is intuitive. What is the proof given there they are dependent?  Formally, $$ E(X+Y)=E(X)+E(Y)$$ where $X$ and $Y$ are dependent random variables.  The proof below assumes that $X$ and $Y$ belong to the sample space. That is, they map from the sample space to a real number line. Is that also a condition for linearity of expectation?  Proof: $$E(X+Y) =\sum\limits_{s}(X+Y)(s) P({s})    $$ $$E(X+Y) =\sum\limits_{s}(X(s)+Y(s)) P({s})    $$ $$E(X+Y) =\sum\limits_{s} X(s)P({s}+ \sum\limits_{s} Y(s)P({s})     $$ $$E(X+Y) =E(X)+E(Y)$$ Here $S$ is the sample space and $s$ is an event in the sample space.   for proof.  Also, more reasoning for step 2 would be helpful. I don't understand it completely.
78884
I am trying to simplify $E[YE[Y|X]|X]$ can I use this property:   $$E[E[Y|X]|X]= E[Y|X]$$  If yes I have never seen a Proof of this property (that seems very reasonable), could I have a reference? If I am right the correct result would be $$E[YE[Y|X]|X] = E[(Y|X)]^2$$
116686
Suppose $X_n,X$ are random variables with $X_n \overset{L^2}{\rightarrow} X$. Does this imply $X_n^2 \overset{L^1}{\rightarrow} X^2$?  $X_n \overset{L^2}{\rightarrow} X$ means $E(|X_n-X|^2) \rightarrow 0$ and what we have to show is that $E(|X_n^2 - X^2|) \rightarrow 0$.
108256
I want to prove very simple equations, but am having some trouble because I have no clue.  Definition of Conditional Probability is given as follows:  Definition $E(Y|X)$ is a conditonal expectation of $Y$ given $X$ if it is a $\sigma (X)$-measurable random variable and for any Borel set $S \subseteq R$, we have $E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})$  I want to solve those two problems based on above definition.   Suppose $X$ and $Y$ are independent.   (a) Prove that $E(Y|X)=E(Y)$ with probability 1.  (b) Prove that $Var(Y|X)=Var(Y)$ with probability 1.  (c) Explicitly verify the following theorem.(with $\mathcal{G}=\sigma (X)$ in this case)  Theorem: Let $Y$ be a random variable, and $\mathcal{G}$ a sub-$\sigma$-algebra. If $Var(Y)&lt;\infty$, then $Var(Y)=E(Var(Y|\mathcal{G}))+Var(E(Y|\mathcal{G}))$   Let X and Y be jointly defined random variables.   (a) Suppose $E(Y|X)=E(Y)$ with probability 1. Prove that $E(XY)=E(X)E(Y)$  (b) Give an example where $E(XY)=E(X)E(Y)$, but it is NOT the case that $E(Y|X)=E(Y)$ with probability 1.  1.  For this part, I think I could deal with this by myself if I understand how to prove (a). So please help me with just (a).  In order to prove that $E(Y|X)=E(Y)$ with probability 1, I need to show that  \begin{equation} E(E(Y|X)1_{X})=E(E(Y)1_{X}). \end{equation}  if $X$ and $Y$ are independent.  From the above definition, conditional expectation $E(Y|X)$ is defined such that $E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})$.  So maybe it suffices to show that $E(Y1_{X \in S})=E(E(Y)1_{X \in S})$. But it looks too obvious. My guess is that there is more educated way of showing why $E(Y1_{X \in S})=E(E(Y)1_{X \in S})$ holds if $X$ and $Y$ are independent.  2.  I can do (b) on my own. So for (a), "$E(Y|X)=E(Y)$ with probability 1" means that $E(E(Y|X)1_{X})=E(E(Y)1_{X})$. But, how can I prove $E(XY)=E(X)E(Y)$ with that information? I know how to prove this equation with integral sign or summation sign. But how does information about a.s. convergence give any implication about that equation?
2479
Im doing a course on measure theory and I'm stuck on one of the exercises.  Take $\{Y_{\gamma}:\gamma \in C\}$ as an arbitrary collection of random variables and $\{X_{n}: n \in N\}$ to be a countable collection of random variables  I now want to show that $$\sigma \{Y_{\gamma} : \gamma \in C\}=\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$$  So i need to show that both $\sigma \{Y_{\gamma} : \gamma \in C\}\subset\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$ and $\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}\subset\sigma \{Y_{\gamma} : \gamma \in C\}$ hold.  I know that for a single random variable X you have $\sigma(X)=\{X^{-1}(B): B \in Borel\}$. How do I expand this for the collection of random variables?  I then also have to show that if $Fn=\sigma\{X_{1},...,X_{n}\}$ and $A=\cup_{n=1}^{\infty}F_{n}$ then $$\sigma(A)=\sigma\{X_{n}: n \in N\}$$  Can anyone help me with this?
73965
I understand how to define conditional expectation and how to prove that it exists.  Further, I think I understand what conditional expectation means intuitively. I can also prove the tower property, that is if $X$ and $Y$ are random variables (or $Y$ a $\sigma$-field) then we have that  $$\mathbb E[X] = \mathbb{E}[\mathbb E [X | Y]].$$  My question is: What is the intuitive meaning of this? It seems quite puzzling to me.  (I could find similar questions but not this one.)
36625
If random variables $X_n$ are dominated by an integrable random variable then $E[X_n]\to E[X]$ follows if $X_n$ converges to $X$ in probability.  Proof: Take any subsequence $\{E[X_{n_k}]\}$. I need to show that it converges to $E[X]$. Now I know that $X_{n_k}$ contains a further subsequence $X_{n_k(i)} $where almost sure convergence is guranteed. and so, $E[X_{n_k(i)}]\to E[X]$. We can say upto this much. But, I have to prove $\{E[X_{n_k}]\} \to E[X]$ . How to proceed from here.
28282
Assuming that $e&lt;y&lt;x$, prove that $ x^y &lt; y^x$.  I think this must be easy, but I can't work it out. Thanks in advance for any kind of help.
9420
If $X_1,\ldots,X_n\sim Uniform(0,1)$ and $Y=\max[X_1,\ldots,X_n]$, find $E(Y)$.   I don't really have any good starting point. I guess Y is a function of the X's and so I might be able to apply the theorem about the expectation of a function, but I don't know how to deal with this since max is a discrete function.  Can anybody help me please?