Analytical formula for discounted exposure of a European Put on a stock in Real-World measure

External question — Quantitative Finance Stack Exchange Author: Rhoyourway Original post: https://quant.stackexchange.com/questions/76175 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. Is there an analytical formula to approximate the discounted exposure for a European Put on a Stock in the Real-World measure? This is just an initial phase to be able to assess the accuracy of using Longstaff-Schwartz regression method, using a simple example. I would like to compare the regression results with analytical solution for discounted exposure for a European Put on a Stock. Also, is it fine to calculate the expected stock price at future points as $E[S(T_{k})] = S(T_{0}) * exp (\mu * T_{k})$ where $T_{k}$ are future time points for $k = 1, 2, .. , M$ , with $\mu$ being the real-world drift of the stock, and subsequently, use Black-Scholes analytical formula for valuing a put using $S(T_{k})$ calculated above - this is in order to calculate the approximate expected exposure at a future time point, $t_{k}$ ? Thanks in advance for any insight into this.

External answer — Quantitative Finance Stack Exchange Author: Arshdeep Original post: https://quant.stackexchange.com/a/76182 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. The put price at any point in the future $t$ , with stock price $S(t)$ , is just $BS(T-t,S(t),K)$ , i.e. the black-scholes price. In american options context the "continuation value" is always the black-scholes price. This is the value I would want the longstaff algorithm to be able to approximate. The expected stock price at any point in the future $t$ is $S(t)=S(0)*exp(mu+vol^2/2)$ . To answer your comment, "can I use the expected stock price at a future time t to plug into BS to get the put price at this future time, t? Is it correct to do so?" Short answer, no. You need $E(BS(T-t,S(t))$ which is different than $BS(T-t,E(S(t))$ . The magnitude of this difference depends on how convex BS function is w.r.t S(t), moneyness, time to expiry, pretty much everything, so I don't think I would be comfortable with this approximation. In the risk neutral measure the former expectation is a martingale so it's (obviously discounted) expectation equals the BS price today! To get it in the real world measure you can integrate this BS price against the real world density maybe numerically, I'm not sure there's a closed form solution available.

External answer — Quantitative Finance Stack Exchange Author: Parag Biswas Original post: https://quant.stackexchange.com/a/81394 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. [Editing my answer to make it more precise] Well, the closed-form solution of Black-Scholes equation is the analytical solution for pricing vanilla options. For the future stock prices, you are simply trying to drift the prices forward in time. It is ok to model a stock price in that way. However, I do not understand why you wrote the expression as an expectation. You can simply try it like this: - S(T_k) = S(T_0) * exp(mu * T_k) These will be your simulated stock prices. Since, you are not using random numbers for simulation, you have a single, defined path when you are drifting the prices into the future. Thus, no need for calculating expected stock prices. Is all above a good approximation in real-world measure? No. It is rare for users to model the underlying with just historical drift, even with frequent (monthly or weekly) updates to the parameter. We try to incorporate as much information available to model the underlying. The regression results should be able to point it out.