I am reading David Williams' "Probability with Martingales" and in this book he supposes that you are given the following information about some 'outcome' $\omega \in \Omega$: $$X_n(\omega) \ \text{s.t.} \ n\in \mathbb{N}$$ for a series of random variables $X_n:\Omega \rightarrow \mathbb{R}$. He then says, based on this information, $\sigma(X_n : n \in \mathbb{N})$ consists of 'events' $F$ such that for each point $\omega$, you can tell whether the event $F$ has occurred or not (that $\omega \in F$).  However, no matter how hard I try to wrap my head around this vague idea, I can't seem to understand it. I don't understand how the 'information' provided helps to understand whether some event has occurred or not and what this means in terms of a sigma algebra generated by random variables (which also confuses me). Is there any easier way to understand what a sigma algebra generated by these functions actually is?
123759
I've been trying to understand $\sigma$-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture.  I'll try to explain the counter-intuition I get with the classical example of : the probability space $\Omega = \{ HH, HT, TH, TT \}$ and a r.v. $X(\omega)$ equal to the number of heads.  At times $0$, $1$ and $2$ the available information is represented using $\sigma$-algebras $\mathcal{F}_0=\{\emptyset,\Omega\}$, $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}$ and $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}$.  One can notice that $X(\omega)$ is not measurable with respect to $\mathcal{F}_0$ and $\mathcal{F}_1$, because $X^{-1}((\frac{3}{2}; +\infty))=\{HH\}$. To me it is quite surprising: intuitively $X$ makes perfect sense at all times. In particular it has an expected value at time $0$, which I interpret as that the probability and value of all outcomes $\{\omega\}$ can be computed. How do you think of a non-measurable function?  Here's another way of expressing the same confusion. The most natural choice of $\sigma$-algebra in a finite discrete case is $\mathcal{F}=2^\Omega$, and it is implicitly used in all elementary probability problems. However, this choice of $\mathcal{F}$ does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "$\sigma$-algebra is known information" make sense only in conditioning? Why is it convenient then?
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 $\min\{X, Y\}$?
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
44349
I just started to working through Karatzas and Shreve "Brownian Motion and Stochastic Calculus". Solving an exercise there's some questions around. I want to prove.     Let $X$ be a stochastic process, every sample path of which is RCLL. Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A\in \mathcal{F}_{t_0}^X$, where the latter is defined as, $\mathcal{F}_{t_0}^X:=\sigma{(\{X_s;0\le s\le t_0\})}$   What I did (sketch):  Write $A$ as,   $$ A=\{\omega \in \Omega; \lim_{s\uparrow t}X_s(\omega)=\lim_{s\downarrow t}X_s(\omega)\}$$  for $t\in [0,t_0)$. Since the Limit of a sequence of measurable functions is measurable, this is in $\mathcal{F}_{t_0}^X$. For more details see the answer of Byron Schmuland below.   Now here's the part which is unclear. There's a second claim, very similar to the first one:       Let $X$ be a stochastic process whose sample paths are RCLL a.s. Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A$ can fail to be in $\mathcal{F}_{t_0}^X$. But if $\{\mathcal{F_t;t\ge 0}\}$ is a filtration satisfying $\mathcal{F}^X_t\subset\mathcal{F}_t,t\ge 0$ and $\mathcal{F}_{t_0}$   is complete under P, then $A \in \mathcal{F}_{t_0}$.   There's a sample solution to this second question. But what I don't get is the difference between the two questions. What exactly is the difference? As mentioned in my comment after Byron Schmulands answer I think this is more a linguistic problem.    cheers  math
94396
If the value of a measure on any subset in a generator of a sigma algebra is known, will the measure for the sigma algebra also be uniquely determined? Thanks!
