2
$\begingroup$

Let $W_t$ be a Wiener process with $W_0=0$, and let $L=\{at+by=c\}$ be a line with $c/b<0$ (i.e. the line crosses the $Y$-axis below $0$).

Assume that $W_t$ stayed above $L$ up to time $T$. What is the PDF of $W_T$ under this assumption? Does it have a closed form?

Improve this question
$\endgroup$
2
  • $\begingroup$ Is W a (standard) one-dimensional Brownian motion? If yes, it would be clearer to give the equation of $L$ in terms of $t$ (time variable) and $x$ (space variable), instead of $t$ and $x$. Or is it a two dimensional Brownian motion? $\endgroup$
    – Christophe Leuridan
    Commented Jan 11, 2023 at 21:11
  • $\begingroup$ @ChristopheLeuridan you're right, I changed the x variable to t. $\endgroup$
    – user2520938
    Commented Jan 12, 2023 at 7:03

2 Answers 2

Reset to default
3
$\begingroup$

The problem can be restated as follows:

For a real $a>0$ and a real $b$, let $$X_t:=a+bt+W_t$$ for real $t\ge0$, where $W$ is a standard Wiener process. Let $$\tau:=\inf\{t>0\colon X_t=0\}.$$ For a real $t>0$, find the joint distribution of $X_t$ and $\tau$.

The answer is well known: $$P(X_t\in dx,\tau>t)=f_t(x-a-bt)(1-e^{-2ax/t})\,1(x>0)\,dx, \tag{1}\label{1}$$ where $f_t$ is the pdf of $W_t$. A derivation of this formula can be found here.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks. Are you sure about this formula? If I'm not mistaken, the following should produce 1: differentiate w.r.t. $t$, integrate over $[0,\infty)\times[0,\infty)$ in the $x,t$-plane. However, Mathematica computions suggest that this does not produce $1$. Am I misunderstanding something? $\endgroup$
    – user2520938
    Commented Jan 12, 2023 at 20:21
  • $\begingroup$ @user2520938 : Please recheck your calculations. Using Mathematica, I have just integrated $f_t(x-a-bt)(1-e^{-2ax/t})\,1(x>0)$ in $x$ to get $P(\tau>t)$, and then the limit of the integral is $1$ as $t\downarrow0$, to complete the $=1$ check. $\endgroup$
    – Iosif Pinelis
    Commented Jan 12, 2023 at 21:26
  • $\begingroup$ @user2520938 : It just occurred to me that the cause of the incorrect result you got may be that Mathematica, weirdly, parametrizes normal distributions by their standard deviations, rather than the variances. So, for Mathematica, the pdf of $W_t$ is PDF[NormalDistribution[0, Sqrt[t]]]. $\endgroup$
    – Iosif Pinelis
    Commented Jan 13, 2023 at 4:59
  • $\begingroup$ Denoting by $g(x,t)$ your formula, it's correct that we would expect, that $\int_{0}^\infty dt\frac{\partial}{\partial t}\int_{0}^{\infty}g(x,t)dx=\lim_{t\to \infty}\int_{0}^{\infty}g(x,t)dx-\int_{0}^{\infty}g(x,0)dx$ to be equal to $(\pm)1$, right? As you note, the second term indeed is $\pm1$, but the first term is not equal to $0$, so the total integral is not $1$. But again, maybe I'm just confusing some things. $\endgroup$
    – user2520938
    Commented Jan 13, 2023 at 6:47
  • $\begingroup$ @user2520938 : Let $G(t):=P(\tau>t)$, which equals $H(t):=\int_0^\infty g(x,t)\,dx$, according to my answer. The $=1$ check in my previous comment was that $H(0+)=1=G(0+)$. Now you suggest that $G(\infty-)=0$. This is indeed true if $b\le0$. However, if the drift $b>0$, then $P(\tau=\infty)=G(\infty-)=H(\infty-)=1-e^{-2ab}>0$ -- that is, with the positive probability $1-e^{-2ab}$ the positive-drifted process $(X_t)$, starting at $a>0$, will never reach $0$ -- which makes sense, in view of the strong law of large numbers (say) for the Brownian motion. $\endgroup$
    – Iosif Pinelis
    Commented Jan 13, 2023 at 16:48
2
$\begingroup$

As I mentionned in my comment, there is an ambiguity in the statement of you question.

Anyway, if $B$ is a standard one dimensional Brownian motion, if $\lambda$ is a real number, then $(B_t-\lambda t)$ is a diffusion. The joint distribution of its position and its current minimum at time $T$ can be derived from http://www.numdam.org/item/AIHPB_1987__23_2_179_0.pdf

And there are probably more elementary methods to get this distribution.

This provides an answer.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .