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Let $B_t$ be a standard Brownian motion. Let us introduce a stopping time $\tau$ as $$\tau \ = \ \ \inf\Big\{t:\ |B_t|=1\Big\}.$$ Now, what I am interested in are random variables $\tau_x$ defined as $$\tau_x(\omega) \ = \ x\cdot\tau(\omega),$$ for $x> 0$. Obviously, $\tau_x$ is not a stopping time anymore. Nevertheless, we can still consider a random variable $$B_{\tau_x}(\omega) \ = \ B_{\tau_x(\omega)}(\omega),$$ and it is (if I understand correctly) still well defined, i.e. measurable - as claimed in

The strong Markov property with an uncountable index set.

My question is, are basic quantities such as $$\mathbb{E}\Big(B_{\tau_\frac{1}{2}}\Big|B_{\tau}=1\Big)$$ computable? How can we calculate it? My crude intuition is, that $$\mathbb{E}\Big(B_{\tau_x}\Big|B_{\tau}=1\Big) \ = \ x,$$ as $\Delta:=B_{\tau}-B_{0}=1$ and on average we should make $x$ of it in the first $x$ part of random $\tau$ time.

I would be glad for any help and insight.

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  • $\begingroup$ If $x \geq 1$, then $\tau_x$ is a stopping time. $\endgroup$
    – nullUser
    Feb 10, 2021 at 1:57
  • $\begingroup$ Though it does seem like you are primarily interested in $x \in (0,1)$. $\endgroup$
    – nullUser
    Feb 10, 2021 at 1:57
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    $\begingroup$ Another comment about your intuition. What if $\tau$ were the inf time $t$ such that $B_t = 1$ or $B_t = -\epsilon$ for some small $\epsilon \in (0,1)$? If one is given $B_\tau = 1$, then to me it seems that we should have $E(B_{\tau_{1/2}} | B_\tau = 1) > x$. To me your intuition seems to fit better with $\tau$ being the inf $t$ such that $B_t = 1$. Of course this is all just intuition, but I wanted to ask your perspective. $\endgroup$
    – nullUser
    Feb 10, 2021 at 2:03
  • $\begingroup$ Yes, I now see that You are right. Your intuition is better. $\endgroup$
    – user617199
    Feb 10, 2021 at 10:22
  • $\begingroup$ Let $\bar{\tau}$ be the first hitting time of $1$. Let $x\in(0,1)$. Recall that by properties of Brownian bridges, for any $T>0$ the distribution of $B_{xT}$ given $B_T = 1$ is Gaussian with mean $x$ and variance $(1-x)xT$. I think this is the connection to the intuition. But to apply this result directly, one would need the conditional distribution of $B_{x\bar{\tau}}$ given $\bar{\tau} = T$ to correspond to the conditional distribution of $B_{xT}$ given $B_T = 1$. I'm quite skeptical about this. (The same considerations apply for the slightly less simple first passage time $\tau$.) $\endgroup$
    – user350942
    Feb 10, 2021 at 13:34

1 Answer 1

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The following R-code serves to compute the mean of $B_{\tau/2}$ conditionally on $B_\tau = 1$. Paths of Brownian motions on $[0,1]$ are simulated via independent Gaussian increments. For the subsample where $\tau \leq 1$ and $B_{\tau} = 1$, the values $B_{\tau/2}$ are computed, if necessary via linear interpolation. The mean is then approximated by taking the average of these values.

(Intuitively, conditioning on $\tau \leq 1$, shouldn't be too problematic; but a rigorous argument escapes me.)

I find that $\mathbb{E}[B_{\tau/2} \, | \, B_\tau = 1] \approx 0.26$. Replacing $\tau$ by $\bar{\tau} := \inf\{t > 0 : B_t = 1\}$, one also finds $\approx 0.26$. I find these numerical results to be somewhat consistent with an earlier comment of mine.

set.seed(2)

N <- 5000
n <- 5000

h <- 1/n

X <- apply(rbind(rep(0,N),matrix(rnorm(n*N, mean = 0, sd = sqrt(h)), nrow = n, ncol = N)), 2, cumsum)
    
tau <- apply(X, 2, function(y){Position(function(x){x>=1 | x<=-1}, y)})

B_tau <- vector()
B_tau_half <- vector()
for (i in 1:N){
  if(!is.na(tau[i]) & X[tau[i],i] >= 1){
    B_tau <- c(B_tau,X[tau[i],i])
    if(tau[i] %% 2 == 0){
      B_tau_half <- c(B_tau_half,X[tau[i]/2,i])
    }
    else{
      B_tau_half <- c(B_tau_half,X[(tau[i]-1)/2,i]/2 + X[(tau[i]+1)/2,i]/2)
    }
  }
}

mean(B_tau_half)
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  • $\begingroup$ Thank You very much, this is very informative. $\endgroup$
    – user617199
    Feb 10, 2021 at 16:12
  • $\begingroup$ You're welcome. $\endgroup$
    – user350942
    Feb 10, 2021 at 16:18
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    $\begingroup$ In my own simulation I'm finding the restriction $\tau \leq 1$ plays a significant role. Using simulating up to a max time of $100$ I find $E[B_{\tau/2}] \approx .20$, but simulating up to $1$ I find closer to what you had $\approx .27$. $\endgroup$
    – nullUser
    Feb 10, 2021 at 19:20
  • $\begingroup$ Further numerical studies are needed to really understand what is going on, agreed. $\endgroup$
    – user350942
    Feb 10, 2021 at 22:14

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