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Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $W:[0,\infty)\times \Omega\rightarrow\mathbb{R}$ be a standard Brownian motion. Is it true that for all $x\in\mathbb{R}$ there exists $\omega\in\Omega$ such that $W(1,\omega)=x$?

I do not know if the question has a positive or negative answer. I know that $P(W(1)\in [a,b])>0$ for all real numbers $a<b$, but I do not think that suffices. I have no idea on how to prove the existence of such an $\omega$, since it seems to me that the proof may use an explicit form of $\Omega$.

EDIT: As I read in a comment below, this statement is not true. My doubt came from reading the proof of Lemma 5.22 in An Introduction to Computational Stochastic PDEs. The step I do not completely understand is the following: from $E[W(t)|W(1)]=t \,W(1)$, where $W$ is a Brownian motion and $0\leq t\leq 1$, it is stated that $E[W(t)|W(1)=0]=0$.

One of the possible definitions that I studied for $E[W(t)|W(1)]$, and in general for $E[X|Y]$, where $X$ and $Y$ random variables, is $E[X|Y](\omega):=E[X|Y=Y(\omega)]$, where $E[X|Y=y]=\int_{\mathbb{R}} x\,P_{X|Y=y}(dx)$, being $P_{X|Y=y}$ the $P_Y$-unique probability satisfying $P(X\in A,Y\in B)=\int_B P_{X|Y=y}(A)\,P_Y(dy)$.

In this example, $E[W(t)|W(1)=W(1)(\omega)]=E[W(t)|W(1)](\omega)=t\,W(1)(\omega)$. I understand that, if $W(1)(\omega)=0$ for some $\omega\in\Omega$, then $E[W(t)|W(1)=0]=t\cdot 0=0$. But if for our particular probability space $(\Omega,\mathcal{F},P)$ and our particular Brownian motion $W$ there is no $\omega$ satisfying $W(1)(\omega)=0$, then I do not see that $E[W(t)|W(1)]=t\,W(1)$ implies that $E[W(t)|W(1)=0]=0$.

Could you explain this to me using probability theory, and not just "intuition"?

