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Consider a Black-Scholes formula for option pricing on conditional interest rate $r=0$ for European call option. Let's write it as $BSCall(\sigma)$ and investigate it as a function of unknown non-random constant volatility.

We can look at historical asset price data and compute sample variance $s$ which distribution law depends on unknown true $\sigma$. It has an important property of $\dfrac{(n-1)\text{s}^2}{\sigma} \sim \chi_{n-1} ^2$.

Due to MLE plug-in property we know that if $\hat{\sigma}$ is an MLE for volatility $\sigma$ then $BSCall(\hat{\sigma})$ is an MLE for Black-Scholes price.

But this is just a point estimator. Is there any approach to compute the variance or confidence interval (HDI) for option price under volatility estimated with sample variance?

I tried to use a trick from expected-value-of-black-scholes but it seems to be incorrect since rewriting $\sigma = \dfrac{(n-1)\text{s}^2}{\chi_{n-1} ^2}$ brings some exogeneous uncertainty by sampling Chi-squared.

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I think the easiest way to to what you want is to use confidence intervals (statistical inference).

In other words, assuming the population has a true variance $\sigma$, the sampling distribution of the variance $s^2$ of an $n$-sample verifies: $$ \frac{s^2(n-1)}{\sigma^2}\sim \chi^2_{n-1}$$

You can exploit this result to build an $1-\alpha$ confidence interval for the population variance ($\alpha \in [0,1]$, typically $\alpha=5\%$).

Indeed, for a confidence level $1-\alpha$, the following equality holds: $$ z_{\alpha/2} \leq \frac{s^2(n-1)}{\sigma^2} \leq z_{1-\alpha/2} $$

where $z_q$ figures the quantile $q$ of a chi-squared distribution with $n-1$ degrees of freedom i.e. $$ X \sim \chi^2_{n-1},\quad \Bbb{P}(X \leq z_q) = q $$

Given a sample variance $\tilde{s}^2$, one can therefore turn the inequality on its head, to write, for a confiedence level $1-\alpha$: $$ \frac{\tilde{s}^2(n-1)}{z_{1-\alpha/2}} \leq \sigma \leq \frac{ \tilde{s}^2(n-1)}{z_{\alpha/2}} $$ Hence the upper and lower bounds of your $1-\alpha$ confidence interval for the (unobserved) population variance: \begin{align} \sigma^+ = \frac{ \tilde{s}^2(n-1)}{z_{\alpha/2}},\quad \sigma^- = \frac{\tilde{s}^2(n-1)}{z_{1-\alpha/2}} \end{align}

This could then help you construct $1-\alpha$ confidence bounds on the BS option price given the measure sample variance $\tilde{s}^2$: $$ V^+ = \text{BSCall}(\sigma^+),\quad V^- = \text{BSCall}(\sigma^+)$$


[Edit]

Given your desire to obtain a full distribution, why not opt for a Bayesian approach?

Assume the true population variance $\sigma^2$ follows a certain prior distribution with hyperparameter $\alpha$, $p(\sigma;\alpha)$ over $\Bbb{R}^+$.

Suppose that, for a specific sample, you measure a sample variance $s^2$ and wish to compute the posterior of the population variance. Bayes' rule gives: $$ p(\sigma^2 \vert s^2, \alpha) = \frac{p(s^2 \vert \sigma^2)}{\int_0^\infty p(s^2 \vert \sigma^2) p(\sigma^2;\alpha) d\sigma^2 } p(\sigma^2; \alpha) $$

Now you know:

  • The prior distribution $p(\sigma^2; \alpha)$: you postulated it.
  • The sampling distribution $p(s^2 \vert \sigma^2)$: $\quad s^2 \sim \sigma^2/(n-1) \chi^2_{n-1}$

Hence you have everything you need to compute the posterior distribution.

Obviously, if you stick with the Maxium A Posteriori (MAP) estimator, once again you'll have a pointwise estimate, so I suggest you to perform the full integration. Off the top of my head chi-squared distributions does not allow for conjugate priors so you might have to resort to numerical integration (e.g. adaptive quadrature and the likes).

Finally, the choice of hyper-parameter $\alpha$ will have an impact on the resulting posterior: you might want to set $\alpha$ so that the prior distribution is centered around the sample variance for instance?

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  • $\begingroup$ In this case $[V^-, V^+]$ has exactly wanted probability but it may be not a Highest Density Interval. And unfortunately this approach doesn't let to compute variance. $\endgroup$ – Denis Korzhenkov Sep 30 '16 at 12:33
  • $\begingroup$ I see, since your question mentioned "confidence interval" I thought you only needed this. $\endgroup$ – Quantuple Sep 30 '16 at 12:38
  • $\begingroup$ @DenisKorzhenkov, I've also edited my answer to include a word on the Bayesian approach? $\endgroup$ – Quantuple Sep 30 '16 at 12:58
  • $\begingroup$ I've already thought about Bayesian approach and looked through some articles, e.g. this one to see how to choose prior for $p(\sigma^2)$. I'm only confused that the result depends on prior. But now MCMC packages seem to be the best solution. $\endgroup$ – Denis Korzhenkov Sep 30 '16 at 13:19
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You have bigger problems to worry about than fiddly mathematics for confidence intervals. To wit, the sample variance is an estimate of past variability in the underlying. The volatility in option pricing models is future volatility.

It is well know that there are structural reasons past variability is biased smaller than future variability under the Black-Scholes model. We actually observe that in the financial markets. Let's look at a simplified explanation.

Consider a more financially plausible model than Black-Scholes: one where the stock can suddenly go bankrupt due to fraud, and the volatility varies over time. Neither model is perfect, but the new one (call it SVJ) will be "less wrong".

Mathematically, we no longer have the Black-Scholes SDE based on a single stochastic generator $W$

$$ \frac{dS}{S} = \mu dt + \sigma dW $$

but rather an SDE with 3 generators: $W,Z$ and a jump process $J$

$$ \frac{dS}{S} = \mu dt + \sigma dW - dJ \\ d\sigma^2= \kappa(\bar{\sigma}^2-\sigma^2) dt + \eta \sigma^2 dZ $$

If we were to back out Black-Scholes volatility $\sigma_{BS}$ from option prices generated from the SVJ world,

  • there would be skew, and
  • we would find $\sigma_{BS} > \sigma$.

Even if $\eta=0$, since no defaults have been observed our observations all come from the cases where we have only followed $W$. Therefore observations of historical variance would cause us to underestimate $\sigma_{BS}=\sigma$.

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