# Ultra Powerfull Vibrato Montecarlo for delta sensitivities of a not regular payoff

Ciao,

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

$$\sqrt{(S_T - K)^+}$$

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 $$h$$.

• When trating $S_T$ as a random variable, $f(S_T)$ does not depend on $S_0$, but $p(S_T)$ does. – Gordon Feb 15 '19 at 15:37
• @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 – clarkmaio Feb 16 '19 at 20:34
• By treating $S_T$ as a random variable, and then derive its density $f$, you will notice the density depends on $S_0$. Then $E(f(S_T) = \int_0^{\infty} x f(x, S_0) dx$. – Gordon Feb 16 '19 at 21:23
• @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 – clarkmaio Feb 16 '19 at 21:40
• The density is a smooth function of $S_0$. – Gordon Feb 16 '19 at 23:47