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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 $\beta=0$?

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    $\begingroup$ 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. $\endgroup$ – dm63 Sep 16 '18 at 12:55
  • $\begingroup$ Could you provide a reference for this statement? $\endgroup$ – LocalVolatility Sep 16 '18 at 13:51
  • $\begingroup$ 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) $\endgroup$ – dm63 Sep 16 '18 at 14:24
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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 $\sigma_t^2(T-t)$: $$ S_T \sim N(S_t,\sigma_t^2(T-t))$$

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Given (conditional on) a realisation of the volatility, it is normal for $\beta = 0$ and lognormal for $\beta = 1$

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The longer term process is not (unless alpha=0).

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