I am reading Taleb "Fooled By Randomness", and the author says that a 15% return with 10% volatility translates to 93% success in a year and 50.02% success in any given second.
Could someone help me understand this calculation?
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$\begingroup$ =1-NORM.DIST(0;0.15;0.1;1) is something around 93.3 %, scaling the mean by 1/365*24*60 and the volatility by =1/SQRT(365*24*60) gives you something around 50.08% $\endgroup$– T123Commented Dec 5, 2023 at 13:49
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$\begingroup$ Cross-posted on Cross Validated here. $\endgroup$– Richard HardyCommented Dec 5, 2023 at 21:17
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1$\begingroup$ I’m voting to close this question because it is cross-posted without a good explanation. $\endgroup$– AlperCommented Dec 6, 2023 at 13:23
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1 Answer
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It's probably a simple textbook example to illustrate Taleb's point. Suppose $\text{d}S_t=\mu S_t \text{d}t+\sigma S_t \text{d}W_t$ with $\mu=0.15$ and $\sigma=0.1$.
By Itô's lemma, the log return follows a normal distribution, $$R=\ln(S_T/S_t)\sim N\left(\left(\mu-\frac{1}{2}\sigma^2\right)(T-t),\sigma^2 (T-t)\right).$$
Then, your success probability is $$\mathbb{P}[R>0]=1-\mathbb{P}[R\leq0]=1-\Phi\left(-\frac{\left(\mu-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right)=\Phi\left(\frac{\mu-\frac{1}{2}\sigma^2}{\sigma}\sqrt{T-t}\right),$$ where $\Phi$ is the cdf of a standard normal random variable. Note that $\frac{0.15-\frac{1}{2}0.1^2}{0.1}=1.45$.
- The probability of earning money over one second is $\Phi\left(1.45\cdot \sqrt{\frac{1}{365\cdot24\cdot60}}\right)=50.08\%$.
- The probability of a positive return over one year is $\Phi(1.45\cdot\sqrt{1})=92.65\%$.