user39756
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    If $X$ is a random variable and $\tilde{\Omega}$ is an event satisfying $\mathbb{P}(\tilde{\Omega}) = 1$, then you have an obvious way of modifying the underlying probability space such that the resulting random variable is defined on $\tilde{\Omega}$ and has the same distribution as $X$. – Sangchul Lee Sep 08 '17 at 11:13
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    @SangchulLee If I understood your idea correctly, you mean the following: Let $(\Omega,\mathcal{F},P)$ be a probability space and $W:[0,\infty)\times\Omega\rightarrow\mathbb{R}$ be a Brownian motion. Fix $x\in\mathbb{R}$. Define $\tilde{\Omega}=\Omega\backslash {\omega\in\Omega:,W(1,\omega)=x}$ and let $\tilde{W}$ be $W$ restricted to $\tilde{\Omega}$. Since $P(\tilde{\Omega})=1$, we have that $\tilde{W}$ is a Brownian motion such that $\tilde{W}(1,\omega)\neq x$ for all $\omega\in\tilde{\Omega}$. – user39756 Sep 08 '17 at 14:01
  • OP, I'm not understanding this. $W_{t=1}$ is a normal distribution on the real numbers with mean 0 and variance 1. So it could take any real value. Isn't the question then equivalent to "can a unit normal a unit normal variable take any real value?"? – Mathemagical Sep 08 '17 at 14:50
  • @Mathemagical Look at the following example. Let $X$ be a normal random variable on $(\Omega,\mathcal{F},P)$. Define $\tilde{\Omega}=\Omega\backslash{X=0}$ and define $\tilde{X}$ as $X$ restricted to $\tilde{\Omega}$. Since $P(X=0)=0$, we have $P(\tilde{\Omega})=1$ and $\tilde{X}$ normally distributed. But $\tilde{X}(\omega)\neq0$ for all $\omega\in\tilde{\Omega}$. – user39756 Sep 08 '17 at 14:56
  • Sure, but it's not about whether $P(W_1 \neq 0)=1$ or less than one. We don't have to infer anything about $W_1$ at all since we already know that $W_1$ is a normal distribution. I guess what I'm not following is why does the question "can it take a certain value (like 0)" have to be answered in terms of the measure, i.e."is there a non-zero proability of taking that value?". A normal R.V. or any continuous RV can take any value, but the probability for each value is zero. – Mathemagical Sep 08 '17 at 15:46
  • @Mathemagical In my comment, $\tilde{\Omega}$ is a probability space and $\tilde{X}$ is a normal random variable. But $\tilde{X}(\omega)\neq0$ for all $\omega\in\tilde{\Omega}$. One thing is the distribution, and another different thing is the pointwise behavior. As we saw, a normal random variable $Y$ may not take every real number, although $P(Y\in [a,b])>0$ for all $a<b$. – user39756 Sep 08 '17 at 15:55
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    The conditional expectation $E[W(t) \mid W(1) = x]$ is defined modulo sets negligible with respect to the distribution of $W(1)$ (= modulo sets of zero Lebesgue measure). Therefore, the expression $E[W(t) \mid W(1) = 0]$ is meaningless, this could be anything. In contrast, note that $E[W(t) \mid W(1) = x] = tx$ is meaningful as an equality of functions a.e. And, of course, this equality does not "imply" that $E[W(t) \mid W(1) = 0] = 0$. – zhoraster Sep 08 '17 at 19:42
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    And it does not matter whether there is a particular $\omega$ such that $W(t,\omega) = 0$ or not. What does matter in the definition of conditional expectation is whether $P(W(1)=0) > 0$ or not. – zhoraster Sep 08 '17 at 19:46
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    The conditioning is indeed subject to interpretation and does depend on the choice of a version. For the standard Brownian motion, however, we can indeed choose a very nice version of $\mathbb{P}(\cdot \mid W(1))$ in view of the following decomposition: for $0 \leq t \leq 1$, we have $$ W(t) = B(t) + tW(1) $$ where $B(t) = W(t) - tW(1)$ is the Brownian bridge between $0$ and $1$, and $(B(t) : 0\leq t\leq1)$ is independent of $W(1)$. So we may choose the version $$\mathbb{E}[W(t) \mid W(1) = x] = B(t) + tx. $$ – Sangchul Lee Sep 08 '17 at 23:07
  • @zhoraster I understand $E[W(t)|W(1)=x]=t,x$ a.e., so it makes no sense to assert $E[W(t)|W(1)=0]=0$. In the book, it is said that the Brownian bridge $B(t)$, $0\leq t\leq1$, is a Brownian motion $W(t)$ conditioned to $W(1)=0$. Then, it is proved that $E[B(t)]=E[W(t)|W(1)=0]=0$, but, as I asked in the question, this assertion makes no sense for me. $E[W(t)|W(1)=x]$ is defined for a.e. $x$, so saying something about $x=0$ is not correct. – user39756 Sep 09 '17 at 07:27
  • @SangchulLee In fact, my question arised when studying about the Brownian bridge in the book I cited. Reading Wikipedia I have the same confusion.It is said that the Brownian bridge, $B(t)$ for $0\leq t\leq 1$, is $W(t)|W(1)=0$. But the law of $W(t)|W(1)=0$, $P_{W(t)|W(1)=0}$, is not uniquely defined. What is defined is $P_{W(t)|W(1)=y}$ for a.e. $y\in\mathbb{R}$. So this definition of the Brownian bridge makes no sense, I think. – user39756 Sep 09 '17 at 07:40
  • @SangchulLee In the last equality of your comment, $E[W(t)|W(1)=x]=B(t)+tx$, the equality is understood for a.e. $x$, so we cannot say that $E[W(t)|W(1)=0]=B(t)$, I think. – user39756 Sep 09 '17 at 07:44
  • I am not claiming that this equality is true for any version of the regular conditional probability, and this is why I am saying that it is subject to interpretation. Rather, you may regard that this identity defines our choice of $\mathbb{P}(W\in\cdot\mid W(1))$. Notice that it is also the unique choice of version that is jointly continuous w.r.t. both $W$ and $x$. – Sangchul Lee Sep 09 '17 at 08:01
  • @SangchulLee Ok, I understood that. But to define the Brownian bridge $B(t)$ for $0\leq t\leq 1$ in the book I cited or in Wikipedia, it is defined as $W(t)|W(1)=0$. However, the law $P_{W(t)|W(1)=0}$ is not uniquely defined, so this definition of the Brownian bridge makes no sense. Do you agree with this? This is my last doubt. – user39756 Sep 09 '17 at 08:08
  • @user39756, wikipedia is different from mathematical books in that you don't need to write explicit definitions. In this case, they don't write what is meant by "conditional". If one uses the general definition (and one usually does use it if nothing else is said), then this does not make sense. But one may use other definitions (in which case this must be written explicitly). For example, one might define the conditional distribution given $W(t) = 0$ as limit of conditional distributions given $W(t) \in (-\varepsilon, \varepsilon)$ as $\varepsilon\to 0+$. I don't like this one. – zhoraster Sep 09 '17 at 10:07
  • Alternatively, one may recall that $W$ has a Gaussian distribution, and there are standard facts about what is conditional distribution of one Gaussian variable given another (provided that the joint distribution is Gaussian). Alternatively, one may define a pinned version of Brownian motion. With these alternatives, one arrives at a Brownian bridge. But, I repeat, it should be explicitly written that one uses one of these alternatives to the general definition of conditional expectation (probability) whenever one does. – zhoraster Sep 09 '17 at 10:13
  • @zhoraster I realized that, in the book, conditional expectation $E[X|Y=y]$ is defined in Definition 4.47. It is said that $E[X|Y=y]:=\phi(y)$, where $\phi:\mathbb{R}\rightarrow\mathbb{R}$ is a measurable function satisfying $E[X|Y]=E[X|\sigma(Y)]=\phi(Y)$. In the case $E[W(t)|W(1)]=t,W(1)$, we have $\phi(y)=t,y$, but a.e. So $E[W(t)|W(1)=x]=t,x$, but only a.e. So with the definition of the book, I think that the definition of Brownian bridge of the book and the proof of Lemma 5.22 are not correct. However, this book is not a probability theory book. Thank you for all your comments. – user39756 Sep 09 '17 at 10:58

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