volatility - Distribution of Black-Scholes option price

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.

Answer

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 $\tilde{s}^2$: $$ V^+ = \text{BSCall}(\sigma^+),\quad V^- = \text{BSCall}(\sigma^+)$$

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[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?