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Quick question regarding the conditional distributions (SABR is just an example here)

Consider $$dS_t = \sigma_tS_tdW_t$$ $$d\sigma_t = \alpha\sigma_tdV $$ $$dW_tdV_t=\rho dt$$

Hence a SABR process with $\beta=1$. The volatility process is a GBM and so we can implement the exact solution and simulate $\sigma_{i+1}$ from $\sigma_i$. Now I dont't know what mathematical terminology to use on $S_t$

When $V_{i+1}$ and $V_{i}$ is KNOWN we know the exact value of $\sigma_{i+1} $ from $\sigma_i$.

With $\sigma_t$ known, we can simulate $W_{i+i}-W_{i}$ and a proper way to compute $S_{i+1}$ from $s_i$ is as a GBM with volatility $\sigma_i$. This is how it is done in practice with these parameter.

My Question: To which extend can we call $S_{i+1}$ as exact?

My own personal take: This is not exact at all because we have to know the whole path of $[\sigma_i,\sigma_{i+1}]$ to call it exact.

The reason for my confusion is that people call SABR for $\beta = 1$ for log-normal for a realized volatility.

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  • $\begingroup$ I think I understand your point, but consider Black-Scholes with a deterministic volatility term structure. To what extent is a discretisation of BS with vol term structure exact? It's not exact either if you apply your logic above (which involves discretisation) to it. But of course BS is exact in continuous time, just as SABR is lognormal for $\beta = 1$, conditional on a realized vol path. $\endgroup$
    – user34971
    Jul 15, 2019 at 15:49

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No, the simulation is not exact in general, precisely for the reason you mentioned. By "exact", it is meant that there is no discretization error in time. Of course, there will always be a Monte-Carlo sampling error.

For the Black-Scholes model, the simulation is exact if you simulate the log asset, as it is a standard arithmetic Brownian motion, and then you just compute the exponential of the log asset on each path. There is no discretization error, the integral over a time interval is computed exactly.

For the lognormal SABR model ($\beta=1$), using the log-asset formulation, you can compute the integral $\int_{0}^{t} \sigma^2(u) du$ exactly, but you will still have the term $\int_{0}^{t} \sigma(u) dW(u)$ to compute. In general, one will use an approximation for this.

Now, in reality, I believe there is actually a way to compute this distribution exactly, and this is used to compute the closed form formula for the price of a vanilla option with $\beta=1$, but this closed-form formula involves a double integral over non trivial functions (this can be found in Pierre Henry-Labordère book "Analysis, Geometry, and Modeling in Finance"). There are also mathematical papers around this stochastic integral. And for a Monte-Carlo simulation, it may not be a good idea to use such a complex formula as it will be very slow in general.

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