# 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 $\beta=0$?

• I don't think it's true that the underlying distribution is perfectly lognormal/normal in those cases. The presence of stochastic vol will give the distribution fat tails, for example. – dm63 Sep 16 '18 at 12:55
• Could you provide a reference for this statement? – LocalVolatility Sep 16 '18 at 13:51
• Hagan's paper Managing Smile Risk has a chart of the smile for Beta=0 and 1 (if the implied vol has a smile then the distribution must be fat tailed) – dm63 Sep 16 '18 at 14:24

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 $$\sigma_t^2(T-t)$$: $$S_T \sim N(S_t,\sigma_t^2(T-t))$$
Given (conditional on) a realisation of the volatility, it is normal for $\beta = 0$ and lognormal for $\beta = 1$