69082
In , $\exists$ this exercise.  Let $s &gt; 1$ and let $\zeta(s) = \sum_{n=1}^\infty n^{-s}$.  Let $X$ and $Y$ be independent $\mathbb{N}$-valued random variables with $P(X=n)=P(Y=n)=\dfrac{n^{-s}}{\zeta(s)}$.  Facts:  1 The events ($E_p$: $p$ prime) where $E_p =$ ($X$ is divisible by $p$) (This time $p$ may not be prime) are independent.  2 The events ($F_p$: $p$ prime) where $F_p =$ ($Y$ is divisible by $p$) (This time $p$ may not be prime) are independent.  3 $P(E_p) = p^{-s}$  4 This is Euler's formula $1/\zeta(s) = \prod_p (1-1/p^s)$ where $p$ is prime.  5 $P(X=1) = 1/\zeta(s) = \prod_p (1-1/p^s)$  6 $P(\text{no square other than 1 divides }X)= \dfrac{1}{\zeta(2s)}$  Here is the exercise:  Let $H$ be the highest common factor of $X$ and $Y$. Prove that $P(H=n) = \dfrac{n^{-2s}}{\zeta(2s)}$.  How do I go about doing this? I know $H$ divides both $X$ and $Y$ but that's just making use of the fact that $H$ is a common factor. Since it is the highest, all the natural numbers greater than $H$ do not divide $X$ and $Y$ or something like that?
65000
I'm really struggling to grasp why $\sigma$-fields are not necessarily closed under finite union.  If the above statement is true then;  Let $(\Omega, \mathcal{F},P) $ be our probability space such that $(A,B \in \mathcal{F} \Rightarrow A \cup B \not\in \mathcal{F})$.  Surely $P(A \cup B) $ then cannot be evaluated.  Any advice on how I could move forward would be greatly appreciated.
118620
Let $Y_0, Y_1, ...$ be independent and identically distributed random variables with  $P(Y_n = 1) = P(Y_n = -1) = 1/2$ for n = 0, 1, 2 ...  Define random variables $X_n = Y_0Y_1Y_2...Y_n = \prod_{i=0}^{n} Y_i$ for n = 0, 1, 2 ...  It can be shown that .  Define the $\sigma$-algebras:  $\mathscr{Y} \doteq \sigma(Y_1, Y_2, ...)$  $\mathscr{T_n} \doteq \sigma(X_r | r &gt; n) = \sigma(X_{n+1}, X_{n+2}, ...)$  $\mathscr{R} \doteq \sigma(\mathscr{Y}, \bigcap_n \mathscr{T_n})$  Prove $Y_0$ and $\mathscr{R}$ are independent.
55906
So assume we have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X : \Omega \rightarrow \mathbb{R}^*$. We can derive from this a distribution $P_X$, and a distribution function $F$. My book says that when working with distribution function directly we have no need to know the underlying probability space since everything is well determined. We just need to show that we can find one. So here is my attempt :   Here we start with the distribution function $F$. increasing and right continuous, st. $F(\infty) = 1$ and $F(-\infty) = 0$. We define :   \begin{align} P_I : I &amp;\longmapsto [0,1] \quad I \in \mathbb{R}^2 ( interval) \\ [a,b] &amp;\longmapsto F(b) - F(a) \end{align}   We can define $P_X$ on the borel sets by decomposing the set into disjoint intervals :    \begin{align} P_X : \mathcal{B}(\mathbb{R}^*) &amp;\longmapsto [0,1] \quad B = \dot \bigcup B_i \quad B_i \in \mathbb{R}^2 \\ B &amp;\longmapsto \sum_i P_I[a_i,b_i]  \end{align}   By taking $X$ as the identity function we have $\Omega = \mathbb{R}^*$ and $P = (P_X)_{\Omega}$   I suspect that I can use the fact that F is increasing and right continuous to say that we can find a measure $\mu$ s.t $\mu(a,b) = F(b)-F(a)$ and then use this measure as probability measure. But I'm scared to miss any vital step along this way.  I would like to know if there is any flaw in my reasoning. Or anything missing. Thanks for any input !
