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I have been asking this similar question before. However, I really want to be concrete and get and concrete explanation.

I have been reading the paper by Moreni and try to implement the same methodology based on least-square monte carlo for both European and American put options. They payoff function for both these puts is given by $f(t,S_t)=\max(K-S(t),0)$. The only difference is that for American case, we have payoffs in different times from simulations depending on the optimal stopping time.

As usual we have the underlying stock price as geometric brownian motion:

$dS_t = r S_t dt + \sigma S_t dW_t$.

Moreni introduces the change of drift in the underlying:

$dS_t = (r+\theta\sigma )S_t dt + \sigma S_t dW_t$.

When simulating from the drifted process, we have to multiply the PAYOFF functions with the likelihood ratio defined as $L_\theta=\exp\{-\theta W_t+\frac{1}{2}\theta^2 t\}$

so we have $f(t,S_t)=\max(K-S(t),0)\cdot\exp\{-\theta W_t+\frac{1}{2}\theta^2 t\}$

instead of just $f(t,S_t)=\max(K-S(t),0)$

The idea is of course to use negative values for $\theta$ so we get more "in-the money paths".

But when I try to use these modified payoffs with the Longstaff and schwartz algorithm, I don't get any relevant results. The price gets extremly biased whichever value i use for $\theta$.
My question is, should the estimated continuation values in least-square method also be modified with the likelihood ratio? In that case how? Or as mentioned in the article, ONLY the payoff functions be multiplied by the likelihood ratio??

Thank you for your patience

PS* Added by request the steps that I use:

  1. Generate $N$-scenarios of the drifted with $\theta$ geometric brownian motion
  2. Save all the standard normal values $Z_t$ from the GBM for use in the likelihood ratio.
  3. Create the likelihood ratio using the $Z_t$'s and for some value $\theta$
  4. Use the likelihood ratio and set $f(t,S_t)=\max(K-S(t),0)\cdot\exp\{-\theta W_t-\frac{1}{2}\theta^2 t\}$ for all paths
  5. Now run the Least-square algorithm from $t=T$ down to $t=0$ to calculate all the continuation values and optimal stopping times.
  6. Take average of all the scenarios.
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  • $\begingroup$ Have you tried applying your IS scheme to the pricing of a standard European option. If so, do you observe the same behaviour? $\endgroup$ – Quantuple Jun 14 '16 at 6:34
  • $\begingroup$ Yes I have, and indeed I get for some values for $\theta$ a correct price with variance reduction. It is just simulating everything at time $T$... $\endgroup$ – Elekko Jun 14 '16 at 7:03
  • $\begingroup$ Ok I assume that you've also checked your IS scheme when generating a full path and not simply sampling $S_T $? Longstaff-Schwartz consists in estimating the continuation value (which is a conditional expectation) using cross sectional info. IMHO you should not involve the likelihood ratio at that stage. $\endgroup$ – Quantuple Jun 14 '16 at 7:29
  • $\begingroup$ Also, have you checked that without IS your LS Monte Carlo works as expected (price an American put without dividend for instance?) $\endgroup$ – Quantuple Jun 14 '16 at 7:35
  • $\begingroup$ The LS works for ordinary case since I am using the LSM library in R. $\endgroup$ – Elekko Jun 14 '16 at 7:54
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Assuming deterministic interest rates, the price of an American call option struck at $K$ and expiring at $T$ is given by $$ V_0 = \text{sup}_{\tau \in \mathcal{T}[0,T]} \mathbb{E}_0^\mathbb{Q}\left[ e^{-r\tau} \max(S_{\tau}-K, 0) \right] $$ where $\mathcal{T}[0,T]$ denotes a family of stopping times with values in $[0,T]$ and where, under the risk-neutral measure $\mathbb{Q}$, we assume $(S_t)_{t\geq 0}$ is a GBM $$ \frac{dS_t}{S_t} = r dt + \sigma dW_t^{\mathbb{Q}} $$

Doing importance sampling boils down to computing the above price by simulating paths and computing the expectation not under the risk-neutral measure $\mathbb{Q}$, but rather, under a modified measure $\mathbb{Q}^\theta$ which you have in your case chosen such that $$ \frac{dS_t}{S_t} = (r+\sigma\theta) dt + \sigma dW_t^{\mathbb{Q}^\theta} \tag{1}$$

which implicitly shapes the Radon-Nikodym derivative (see Girsanov theorem) $$ \left. \frac{d\mathbb{Q}}{d\mathbb{Q}^\theta} \right\vert_{\mathcal{F}_t} = \mathcal{E}\left(-\theta W_t^{\mathbb{Q}^\theta}\right) = \exp\left(-\theta W_t^{\mathbb{Q}^\theta} - \frac{1}{2}\theta^2 t \right) := L(\theta,t) \tag{2}$$ which defines what you refer to as the likelihood ratio.

