# Estimating optimal exercise boundary for an American call by LSM method

I'm trying to derive optimal exercise boundary using LSM method and got some weird outcome. So, I evaluated an American call option by LSM method and now need to find the optimal exercise curve.

Do I understand it correct that it is necessary to do the following: on each time step I have to solve non-linear equation (Continuation value which is approximated by basis polynomials of some unknown price S_opt = S_opt - Strike)? And the coefficients of these basis polynomials are the same that I got previously (while I was regressing the discounted payoffs when I was evaluating the option)?

Could you please give me the cue?

Thank you

With $$V$$ American option value, $$H$$ holding (aka continuation) value, and $$B$$ bank account value, we have:

$$V_N(S_N) = (K-S_N)^+$$

and for $$i$$ backwards from $$N-1$$ down to 0, we have:

$$H_i(S_i) = \mathbf{E}\left[B_{i}B_{i+1}^{-1}V_{i+1}(S_{i+1})|S_i\right]$$

$$V_i(S_i) = \max (K-S_i, H_i(S_i))$$

The algorithm result is $$V_0(S_0)$$.

The different regression functions $$r_i$$ (linear combinations of basis functions) give the conditional expectations needed at every step $$i$$:

$$\mathbf{E}\left[B_{i}B_{i+1}^{-1}V_{i+1}(S_{i+1})|S_i\right] = r_i(S_i)$$

Optimal stopping time index $$\eta$$ taking values in $$\{1,..., N \}$$ is defined as: $$\eta = \min \{k\geq 1 \mid K-S_k \geq r_k(S_k) \} \wedge N$$

Edit: What is the exercise boundary in the probabilistic (Monte Carlo) framework?

When solving for the optimal stopping time index $$\eta$$ above, one needs to introduce the probability state set $$\Omega =\{\omega^1,...,\omega^J\}$$, which in Monte Carlo context represents the labels of the primitive market variables fully (all the way to the end of the financial contract) simulated paths, as $$\eta$$, like all variables here, are functions on it:

$$\eta (\omega^j)= \min \: \{k\geq 1 \mid K-S_k(\omega^j) \geq r_k(S_k(\omega^j)) \} \wedge N$$

The underlying price on path $$\omega_j$$ where one stops is $$S_{\eta{(\omega_j)}} (\omega_j).$$

If we introduce a new variable $$\Gamma: \{1,...,n\}\times \Omega \rightarrow \{0,1\}$$ defined as:

$$\Gamma (i, \omega) = 1 {\rm \: if \:} i > \eta(\omega),$$

and $$0$$ otherwise, then the (topological) boundary of the set $$\Gamma^{-1}(1) = \{(i,\omega) | i > \eta(\omega) \}$$ is called the exercise boundary.

Edit2: If you are referring to the PDE framework, then indeed the exercise boundary is defined as the deterministic curve $$S^{\rm opt}_k$$

$$S^{\rm opt}_k = \inf \: \{x | (K - x)^+ = V_k(x) \}$$

Edit3: If one uses the conditional density $$\phi_i(y\mid x)$$ (transition from $$S_i$$ to $$S_{i+1}$$) to compute the conditional expectation for holding value by integration (rather than using regression function $$r_i$$), we have $$H_i(x) = \mathbf{E}\left[B_{i}B_{i+1}^{-1}V_{i+1}(S_{i+1})|S_i =x\right]$$ $$= B_{i}B_{i+1}^{-1}\int_0^\infty \max (K-x, H_{i+1}(x))\phi_i(y\mid x)dx$$ The integral calculation can be improved if we split its integration domain at $$x^*$$, the root of the equation: $$K-x = H_{i+1}(x)$$

• In your formula of $H_i(S_i)$, why do we need additional $B_i$? I thought just discounting will do? Jun 13 '20 at 4:53
• I dont think this is the case. You described the way we determine the stopping rule. But I need something different. I need to find the optimal exercise boundary i.e. this is the minimal price S_opt such that if S >= S_opt we should early exercuse the option, if S < S_opt we should wait and shouldnt exercise the option. So seems we need to solve such equation: Continuation value of S_opt = S_opt - strike. And the question is: in this equation Contunation value of S_opt should be approximated with the same coefficients we got when we were evaluating the option? Jun 13 '20 at 10:55
• @Idonknow My Bi is the wealth factor (bank account value accumulation) from 0 to Ti, not discount factor. You need two of them to get the discount factor from Ti+1 down to Ti.
– ir7
Jun 13 '20 at 13:51
• @Kevin To compute the optimal strategy eta you must go through Sk states of course. That’s where the optimal Sopt will surface, by simulated path, with k being the first time on that path where the inequality takes place (the elements in the inequality are clear, including which regressor function). I’ll formalize this in the answer.
– ir7
Jun 13 '20 at 14:07
• @Kevin I added two edits along the lines in my previous comment. Hope they help.
– ir7
Jun 13 '20 at 16:45