I am working on a derivative with the following payoff at time $T$:

$$ \sqrt{(S_T - K)^+} $$ enter image description here

where $S_T$ is the value of the stock at the expiring date. As usual we will assume $S_t$ to be a GBM:

$$ dS_T = \sigma S_T dW_T $$

I am interested in computing greeks, in particular $\delta$ sensitivities and I decided to do it by using Vibrato Montecarlo methods and LRM techinque.

By doing a discretization of time in $N$ we can associate $T$ to $N-th$ step and $T-dt$ to $N-1 th$ step.

I've studied the structure of conditional expectation at time $T-dt$. By standard integration we can easily write:

$$ S_N = S_{N-1} \exp\left( -\frac{\sigma^2}{2}dt + \sigma \sqrt{dt}Z\right) $$

where $Z \sim N(0, 1)$.

Of course we can also express $S_N$ in terms of $S_0$:

$$ S_N = S_0 \exp\left( -\frac{\sigma^2}{2}T + \sigma \sqrt{T}Z\right) $$

Depending on choice we do we get different distribution for $S_N$. In the first case we have: $$ p_{S_{N}, S_{N-1}}(x) = \frac{1}{\sqrt{2\pi} \sigma \sqrt{dt} x}\exp \left( - \frac{ \left(\ln(x) + \frac{\sigma^2}{2}dt - \ln(S_{N-1})\right)^2}{2\sigma^2 dt} \right) $$ i.e. a log normal distribution with mean $-\frac{\sigma^2dt}{2} + \ln(S_{N-1})$ and standard deviation $\sigma \sqrt{dt}$.

In the second case we have:

$$ p_{S_{N}, S_{0}}(x) = \frac{1}{\sqrt{2\pi} \sigma \sqrt{T} x}\exp \left( - \frac{ \left(\ln(x) + \frac{\sigma^2}{2}T - \ln(S_0)\right)^2}{2\sigma^2 T} \right) $$

Of course the distribution is smooth wrt $S_0$ and but the payoff is not. More over the payoff function $$ f(x, K) = \sqrt{ (x-K)^+ } $$ is different from usual call option pay off $(x-K^+)$ since its derivative has integrability issue due to the square root. That's why I am trying to do the computation via Vibrato Montecarlo, but I have still trouble as I write below

Giles (the Great) explains that for a given parameter $\theta$ we can compute the sensitivity even if payoff is discontinuous by using LRM:

$$ \partial_\theta \mathbb{E} \left[ f(S_T) \right] = \partial_\theta \int f(S_T) p(S_T, \theta) dS_T = \int f(S_T) \partial_\theta \log\left( p(S_T, \theta)\right) p(S_T, \theta) dS_T = \mathbb{E} \left[ f(S_T) \log\left( p(S_T, \theta)\right) \right] $$

Until know I've just used the usual procedure of Vibrato Montecarlo. Now my question:

How should I behave in case I have to compute delta sensitivity? Infact in this case the derivative can not "pass over" $f(S_T)$ term since it depends on the paramter (which is $S_{N-1}$).

The problem here is that $f$ is not differentiable so that I am messed up at this point.

Notice that I've asked this question few time ago but in the answer Quantuple asked for some regularity for payoff function $h$.

  • $\begingroup$ When trating $S_T$ as a random variable, $f(S_T)$ does not depend on $S_0$, but $p(S_T)$ does. $\endgroup$
    – Gordon
    Feb 15 '19 at 15:37
  • $\begingroup$ @Gordon I do not understand. In my case both payoff and distribution depend on $S_{T-1}$ so that both depend on $S_0$. Can you give details about what do you mean? Thx $\endgroup$
    – clarkmaio
    Feb 16 '19 at 20:34
  • $\begingroup$ By treating $S_T$ as a random variable, and then derive its density $f$, you will notice the density depends on $S_0$. Then $E(f(S_T) = \int_0^{\infty} x f(x, S_0) dx$. $\endgroup$
    – Gordon
    Feb 16 '19 at 21:23
  • $\begingroup$ @Gordon I see what are you saying. $S_T$ can be of course expressed in terms of $S_0$ in the same way I have wrote it in term of $S_{T-1}$. But if I do as you said, i.e. the standard integration, I came up with a not differential function in the integral. Remember that I have to integrate the expected value wrt $S_0$. This is why I am trying to use vibrato Montecarlo $\endgroup$
    – clarkmaio
    Feb 16 '19 at 21:40
  • $\begingroup$ The density is a smooth function of $S_0$. $\endgroup$
    – Gordon
    Feb 16 '19 at 23:47

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