Using a change of measure argument (+ optimal sampling theorem), one can then claim that \begin{align} V_0 &= \text{sup}_{\tau \in \mathcal{T}[0,T]} \mathbb{E}_0^\mathbb{Q}\left[ e^{-r\tau} \max(S_{\tau}-K, 0) \right] \\ &= \text{sup}_{\tau \in \mathcal{T}[0,T]} \mathbb{E}_0^\mathbb{Q^\theta}\left[ e^{-r\tau} \max(S_{\tau}-K, 0) L(\theta,\tau) \right] \end{align} where in the last equality above, $S_{\tau}$ (and the expectation) are simulated (taken) under $\mathbb{Q}^\theta$.

So what you should do is:

  1. Generate $N$ stock price paths under the modified measure $\mathbb{Q}^\theta$ using the SDE (1)
  2. Apply the Longstaff-Schwartz algorithm (Least Squares Monte Carlo) to obtain, for each simulated path $n=1,...,N$ the value of the discounted exercise value at the optimal stopping time $\tau^{(n)}$ $$ e^{-r\tau^{(n)}} \max( S_{\tau^{(n)}}-K, 0 ) $$ (under the hood, the LS algorithm identifies optimal stopping times by approximating continuation values through polynomial regressions relying on cross-sectional information and working backwards in time).
  3. Once you have that, it is straightforward to compute the Monte Carlo estimator: \begin{align} V_0 &= \text{sup}_{\tau \in \mathcal{T}[0,T]} \mathbb{E}_0^\mathbb{Q^\theta}\left[ e^{-r\tau} \max(S_{\tau}-K, 0) L(\theta, \tau) \right] \\ &\approx \frac{1}{N} \sum_{n=1}^N \left( e^{-r\tau^{(n)}} \max( S_{\tau^{(n)}}-K, 0 ) \right) \times L(\theta, \tau^{(n)}) \end{align} with see (2) $$ L(\theta,\tau^{(n)}) = \exp\left(-\theta W_{\tau^{(n)}}^{\mathbb{Q}^\theta} - \frac{1}{2}\theta^2 \tau^{(n)} \right) $$ and by definition of a Wiener process (i.e. standard Brownian motion) $$ W_{\tau^{(n)}}^{\mathbb{Q}^\theta} = \sum_{ i : 0 \leq t_i \leq \tau^{(n)} } Z_i \sqrt{t_{i+1}-t_i},\ \ \ Z_i \sim N(0,1)\ \ \text{i.i.d.} $$

notice how the likelihood ratio $L(\theta,n)$ should be computed for each individual path $n=1,...,N$ depending on the optimal stopping time for that path $\tau^{(n)}$.

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  • $\begingroup$ Thanks, just to clarify. So winer process for the likelihood function is a sum of the standard normals up to the optimal stopping time for each individual path, correct? $\endgroup$ – Elekko Jun 14 '16 at 10:16
  • $\begingroup$ No Weiner process is not a sum of standard normals, it is important that the variance scales correcly. Because $\text{Var}(W_t) = t$, you should weigh the samples from your standard normals by the square root of the time step as explained in my answer. But yes, since you want to compute $L(\theta,\tau^{(n)})$ why would you go further than $\tau^{(n)}$ ?? $\endgroup$ – Quantuple Jun 14 '16 at 10:21
  • $\begingroup$ Yes, I got it. But equation (3), I think the index on sum should go from $n=1$ to $N$ instead from $i=1$ to $N$ $\endgroup$ – Elekko Jun 14 '16 at 10:27
  • $\begingroup$ Note that for an American option, what you call the "payoff" can happen before maturity, due to the early exercise possibility. This is represented by $\tau^{(n)}$ and $L(\theta,\tau^{(n)})$ in the above. So when you say "we have to multiply the PAYOFF functions with the likelihood ratio", beware that this payoff does not necessarily happen at $T$. $\endgroup$ – Quantuple Jun 14 '16 at 10:28
  • $\begingroup$ Yes, it is, I just saw that and corrected in some seconds ago ;) Please tell me if it works now. And if the answer helped you don't hesitate to accept it. Bye. $\endgroup$ – Quantuple Jun 14 '16 at 10:28
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I would suggest to use :

$$f(t,S^\theta_t)=\max(K-S^\theta_t,0)\exp(-\theta W_t\color{red}{\mathbf{-}}\frac{1}{2}\theta^2t)$$

where $dS^\theta_t=(r+\sigma\theta)S^\theta_t dt + \sigma S^\theta_t dW_t$

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