I did a test about quantitative finance. One of the question was :

What is the probability, in the Black-Scholes world, that the realized volatility is larger the implied volatility ? And why ?

I couldn’t find any answer on the internet. Could you help me please ?

Thank you very much for your help

  • 1
    $\begingroup$ I would say that in a black-scholes world the expected variance risk premium is zero. So the ex-ante probability that realized volatility is larger than implied volatility is zero. $\endgroup$
    – phdstudent
    Commented Mar 31, 2016 at 19:48
  • $\begingroup$ I do not understand why the probability would be zero. Could you be more specific please ? Can't we use the fact the r_t follows a normal law with mean (µ-vol^2/2)t and variance vol^2*t in order to have the law of r_t^2 and then compute the probability ? $\endgroup$ Commented Mar 31, 2016 at 20:33
  • 1
    $\begingroup$ How is "realized volatility" defined? If it is just the sample standard deviation then the probability follows a chi-distribution. $\endgroup$
    – user9403
    Commented Mar 31, 2016 at 21:25
  • $\begingroup$ Yes I think the realised volatility is the standard deviation. So, what would be the answer ? and why ? $\endgroup$ Commented Mar 31, 2016 at 22:04
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    $\begingroup$ In the BS world the implied vol is not greater than the realized, not smaller either, but equal. "In the theory of general relativity what is the probability that $E>M c^2$" LOL! $\endgroup$
    – Alex C
    Commented Mar 31, 2016 at 22:35

2 Answers 2


I succeeded to have the expected answer, and I was on the right way :

The implied volatility is the ATM IV The realised volatility is defined as the absolute value of the log return

We know that a gaussian random variable is in the range +- 1*sigma with a probability of 68%. So, the correct answer is approximately 32%

Should you have any comment I am interested

  • $\begingroup$ Ex-ante the expected probability is still zero (which is what we care about anyway). Ex-post, your argument is correct. $\endgroup$
    – phdstudent
    Commented Apr 1, 2016 at 17:12
  • $\begingroup$ @unkownhuman Your answer does not answer the question. You just adress the change of the stock price from start till maturity 1, that is $\ln(S_1) - \ln(S_0)$. Hence, you omit the variance from $0 \rightarrow 1$ and just concentrate on the first moment. $\endgroup$
    – Phun
    Commented Apr 2, 2016 at 23:32

It's clear that over time $t\rightarrow t+dt$ the (expected) implied variance in the BS world (under the risk-neutral measure $\mathbb{Q}$) is $\sigma^2 dt$.

The realized variance of an arbitrary single log stock path (under the physical measure $\mathbb{P}$) is $$\sum_{i=1}^\infty \sigma^2(W_{i+1}-W_i)^2 \approx \sum_{i=1}^n \sigma^2(W^{(n)}_{i+1}-W^{(n)}_i)^2 \sim \sigma^2 \frac{dt}{n}\chi_n$$ for large $n$ and $W^{(n)}_{i+1}-W^{(n)}_i \sim \mathcal{N}(0,\frac{dt}{n})$ i.i.d.. Now $$ \text{Prob}(\sigma^2 \frac{dt}{n}\chi_n > \sigma^2 dt) = \text{Prob}(\chi_n > n) = \text{Prob}(\gamma(\frac{n}{2},\frac{1}{2}) > n),$$ where $\gamma(\frac{n}{2},\frac{1}{2})$ is the gamma distribution. The last expression can be solved numerically.

  • $\begingroup$ doesn't this imply that for large n, the probability is 1/2 ? $\endgroup$
    – dm63
    Commented Nov 19, 2016 at 13:54

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