Let $\{\theta_t : \Omega \mapsto \Omega \  | \  t \in T \}$ be a family of measure preserving transformations on the probability space $(\Omega, \mathcal{F}, \mathbb{P}).$  A measurable random dynamical system on the measurable space $(X, \mathcal{B})$ over $(\Omega, \mathcal{F}, \mathbb{P}, \theta_T)$ is a mapping $$\phi : T \times \Omega \times X \mapsto X$$ so that the mappings $\phi(t,\omega, \cdot) : X \mapsto X$ form a cocyle over $\theta(\cdot)$. I admit it: this is over my head. I think I understand the basic idea of a random dynamical systems. We take some point $x_0$, and map it some other point. But, the maps are random so each map is different, hence the word &quot;random&quot; in &quot;random dynamical system.&quot;  (is this correct?) However, I am not seeing that in this definition. I was wondering if someone could explain this in somewhat less rigorous manner, even if the itty gritty details are missed. So, if someone could provide a pretty detailed explanation of this definition, with less rigor, and maybe an example, that would be great! Edit: There is a question similar to this but the answer did not help.
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As of now, the formal definition of a random dynamical system is a bit over my head. And many of the papers and books on this subject are not exactly &quot;introductory-friendly.&quot; I still have a lot of years to go before I fully understand this topic, or rather, fully understand the prerequisites for this topic. I was wondering if someone could provide me a definition of a random dynamical system in layman's terms. For example, how would you describe random dynamical systems to a group of high schoolers?
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$X(t) := 0 \;\; (t \leq \tau),\;\; t - \tau\;\;(t \geq \tau)$  with $\tau$ exponentially distributed.   Then X has the Markov property but not Strong Markov Property. But why ???? Can someone kindly explain in words and maths why the strong markov property does not hold?
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Consider a probability space $(\Omega, \mathcal{F}, P)$ where $\Omega$ is the sample space, $\mathcal{F}$ is the $\sigma$-algebra of $\Omega$, and $P$ is the probability measure. Let $X:\Omega\to\mathbb{R}$ be a random variable, inducing a $\sigma$-algebra $\mathcal{F}_X=\{X^{-1}(B)\mid B\in\mathcal{B}\}$ where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$.   Suppose we have another random variable $Y$ as well as the induced $\sigma$-algebra $\mathcal{F}_Y$, the conditional expectation can be defined as $E[Y|X]\stackrel{\text{a.s.}}{=}g$ where $g$ is determined by $E[YI_A]=\int_A g(s)P(s)\,\mathrm{d}s, \forall A\in \mathcal{F}_X$ according to Radon–Nikodym theorem.  I wonder what will happen if $\mathcal{F}_X=\mathcal{F}$ already? It looks like $E[Y|X]=Y$ so $X$ is kind of independent of any arbitrary random variable but I am not sure. Or, can anybody give an example of this kind of $X$? Thank you.  Some references:  by Mark J. Schervish.
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I'm not new to the concept of random variable and I know the measure theory. Anyway, I started reading the book &quot;Stochastic Differential Equation&quot; by B. Oksendal, and I'm having some problem in understanding.  Given a probability space $(\Omega, \mathcal{F}, P)$, $X : \Omega  \rightarrow \mathbb{R}^n$ is a random variable if $X$ is a $\mathcal{F}$-measurable function, or equivalently, if $$ X^{-1}(U) = \{\omega \in \Omega; X(w) \in U\} \in \mathcal{F} $$ for all the open set $U \subset \mathbb{R}^n$.  My main problem is that I always thought that $$X \in \Omega$$ while the book by B. Oksendal asserts that $X : \Omega \rightarrow \mathbb{R}^n$ and hence $X \in \mathbb{R}^n$. Addition For example, consider a fair dice. Then, $\Omega = \{1, 2, 3, 4, 5, 6\}$. One can be concerned to evaluate the probability that the dice result is even. In this case, we write: $$P(X \in \{2, 4, 6\}) = ...$$ In general, the probability of an event $U \subset \Omega$ ($U \in \mathcal{F}$) is normally written as follows: $$P(X \in U)$$ This notation suggest me that $X \in \Omega$. Again, I don't understand what means that $X : \Omega \rightarrow \mathbb{R}^n$.
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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 !
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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?
