Conditional SABR Distributions at Beta Zero and One
Summary
The document explains two special cases of the SABR model by conditioning on a realized volatility path. With beta equal to one, the asset follows a multiplicative diffusion, so its logarithm has a normal distribution and the asset level is lognormally distributed. With beta equal to zero, the diffusion is additive, giving a normally distributed asset level. The argument uses the corresponding stochastic differential equations and their integral forms.
The key limitation is that these distributional claims are conditional on realized volatility; they do not establish that the unconditional SABR distribution has the same form. A brief response also cautions that the longer-term process need not retain these distributions when volatility itself evolves. The equations in the document contain apparent notation errors in the stated means and variances, so the central lesson is the conditional distinction between multiplicative and additive diffusion rather than those displayed parameter expressions.
Key ideas
- At beta one, the asset follows a multiplicative diffusion and its logarithm is conditionally normal given the volatility path.
- At beta zero, the asset follows an additive diffusion and its level is conditionally normal given the volatility path.
- These special-case distribution claims are conditional and do not automatically describe the unconditional SABR distribution.
- The document's displayed distribution parameters appear inconsistent with its stochastic integral expressions.
Tags
Full text
# How to show that SABR is log-normal for $\beta=1$ and normal for $\beta=0$? # How to show that SABR is log-normal for $\beta=1$ and normal for $\beta=0$? For $\beta = 1$ SABR is log-normally distributed and for for $\beta = 0$ SABR is normally distributed. This is a very common property mentioned in almost every paper about SABR. But I can't find the mathematical derivation (It might be really simple to derive), which leads to my question: How I proof that SABR is log-normal for $\beta=1$ and normal for $\beta=0$? ## Answer by Sanjay (score 3) https://quant.stackexchange.com/a/43147 First and foremost it is important to clarify that the underlying is not necessarily normal/lognormal but for the special cases of $\beta$ the underlying is normal/lognormal Conditioned on a realization of the volatility. As mentioned in the answer by @ilovevolatility. Simple stochastic calculus will show the properties you mentioned. For realized volatility the following holds: $$ dS_t = S_t^\beta\sigma_tdW_t, $$ For $\beta=1$, $dS_t=S_t\sigma_tdW_t$, $S_t$ becomes a geometric brownian motion which means that at time $t$ the distribution of $\log S_t$ is given: $$ \log S_T \sim N(S_t,\sigma_t^2(T-t)) $$ For $\beta=0$: $$dS_t=\sigma_tdW_t$$ which can be written in integral form $$ S_T=S_t + \int^T_t \sigma_t^2 dW_u $$ According to stochastic calculus theory the integral is normally distributed with mean zero and variance $\sigma_t^2(T-t)$: $$ S_T \sim N(S_t,\sigma_t^2(T-t))$$ ## Answer by user34971 (score 1) https://quant.stackexchange.com/a/41781 Given (conditional on) a realisation of the volatility, it is normal for $\beta = 0$ and lognormal for $\beta = 1$ ## Answer by Charles Fox (score 0) https://quant.stackexchange.com/a/41757 The longer term process is not (unless alpha=0).
Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.