100865
Good day,  I have the problem of an old exam without solutions. It is the following:    Let $X_1,X_2,...$ be i.i.d. (independent and identically distributed) random variables on the same probability space such stat $$P\left(X_1=-\frac{1}{4}\right)=P\left(X_1=\frac{1}{4}\right)=\frac{1}{2}$$ For all $n\in \mathbb{N}$ define $M_n:=\prod_{k=1}^n (1+X_k)$.   Show that $(M_n)_{n\in\mathbb{N}} $ is a martingale w.r.t. a suitably chosen filtation (This was no problem for me) Use the law of large numbers to show that $(M_n)_{n\in\mathbb{N}}$ converges almost surely and determine the limit. (Here is the problem for me.)     This is the strong law of large numbers (SLLN) we had in the lecture:  Let $X_i$ be uncorrelated random variables, $E(X_i)=\mu, E(X_i^2)&lt;\infty$, $\exists C$ such that $\text{Var}(X_i)\leq C$ for all $i$ and define $S_n=\sum_{i=1}^n X_i$. Then $$\frac{S_n}{n} \to \mu \text{ almost surely.}$$    Therefore I have to get this $M_n$ into the form of $\frac{S_n}{n}$. I had a problems in the class where I had $$ A_n:=\left( \prod_{i=1}^n X_i \right)^{1/n}.$$ This was pretty elegant since we get the wished form via $$\log A_n = \frac{1}{n} \sum_{i=1}^n \log(X_i)=:\frac{1}{n} \sum_{i=1}^n Y_i.$$  I wanted to use this here:   $$\log M_n = \sum_{k=1}^n \log(1+X_k) = \frac{1}{n} \sum_{k} \log(1+X_k)^n $$  But now this "new" random variable $Y_{n,k}:=\log(1+X_k)^n$ is dependent of $n$. I don't know if this is a problem (now I think this is a huge problem, see below.) Let's try and use the SLLN for this: We would get $\log M_n \to E(Y_{n,k})$ if the assumptions are fulfilled but the limit should not be dependent on $n$. $$E(Y_{n,k})=E(\log(1+X_k)^n)=\frac{1}{2} \left( \log(5/4)^n +\log(3/4)^n \right) = \frac{1}{2} \log (15/16)^n \to -\infty$$  I don't know what to do. I think I could use the Martingale Convergence Theorem to get that $M_n$ converges a.s. but via this method I don't get the limit. Further the exam states that I have to use the SLLN explicitly.  I am thankful for every help/hint.
27479
Given $X_1,...,X_n$ iid. random variables which follow a $\mathcal{N}(\mu, \sigma^2)$ distribution. I already proved that  $T:=\frac{1}{n}\sum_{i=1}^n (X_i-\mu)^2$ is a minimal sufficient statistic for $\sigma^2$. Since, I want to apply the Lehmann-Scheffe Lemma, I have further to show that $T$ is complete, meaning that $\mathbb{E}_{\sigma^2}[g(T)]=0\Rightarrow g(T)=0$ for any measurable function $g$.   However, I am unable to do this, as I don't know how $T$ is distributed. Also there is a theorem which states that in exponential families the statistic is complete if the parameter space contains an open neighbourhood about the origin, but this is clearly not applicable since $\sigma^2\in (0,\infty)$. I would appreciate a little hint.
82231
I've just read in a stochastics textbook:     Let $(\Omega, P)$ be a discrete probability space.      (a) The events $A_i \subseteq \Omega, i=1,2, \dots$ are called independent, if   $$P(A_{i_1} \cap A_{i_2} \cap \dots \cap A_{i_k}) = P(A_{i_1}) \cdot \dots \cdot A_{i_k}$$   for all finite sets $\{i_1, \dots, i_k\} \subsetneq \{1,2, \dots\}$ and all $k \geq 2$.      (b) The events $A_i \subseteq \Omega, i=1,2,\dots$ are called pairwise independent, if   $$P(A_{i_1} \cap A_{i_2}) = P(A_{i_1}) \cdot P(A_{i_2})$$   for all pairs $\{i_1, i_2\} \subsetneq \{1,2,\dots\}$.   After these two definitions it states that pairwise independent events are not always independent.  Do you have an example?
