When I was learning finance, we didn't cover the subject of Brownian Bridges. So I am trying to learn the proper way of generating paths when you have an arithmetic Asian option which has an averaging period at some point in the future, call it 0.5 years. Then a time of expiry, call it 0.6 years. What my dilemma is with Brownian Bridges, most of the literature assumes that you either generate a starting T_timestep1 OR T_expiry, and create the bridge that starts and ends with the same value (the GBM simulated price). Now less commonly taught, is transitioning between the T_timestep1 price and T_expiry price, assuming GBM, with the same "shock", there is higher volatility as sqrt(T) is higher at T_expiry. So call these 2 points a and b, where abs(a < b), but your Brownian Bridge shocks assume a 0 shock starting price and a 0 shock ending price. Although I've read as much as I can find on the internet, I still don't know the "correct" way of transitioning between these 2 simulated points.

Is the proper way a simple slope between a and b to which you apply the Brownian Bridge shocks? Say simulated price T_timestep1 is 20 and simulated price (with the same shock) is 30 at T_expiry, how do you generate the proper Bridge between a and b? If I was to guess an approach that would theoretically work, I would solve for a drift component that yields a change of 10 (30-20) over the time period, for this example 0.1 years (0.6-0.5). But just using a slope is simple, since I'd need to calculate it for every simulation path. And then I would still have the option of using drift (although simple GBM with drift gives crazy results over long time frames without mean reversion).

Just looking for a simple explanation of how the path from a->b is constructed with a Brownian Bridge, when a != b. The Brownian Bridge seems to be explained, wherever I've been able to find an explanation, assuming a = b. Documentation on this subject is severely lacking. So any help is appreciated! Obviously, I haven't taken any financial engineering classes in ages, so excuse my lack of knowledge in this area.

  • $\begingroup$ A formula for the Brownian Bridge when the starting point is $a$ and the end point is $b$ is given for example here en.wikipedia.org/wiki/Brownian_bridge#General_case $\endgroup$
    – nbbo2
    Nov 20, 2021 at 16:41
  • $\begingroup$ I gave a quick look to that formula, and it appears it is only assuming you are still in a bridge with starting and endpoints equal - that just gives the variance and other properties between T1 and T2 on a bridge whereas the start and end are still identical, from what I gather. What may be the solution (possibly), that I don't even need to consider the path between a->b at all, and the fixed start & end points will still generate a valid solution. I just haven't tested the resulting paths in such a manner and compared vs. a reference model. Basically, I'm in the design phase of exotics. $\endgroup$
    – Matt
    Nov 20, 2021 at 20:59


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