3
$\begingroup$
I would like to validate my understanding of IV as a prediction tool. Black-Scholes model is based on the assumption that rate of return of a stock is a Wiener process: $$ \frac{dS_t}{S_t} =\mu \,dt+\sigma \,dW_{t} $$ My take is that the rate of return (ROR) less the drift component has a volatility of ${\sigma}\sqrt{t}$. But many interpretations of IV state that the stock PRICE itself is a normal distribution and has a 68.2% of chance to stay within $[S_0 - \sigma, S_0 + \sigma]$. such as the the article linked here: https://www.ally.com/do-it-right/investing/what-is-implied-volatility/

I found this interpretation has three pitfalls:

1. what the model says is ROR follows a normal disrtibution with $E={\mu}t$ and $Var={\sigma^2}t$, so the price itself is a log-normal distribution which has a different Variance (and confidence interval)

2. the interpretation does not take into account the drift compoent which increases as time increases.

3. the ROR Variance also increases with time and not a constant.

Hence it will be very difficult to correctly use IV as a tool to predict the range of stock price movements. If you see my understanding incorrect, please kindly give your comments. Thanks in advance.

$\endgroup$
4
  • $\begingroup$ You are correct that it's an approximation. If you then know that it's an approximation, you can happily use the results knowing that they are only approximate. $\endgroup$
    – will
    Oct 13, 2018 at 9:09
  • $\begingroup$ To do it accurately, work with the log of stock price. Determine the endpoints of the confidence interval in terms of logs (the log of S has a known mean and variance), then take the exponential of these two numbers to obtain the endpoints for S. $\endgroup$
    – nbbo2
    Oct 13, 2018 at 14:17
  • $\begingroup$ noob2: thanks for the idea. my understanding is that B-S model assumption clearly shows both mean and variance have time as their components. are you suggesting using the current S as instantaneous mean and solved IV as instantaneous standard deviation and assume both have already factored in the time variable? $\endgroup$
    – techie11
    Oct 13, 2018 at 14:51
  • $\begingroup$ You need to know 3 numbers:$\mu,\sigma,t$ then the mean of the log is $\mu t$ and the standard deviation is $\sigma \sqrt{t}$. $t$ is known (the time horizon), $\sigma$ can be estimated from options, the real word $\mu$ is difficult/impossible to estimate accurately, so you have to make an assumption about plausible long run stock returns. $\endgroup$
    – nbbo2
    Oct 14, 2018 at 21:49

1 Answer 1

1
$\begingroup$
  1. The Bachelier model which assumes that price follows a normal distribution is a correct approximation for the Black-Scholes one for short times t. When time is short it's fine to ignore drift because the movements will be more driven by volatility
  2. For longer time intervals you cannot ignore drift any more. But the actual distribution of the stock price at horizon T may be very different from the one implied by the options market because the latter one describes the stock in a very specific measure called the "risk neutral" measure.

So overall I would answer that if your goal is to get insight about the distribution of the stock at a certain date T you first need to express whether you are interested by this distribution in the real world measure or the risk neutral measure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.