95909
I got this problem from Shiryaev's book "Probability-1". I tried to use technique involving $\pi - \lambda$ system but I failed.  Let $\mathcal{F}_1, \mathcal{F}_2$ be two $\sigma$-algebras of subsets of the set $\Omega$. I want to prove  $$d(\mathcal{F}_1,\mathcal{F}_2)=4 \sup_{A_1 \in \mathcal{F}_1,A_2\in \mathcal{F}_2} |P(A_1 \cap A_2)-P(A_1)P(A_2)|$$  satisfies inequality $0\leq d(\mathcal{F}_1,\mathcal{F}_2) \leq 1$.
102059
How is the notation $X\sim N(\mu,\sigma^2)$ read? Is it $X$ follows a normal distribution? Or $X$ is a normal distribution? Or perhaps $X$ is approximately normal..  What if there are several variables that follow (or whatever the words is) the same distribution? How is it written?
70634
Suppose $X,Y$ are random variables. I'm trying to understand why $\max\{X,Y\}$ and $\min\{X,Y\}$ are also random variables. The proof in the book that I'm using states that for each $t$,      $\{ \max\{X,Y\} \leq t \} = \{X \leq t \} \cap \{Y \leq t \}$,   and     $\{ \min\{X,Y\} \leq t \} = \{X \leq t \} \cup \{Y \leq t \}$.   I can't make the mental leap. Could someone explain this more explicitly?  Edit:   We use the following definition for a random variable: A random variable is a function on  $X: \Omega \rightarrow \mathbb{R}$ such that      $X^{-1}((-\infty,t]) := \{\omega \in \Omega : X(\omega) \leq t\} \in \mathcal{F}$   for all $t\in\mathbb{R}$, where $\mathcal{F}$ is the set of all events.  The obvious first step that I made was      $\{ \max\{X,Y\} \leq t \} = \{ \omega : \max\{X(\omega),Y(\omega)\} \leq t \}$.   I just don't see why the next equality follows.  Edit2: Fixed typo.
6122
I am currently reading S. Shreve's book Stochastic Calculus II, and I have a question regarding Example 1.6.4 (p.35-36) which describes a change of measure, but I am puzzled by the notation.  $\Omega=[0,1], \mathbb{P}$ is the uniform measure and  $$\tilde{\mathbb{P}}[a,b]=\int_a^b 2\omega d\omega=b^2-a^2$$ for $0\le a\le b \le 1$.  The first step is:     We may use the fact that $\mathbb{P}(d\omega) = d\omega$ to rewrite the equation above as   $$\tilde{\mathbb{P}}[a,b]=\int_a^b 2\omega d\mathbb{P}(\omega).$$   I have several questions regarding this statement:   $\mathbb{P}$ is a probability measure from $\mathcal{F}\rightarrow\mathbb{R}$, I cannot find a clear definition in the book what $\mathbb{P}(d\omega)$ actually means. Is $d\omega$ actually an event? Why does this $\mathbb{P}(d\omega) = d\omega$ help in rewriting, shouldn't we need $d\mathbb{P}(\omega) = d\omega$? What does $\mathbb{P}(d\omega) = d\omega$ even mean, and why is this true?
53129
Let $\Omega$ be a set.  If $\Omega$ is finite, then any $\sigma$-algebra on $\Omega$ is finite.  If $\Omega$ is infinite and countable, a $\sigma$-algebra on $\Omega$  cannot be infinite and countable.     What about if $\Omega$ is not countable ? Is it possible to find an uncountable $\Omega$ with a $\sigma$-algebra that is infinite and countable ?