# Conditional probability of Brownian motion (with drift and scaling) hitting barrier

I am trying to understand the pricing of barrier options, and am considering the Brownian motion $$\mathrm{d}X_t=a\mathrm{d}t+b\mathrm{d}W_t$$, $$a$$ and $$b$$ constant. I am trying to:

1. derive the distribution of $$X_{T/2}$$ given $$X_T$$ and $$X_0$$;

2i) prove that the probability $$\mathbb{P}\left(\inf_{[t_1,t_2]}X_t; and

ii) find $$\mathbb{P}\left(\sup_{[t_1,t_2]}X_t>U\Big|X_{t_1},X_{t_2}\right)$$.

For 1, is there a better way about it than to calculate the conditional density? I tried using the approach that one would use for a driftless Brownian motion, but ended up with multiple cross-terms that I can't get rid of. For 2i and ii, how would one derive them? The texts I have read mention the reflection principle, and I understand the conditional density can be imagined as the fraction of which paths hit the barrier, but I really don't understand much about it, particularly where the $$(\cdot)^+$$ operators enter. Any help and rigour is much appreciated. Thanks!

For part 1 of your question, the short answer is no, calculating conditional density is a looong way of doing it. Possible but not the easiest. Here is the sketch for a shorter version. We note that $$(X_{T/2},X_{T})$$ is a jointly Gaussian vector with mean $$\mu = (X_0 + aT/2,X_0 + aT)$$, and the variance-covariance matrix $$\begin{pmatrix} b^2 T/2 & b^2 T/2 \\ b^2 T/2 & b^2 T \end{pmatrix}$$

A conditional distribution of one element of a Gaussian vector on another on is Gaussian. So the conditional distribution of $$X_{T/2} \vert X_T$$ is Gaussian. The mean and the variance of this distribution can be expressed in terms of $$X_T, X_0, \mu,\Sigma$$. Details can be found in many places, for example here. The conditional mean in particular is just a linear regression formula so easy to remember

$$\mathrm{E}(X_{T/2} \vert X_T) = X_0 + aT/2 + \beta (X_T - aT - X_0)$$ where $$\beta = \frac{b^2 T/2}{b^2T} = \frac{1}{2}$$ so it simplifies to $$\mathrm{E}(X_{T/2} \vert X_T) = (X_0 + X_T)/2$$ The conditional variance is calculated along the similar lines (see the link above)

Interestingly the conditional mean and, in fact, the whole conditional distribution does not depend on the drift $$a$$ and is the same as for the standard Brownian motion with $$a=0$$, the so-called Brownian bridge.

Q2 is quite a bit more involved, you should look it up in any decent stochastic calculus textbook (Karatzas & Shreve is my favourite). As you can see the right-hand side is independent of the drift $$a$$. Our discussion for Q1 demonstrates (if falls somewhat short of a formal proof) why it is the case -- once you "pin" the start and the end of a Brownian motion, the fact that it has (constant) drift is not relevant anymore.

As to your specific point as to what $$(...)^+$$ terms are doing in the formula. This is just a convenient shortcut to have one formula for different configurations of $$X_{t_1}$$, $$X_{t_2}$$ and $$L$$. For example, if $$X_{t_1} < L$$ then the lhs is trivially 1, and the rhs is $$1$$ as well because $$(X_{t_1) - L)^+ =0$$ in this case.

• Thanks so much for your answer! Before I accept, may I ask which pages in Karatzas and Shreve cover question 2? – user107224 Mar 18 at 0:14
• In my copy it is Karatzas and Shreve, Brownian Motion and Stochastic Calculus, Second Edition, 1991, section 4.3.C, p265, eq (3.40). If you have a different edition (or a different book!) look for "Brownian bridge, Maximum of" in the index – piterbarg Mar 18 at 0:19
• @piterbarg: That was a beautiful answer and made sense. ( once I went to the other for the conditional formulae of the bivariate normal. I always forget them ). But, if you don't mind, what's the intuition for why the answer doesn't depend on $X_0$. Is that because the knowledge of the value of $X_T$ kinds of cancels the usefulness of knowing $X_0$ ? Thanks. – mark leeds Mar 18 at 0:50
• @markleeds ah sorry it does depend on X_0, I assumed X_0 = 0 implicitly. The general case follows if we consider X_{T/2} - X_0, X_T-X_0 instead. I'll fix that when I have a moment – piterbarg Mar 18 at 7:24
• @piterbarg another silly question but does the reflection principle apply even with drift? also for the infimum case do I just plonk in minus times a supremum? – user107224 Mar 18 at 14:21