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There is a famous citation that says "It is evident that the primes are randomly distributed but, unfortunately, we don't know what 'random' means." R. C. Vaughan (February 1990)  I have this very clear but rather broad question that might be answered by different opinions and view points. However, my question is really not targeting an intuitive or philosophical answer, and I beg you for view points with a strength of mathematical foundation.   are primes randomly distributed?  so then what is 'random' in this context?    A posterior I  A possible hint comes perhaps from the theory of complex dynamical systems.  It can be difficult to tell from data whether a physical or other observed process is random or chaotic, because in practice no time series consists of pure 'signal.' There will always be some form of corrupting noise, even if it is present as round-off or truncation error. Thus any real time series, even if mostly deterministic, will contain some randomness. All methods for distinguishing deterministic and stochastic processes rely on the fact that a deterministic system always evolves in the same way from a given starting point.(ref , , , and "") - complying to latter, remind that every prime $p$ can be trivially identified by a sieving that applies prior primes $q&lt;p$ so it is possible to determine that somehow the system evolves in the same way from a given starting point. Of course to take into account that time must be substituted by a walking index as well.    A posterior II  Thank you for all of many the comprehensive answers and discussions. This is a quite classic question on MSE and meanwhile we moved much forward. You are right that primes are not random as per above question. Indeed we could show that they are in their sequence some type of "deterministic chaos". We don't need the Riemann function for this purpose. The primes sequence is a so called "ordered iterative sequence". Meanwhile this has been further elaborated by this source: "The Secret Harmony of Primes" (ISBN 978-9176370001)  Some of you correctly referred to sieving. It is crucial however that we regard sieving procedures as a subset of "interference" (incl. frequencies and amplitudes). We can iteratively apply interference rules in order to gain from the first prime 2, the next ordered sequence. This can be iterative continued in an "ordered" way and within exact boundaries of p-squares (for 100% certainty). Indeed, in order to construct an ordered sequence of primes you just need to begin with 2. The Riemann approach is charming but would raise difficulties since we don't have yet a proof of the hypothesis that connects the order of the non-trivial zeros with the primes. So if you apply Riemann, as some colleagues here suggest, we would need to say at any time in the begin of your argumentation something like "provided the Riemann hypothesis would be true...". Having in mind that the very unique rule that primes follow, is that in an interference scheme all odd prime frequencies dance on the base frequency of 2 (ordered iterative sequence), one may even give it a thought to something of a parallel in the Riemann transformed world, that all non-trivial zeros dance on 1/2. But latter remains not more than a tempting trivial speculation yet.
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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: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
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Itō isometry from :     Let $W : [0, T] \times \Omega \to \mathbb{R}$ denote the canonical   real-valued Wiener process defined up to time $T &gt; 0$, and let $X :  [0, T] \times \Omega \to \mathbb{R}$ be a stochastic process that is   adapted to the natural filtration $\mathcal{F}_{*}^{W}$ of the Wiener   process. Then $$      \mathbb{E} \left[ \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \right)^{2} \right] = \mathbb{E} \left[ \int_{0}^{T} X_{t}^{2} \,  \mathrm{d} t \right], $$ where $\mathbb{E}$ denotes expectation with   respect to  $\gamma$. In other words, the Itō   stochastic integral, as a function, is an isometry of normed vector   spaces with respect to the norms induced by the inner products $$      ( X, Y )_{L^{2} (W)} := \mathbb{E} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) =  \int_{\Omega} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t}  \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) \, \mathrm{d} \gamma  (\omega) $$ and $$      ( A, B )_{L^{2} (\Omega)} := \mathbb{E} ( A B ) = \int_{\Omega} A(\omega) B(\omega) \, \mathrm{d} \gamma (\omega). $$   I was wondering what the two normed spaces are and what their norms are, so that the two normed spaces are isometric wrt their norms?  Thanks and regards!
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Problem Denote by $\lambda_n$ the Lebesgue measure on $\mathbb {R}^n$. Let $h \in \mathbb {R}_{&gt;0}$. Let $A \subset \mathbb {R}^{n-1}$ be Lebesgue-measurable with finite measure. Define the cone $C(A,h)$ as the union of all straight lines connecting a point in $A \times\{0\}$ with the point $(0, h) \in \mathbb {R}^{n-1} \times \mathbb{R}$. Prove that $\lambda_n (C(A,h))=\frac{h}{n}\cdot \lambda_{n-1}(A)$. Where I struggle  Assuming that $C(A,h)$ is (Lebesgue-)measurable, this is a simple application of Fubini. How Fubini is used is all clear to me.  However, why is $C(A,h)$ measurable? It seems one could roughly argue as follows: As $A$ and $[0,h]$  are measurable so is $A \times [0,h]$. As $A \times [0,h]$ is measurable it is the union of a null set and a Borel set. $C(A,h)$ is the image of $A \times [0,h]$ under a certain diffeomorphism (c.f. ). Diffeomorphisms map null sets to null sets, and Borel sets to Borel sets. Hence, $C(A,h)$ is (as a finite union of measurable sets) measurable.  Unfortunately, I am not allowed to use the above properties of diffeomorphisms. What are other ways to prove the claim? Please only give a hint to set me thinking.
