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 $\mathbb{P}\left(\inf_{[t_1,t_2]}X_t<L\Big|X_{t_1},X_{t_2}\right)=\mathrm{exp}\left[-\frac{2(X_{t_1}-L)^+(X_{t_2}-L)^+}{b^2(t_2-t_1)}\right]$; 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!


1 Answer 1


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.

  • $\begingroup$ Thanks so much for your answer! Before I accept, may I ask which pages in Karatzas and Shreve cover question 2? $\endgroup$
    – user107224
    Commented Mar 18, 2021 at 0:14
  • $\begingroup$ 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 $\endgroup$
    – piterbarg
    Commented Mar 18, 2021 at 0:19
  • $\begingroup$ @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. $\endgroup$
    – mark leeds
    Commented Mar 18, 2021 at 0:50
  • $\begingroup$ @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 $\endgroup$
    – piterbarg
    Commented Mar 18, 2021 at 7:24
  • $\begingroup$ @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? $\endgroup$
    – user107224
    Commented Mar 18, 2021 at 14:21

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