# Conditional running maximum of Geometric Brownian Motion (maximum of Brownian Bridge)

I would appreciate help proving a formula that I came across on page 774 of these lecture note slides on pricing barrier options : https://www.csie.ntu.edu.tw/~lyuu/finance1/2015/20150520.pdf

Assume the stock price follows a Geometric Brownian Motion, i.e. $$dS_t = \mu S_t dt + \sigma S_t dW_t$$ for $$t \in [0,T]$$, and $$W_t$$ is a Brownian Motion (seeded at 0) under the measure $$\mathbb{P}$$. Let $$H$$ be a barrier satisfying $$H > S(0)$$ and $$H > S(T)$$.

Then the formula i'm trying to prove is : $$\mathbb{P} \Big[ \max_{t \in [0,T]} S(t) < H \ | \ S(0) = S_0, S(T) = S_T \Big] = 1 - \exp \Big(-\frac{2\ln(H/S_0)\ln(H/S_T)}{\sigma^2 T} \Big).$$

Using the solution to the SDE, namely $$S(t) = S_0 \exp((\mu - \sigma^2/2)t + \sigma W_t)$$, we can rewrite this probability as

$$\mathbb{P} \Big[ \max_{t \in [0,T]} W_t + \theta t < b \ | \ W_0 = 0, W_T + \theta T = a \Big]$$ $$= \mathbb{E}^{\mathbb{P}} \Big[ \mathbb{1}_{ \{\max_{t \in [0,T]} W_t + \theta t < b \}} | \ W_0 = 0, W_T + \theta T = a \Big],$$

where $$\theta := \mu/\sigma -\sigma/2$$, $$b:= \ln(H/S_0)/\sigma$$ and $$a:= \ln(S_T/S_0)/\sigma$$. My understanding is that under the measure $$\mathbb{P}$$, $$W_t$$ is a Brownian Motion, but $$W_t + \theta t$$ isn't. So what I would have done is to apply Girsanov's theorem. Setting $$\tilde{W}_t := W_t + \theta t$$, then we know that $$\tilde{W}_t$$ is a Brownian Motion (also seeded at 0) under $$\tilde{\mathbb{P}}$$, satisfying $$\frac{d\tilde{\mathbb{P}}}{d\mathbb{P}} = \exp(-\theta W_T -\theta^2 T /2) = \exp(-\theta \tilde{W}_T + \theta^2 T /2).$$ Then our calculation becomes equal to

$$\mathbb{E}^{\mathbb{\tilde{P}}} \Big[ \frac{d\mathbb{P}}{d\tilde{\mathbb{P}}} \mathbb{1}_{ \{\max_{t \in [0,T]} \tilde{W}_t < b \}} | \ \tilde{W}_0 = 0, \tilde{W}_T = a \Big]$$

$$=\exp(\theta a - \theta^2 T /2) \mathbb{E}^{\mathbb{\tilde{P}}} \Big[ \mathbb{1}_{ \{\max_{t \in [0,T]} \tilde{W}_t < b \}} | \ \tilde{W}_0 = 0, \tilde{W}_T = a \Big]$$ $$=\exp(\theta a - \theta^2 T /2) \mathbb{\tilde{P}} \Big[ \max_{t \in [0,T]} \tilde{W_t} < b \ | \ \tilde{W}_0 = 0, \tilde{W}_T = a \Big].$$

The latter probability is "well-known" as the probability of the running maximum of a Brownian Bridge. An online derivation is given in : https://eventuallyalmosteverywhere.wordpress.com/tag/brownian-bridge/

This probability is equal to
$$\mathbb{\tilde{P}} \Big[ \max_{t \in [0,T]} \tilde{W_t} < b \ | \ \tilde{W}_0 = 0, \tilde{W}_T = a \Big] = 1 - \exp \Big( \frac{a^2 - (2b-a)^2}{2T} \Big).$$

Plugging in the values of $$a$$ and $$b$$ we find that $$\mathbb{\tilde{P}} \Big[ \max_{t \in [0,T]} \tilde{W_t} < b \ | \ \tilde{W}_0 = 0, \tilde{W}_T = a \Big] = 1 - \exp \Big(-\frac{2\ln(H/S_0)\ln(H/S_T)}{\sigma^2 T} \Big) \ !!$$

But this is supposed to be the final answer to my problem, so it appears that the factor $$\exp(\theta a - \theta^2 T /2)$$ is not supposed to be there.... Any help will be greatly appreciated !

You did not perform the change of measure correctly. You seem to believe that

$$\beta \tilde{\mathbb{P}} = \mathbb{P} \implies \mathbb{E}_{\tilde{\mathbb{P}}}(\beta X \mid A) = \mathbb{E}_{\mathbb{P}}(X \mid A) \tag{1}$$

but this is actually wrong; if we choose, for instance, $$X:=1$$, then the right-hand side of $$(1)$$ equals $$1$$ but the left-hand side does not - this shows that there is something off. The correct formula reads

$$\beta \tilde{\mathbb{P}} = \mathbb{P} \implies \frac{\mathbb{E}_{\tilde{\mathbb{P}}}(\beta X \mid A)}{\mathbb{E}_{\tilde{\mathbb{P}}}(\beta \mid A)} = \mathbb{E}_{\mathbb{P}}(X \mid A). \tag{2}$$

Using $$(2)$$ with $$\beta := \exp(\theta \tilde{W}_T - \theta^2 T/2)$$ we get

$$\mathbb{E}_{\mathbb{P}} \left( 1_{\{\max_{t \leq T} (W_t+\theta t) < b\}} \mid W_0 = 0, W_T+\theta T = a \right) = \frac{\mathbb{E}_{\tilde{\mathbb{P}}}(\beta 1_{\{\sup_{t \leq T} \tilde{W}_t

and so, by the definition of $$\beta$$,

$$\mathbb{E}_{\mathbb{P}} \left( 1_{\{\max_{t \leq T} (W_t+\theta t) < b\}} \mid W_0 = 0, W_T+\theta T = a \right) = \mathbb{E}_{\tilde{\mathbb{P}}}(1_{\{\sup_{t \leq T} \tilde{W}_t

Now you can continue with your computations and you will get the correct result.

• Perfect ! thank you so much! – Marc_Adrien Feb 22 at 18:16
• @Marc_Adrien You are welcome. – saz Feb 22 at 18:20