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If you randomly select a subset of $[0,1]$, what is the probability that it will be measurable?  Edit: This question may be unanswerable as asked.  If additional assumptions could be made to make it answerable, I would be interested in learning about them.
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I recently found out that the expectation of a random variable $X$ in a probability space $(\Omega, \mathcal F,  \mathbb P)$, $\mathbb E(X)$, is just the term used in probability theory for the measure of the function $X$; i.e.:  $$ \mathbb E(X)=\mathbb P(X) $$  or, for a real-valued random variable $X:\Omega\to\mathbb R$:  $$ \mathbb E(X)=\int_\Omega X(x)\mathbb P(dx) $$  The thing is, I am also familiar with the definition of the expectation of a random variable $X:\Omega\to\mathbb R$ with probability distribution function $f$ to be:  $$ \mathbb E (X)=\int_\mathbb R xf(x)dx $$  Here, the integral is with respect to the Lebesgue measure on $\mathbb R$.    We can define the function $f$ in the following way: we have $F(x)=\int_\mathbb R f(x)dx$, where $F(x)=\mathbb P(X^{-1}((-\infty,x]))$.    I can't see any way to show that these two definitions are equal to one another.  The only transformation between integrals with respect to different measures that I know is that $\mu(f^{-1}(g))=\mu(g\circ f)$, which gives that $\mathbb E(g)=\mu_X(g)$, where $\mu_X$ is the image measure $\mathbb P\circ X^{-1}$, but that doesn't seem to help.  Why can we write $\mathbb E(X)=\int_\mathbb R xf(x)dx$?  Why are these two definitions consistent?
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I'm facing the following issue. Let $X$ be an integrable random variable on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be two sigma-algebras. We assume that $X$ is independent of $\mathcal{G}$, i.e. $\sigma(X)$ is independent of $\mathcal{G}$. Can I say (and can I prove) that $$ E(X \mid \sigma(\mathcal{G} \cup \mathcal{H})) = E(X\mid \mathcal{H}) ?$$  Thank you very much for your help!
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Suppose that $(X,\mathcal{E},\mu)$ is a non-atomic finite measure space (i.e. for every $E \in \mathcal{E}$ with $\mu(E)&gt;0$ there exists $F \subset E$ measurable such that $0&lt;\mu(F) &lt;\mu(E)$.)  a) Prove that for every $ \varepsilon &gt;0$ there is a finite partition of $X$ in measurable subsets $X_1,..,X_n$ such that $\mu(X_i)\leq \varepsilon$.  b) Prove that for every $\alpha \in [0,\mu(X)]$ there exists $E \in \mathcal{E}$ with $\mu(E)=\alpha$.    I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from , which proves b)  )  Denote  $$ \Gamma = \{ (X_1,..,X_n) : n &gt;&gt; \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}$$ ordered by componentwise inclusion. Totally ordered parts $(Y_\alpha)$ of $\Gamma$ have an upper bound element $(\bigcup Y_\alpha^i)_{i=1}^n$ which is still in $\Gamma$.   By Zorn's lemma $\Gamma$ has maximal elements. If a maximal $(X_1..X_n)$ element is not a partition, then we can replace it with something better, of the form $(X_1,..,X_n\cup A)$ where $A \subset X \setminus (X_1 \cup..\cup X_n)$ and $\mu(A)&gt;0$.     My questions are:  1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions.  2) What is a simpler solution of a)?
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I often see the sentence "let $X_1, X_2, \ldots$ be a sequence of i.i.d. random variables with a certain distribution". But given a random variable $X$ on a probability space $\Omega$, how do I know that there is a sequence of INDEPENDENT random variables of the same distribution on $\Omega$?
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Could you please verify this (rather simple) proof? I'm a bit new to this kind of reasoning.  The question is:  Let $\mathcal{T}_1$, $\mathcal{T}_2$ be two topologies on some set $X$, when is the identity map $id:X \to X$, $id(x) = x$ continuous?  Here is my reasoning:   If $\mathcal{T}_1 = \mathcal{T}_2$, then for any $S \in \mathcal{T}_2$, $id^{-1} (S) = S \in \mathcal{T}_1$ so then it's continuous.  If $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it's exactly like above, then it's also continouos.  Now:  If  $\mathcal{T}_1 \subseteq \mathcal{T}_2$ then take some $x \in U \in \mathcal{T}_2 \setminus \mathcal{T}_1$, then $id^{-1}(U) \not\in \mathcal{T}_1$, so it's not continuous.  So the identity map is continuous if $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.   Is this OK?  Is there a better way to show this?  Any extra applicable theory?  Thanks in advance
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It is a Dugundji's exercise:  Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.  If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.  For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$.   Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.