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Let $0<t<a$. In this post, @Ian says that $\mathbb P(B_t\in A\mid B_0=B_a)$ is not a random variable, but I don't understand why.

The way I interpret it's as follow : set $f(x)=\mathbb P(B_t\in A\mid B_0=x)$. So $f(x)$ give the probability that a brownian motion starting at $x$ a.s. is in $A$ at time $t$. Then $\mathbb P(B_t\in A\mid B_0=B_a)=f(B_a)$. So, why is this not a random variable ? (it should be a $\sigma (f(B_a))-$measurable random variable, no ? Namely, for almost all $\omega \in \Omega $, $$\mathbb P(B_t\in A\mid B_0=B_a)(\omega )=\mathbb P(B_t\in A\mid B_0=B_a(\omega )),$$

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I agree with Ian. What we have is three dependent random variables ($B_0$ is probably not actually random, but whatever), so let me just write it as $\newcommand{\P}{\mathbb P}\P(X\in A \mid Y=Z)$ instead. Now, $Y=Z$ is an event, so it's a bona fide conditional probability. If we had $\P(Y=Z)>0$, then it would make sense to write e.g. $$ \P(X\in A \mid Y=Z) = \frac{\P(X\in A \wedge Y=Z)}{\P(Y=Z)}. $$ (In the case of Brownian motion I'm sure that $\P(B_0=B_a)=0$, so a more sofisticated definition would need to be used of course).

Let me give a concrete example. Let $Y$ and $Z$ be the outcomes of two independent $6$-sided die throws, and let $X=Y+Z$. Let's consider $A=\{12\}$. So $$ \P(X=12\mid Y=Z) = \frac{\P(Y+Z = 12 \wedge Y=Z)}{\P(Y=Z)} = \frac{1/36}{1/6} = \frac 16. $$ On the other hand, let's try your interpretation. Let $f(z) = \P(X=12\mid Y=z)$. Then $$ f(z) = \P(X=12\mid Y=z) = \frac16\cdot 1_{[6\le z\le 11]} $$ (where $1_{[]}$ is an indicator function). Then we could plug $Z$ into $f$ and get $$ f(Z) = \frac16\cdot 1_{[6\le Z\le 11]} = \frac16\cdot 1_{[Z=6]}. $$ There's nothing wrong with doing this per se, but just notice that it's very different from the previous result, and most importantly not what the notation means.

To hopefully drive the point home, why shouldn't $\P(X\in A \mid Y=Z) = \P(X\in A \mid Z=Y)$ be equal to $$ g(Y) = \P(X\in A \mid Z=y)\bigl\vert_{y=Y} = \frac16\cdot 1_{[Y=6]}, $$ according to your interpretation? This is not(!) the same as before, so that's a big problem. (This would perhaps have been clearer if $Y$ and $Z$ hadn't had the same distribution).

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I don't agree with Ian (or what you say he said that $\mathbb{P}(B_t\in A|B_0 = B_a)$ is not a random variable). Indeed, we have: $$\begin{align} \mathbb{P}(B_t\in A|B_0 = B_a) &= \mathbb{P}(\{(B_t-B_0)+B_a\in A\}|B_0 = B_a)\\ &= \mathbb{P}(\{X+B_a\in A\}|B_0 = B_a)\\ &=\mathbb{E}(\mathbb{I}_{\{X+B_a\in A\}}|B_0 = B_a) \end{align} $$ with $X$ is a random varible independent to $B_a$ (which is also a random variable).

Then, $\mathbb{I}_{\{X+B_a\in A\}}$ is a random variable.

According to the definition of conditionnal expectation, $\mathbb{E}(\mathbb{I}_{\{X+B_a\in A\}}|B_0 = B_a)$ is a random variable because there exists a function $f$ such that $$\mathbb{E}(\mathbb{I}_{\{X+B_a\in A\}}|B_0 = B_a) = f(B_a)$$ and $f(B_a)$ is a random variable. $$$$ Return back to your initial question in this post, the answer should be $$ \begin{align} \mathbb P\left(\sup_{t\in [a,b]}B_t>x \right) &=\mathbb P\left(\sup_{t\in [a,b]}(B_t-B_a) + B_a>x\right) \\ &=\mathbb P\left(\sup_{t\in [0,b-a]}(W_t) + B_a>x\right) \\ &=\mathbb E\left( \mathbb{I}_{\left \{\sup_{t\in [0,b-a]}(W_t) + B_a>x\right\}} \right) \\ &=\mathbb E\left(\mathbb{E}\left( \mathbb{I}_{\left \{\sup_{t\in [0,b-a]}(W_t) + B_a>x\right\}}|B_a \right)\right) \\ &=\mathbb E\left(\mathbb{P}\left( \sup_{t\in [0,b-a]}(W_t) + B_a>x|B_a \right)\right) \\ \end{align} $$ Here, $W_t$ is second Brownian motion which is independent to the Brownian motion $B_t$ (I think this notation $W_t$ is necessary to remove all ambiguity)

And you will find that the conditional probability $\mathbb{P}\left( \sup_{t\in [0,b-a]}(W_t) + B_a>x|B_a \right)$ is a random variable (this probability is equivalent to the one in your question $\mathbb{P}(B_t\in A|B_0 = B_a)$).

Of course, $\mathbb E\left(\mathbb{P}\left( \sup_{t\in [0,b-a]}(W_t) + B_a>x|B_a \right)\right)$ (the expectation of a random variable) is a number, which is consistent with the fact that $\mathbb P\left(\sup_{t\in [a,b]}B_t>x \right) $ is a number.

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  • $\begingroup$ In one answer you answer 2 questions. Thanks a lot. $\endgroup$
    – joshua
    May 30, 2021 at 10:11
  • $\begingroup$ Just one last thing do you agree with $$\mathbb P(B_t\in A\mid B_0=B_a)(\omega )=\mathbb P(B_t\in A\mid B_0=B_a(\omega )),$$ which is the probability that the brownian motion is in $A$ at time t given that it started at $B_a(\omega )$ ? $\endgroup$
    – joshua
    May 30, 2021 at 10:12
  • $\begingroup$ Yes, I think your statement is correct. Because $$\mathbb P(B_t\in A\mid B_0=B_a)(\omega) = f(B_a)(\omega) = f(B_a(\omega)) = \mathbb P(B_t\in A\mid B_0=B_a(\omega ))$$ $\endgroup$
    – NN2
    May 30, 2021 at 10:40

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