# Estimation Risk-Neutral Variance of Returns

I am trying to find a method which allows me to estimate $Var_{\mathbb{Q}}\left(\frac{S_{t_{i+1}}}{S_{t_i}}\right)$ where $S$ denotes the price process of an underlying stock (which has to be assumed to be stationary) and $\mathbb{Q}$ should be the risk-neutral (pricing) measure. A first approach would rely on the Breeden-Litzenberger (1978) result which enables to compute marginal distributions of $\mathbb{Q}$ by observing prices of plain-vanilla options and by using: $$Var_{\mathbb{Q}}\left(\frac{S_{t_{i+1}}}{S_{t_i}}\right)= \mathbb{E}_\mathbb{Q}\left[\frac{1}{S_{t_i}^2}\mathbb{E}_\mathbb{Q}[S_{t_{i+1}}^2|S_{t_i}]\right]-1.$$ Next, I get $\mathbb{E}_{\mathbb{Q}}[S_{t_{i+1}}^2|S_{t_i}]$ by observations of option prices at time $t_i$. The value of $\mathbb{E}_\mathbb{Q}\left[\frac{1}{S_{t_i}^2}\right]$ can be computed similiar by using stationarity. Nevertheless I am not satisfied with this approach. When talking about estimations I will face the problem that these estimations are made under the physical measure. Does anybody know an approach allowin to estimate the variance of the returns under the risk-neutral measure?

• A bit early in the morning here, but I don't see any vanilla prices anywhere in these formulas. How do you get $\mathbb{E}_{\mathbb{Q}}[S_{t_{i+1}}^2|S_{t_i}]$ from vanillas? As for the question, maybe through calibration of some stoch-vol model to the vanilla market? Dec 22 '17 at 13:25
• I assume stationarity. This means -roughly spoken - past value had the same distribution than future values. So, I take a historical sample of option prices and observe vanilla options with maturity $|t_{i+1}-t_i|$. delivering the marginal distribution at $t_{i+1}|$ which allows to compute the desired conditional expectation. Thanks for your comment! I would like to avoid model-depending approachs and really prefer a model-independent approach only relying on historical data. Dec 22 '17 at 14:27
• I can't see how you can do any risk-neutral analysis without assuming some model for the option prices. Dec 22 '17 at 15:55
• The idea is to use market data about option prices. For this one assumes that vanillas are priced under a risk-neutral pricing measure $\mathbb{Q}$, then the second derivative of the option price w.r.t the strike $K$ yields the density of the marginal distribution at maturity at point $K$. See Breeden-Litzenberger (1978) for this result. Dec 23 '17 at 11:32

Just trying (I'm not sure I've well understood the question).

I will assume the ususal risk neutral dynamic for $S_t$: $$dS_t = rS_t dt + \sigma S_tdW_t$$ so that $\forall T>t$ we have:

$$S_T = S_te^{\left(r-\frac{1}{2}\sigma^2\right)(T-t) + \sigma W_{T-t}}$$

At this point the computation is quite easy and straight forward.

$$\mathbb{E}\left[ \frac{S_T}{S_t}\right] = e^{r(T-t)}$$

\begin{align} \mathbb{E}\left[ \frac{S^2_T}{S^2_t}\right] &= \mathbb{E}\left[ e^{(2r - \sigma^2)(T-t) + 2\sigma W_{T-t}}\right] \\ & = e^{(2r - \sigma^2)(T-t)} \mathbb{E}\left[ e^{2\sigma W_{T-t}}\right] \\ & = e^{(2r - \sigma^2)(T-t) + 2\sigma^2(T-t)}\\ & = e^{2r(T-t) + \sigma^2(T-t)} \end{align}

so that we get:

$$Var\left( \frac{S_T}{S_t}\right) = e^{2r(T-t) + \sigma^2(T-t)} - e^{2r(T-t)} = e^{\sigma^2(T-t)}$$

This is the general result. Of course if we put $T = t_{i+1}$ and $t = t_i$ we get the result we were looking for.

• Thanks a lot for your response which yields just the definition of $\sigma$ in the Black-Scholes model (1 year return vol.). Unfortunately I don't want to assume the geometric Brownian dynamics. My aim is to estimate the risk-neutral variance of the returns by using historical stock data. The standard estimator would of course be $\frac{1}{n-1}\sum_{i=1}^n \left( \frac{S_{t_{i+1}}}{S_{t_i}}-1/n\sum_{i=1}^n \frac{S_{t_{i+1}}}{S_{t_i}} \right)^2$ which producs an estimation for the variance under the physical measure. I want to get en estimation under the risk-neutral measure. Dec 22 '17 at 15:14

Let me throw some water on your goal and any nice proof. For my article on this, you can find it at:

Harris, D.E. (2017) The Distribution of Returns. Journal of Mathematical Finance, 7, 769-804

Let us use even weaker assumptions than your assumption that $S_t,\forall{t}$ is stationary. Let us use more Markowitz style assumptions.

Our first assumption is that there are very many buyers and very many sellers. Normally this is to motivate the absence of liquidity costs, but we are going to repurpose it as it has other consequences that no one noticed.

Stocks are sold in a double auction. Because of this, there is no winner's curse. As a consequence, the rational behavior is to bid your expectation. With many buyers and many sellers bidding their expectation, the limit book will converge to the normal so that as the number of bids becomes large enough, the limit book will be normally distributed.

We could also just assume that stock prices are drawn from a normal distribution. The weakness of that assumption is that it does not cover things such as auctions at Christie's which are subject to the winner's curse. See the paper for the solution to that problem.

So, let $R_t=\frac{S_{t+1}}{S_T}$. We will call this the reward for investing. Subtracting one makes it the return on investing. We will ignore the $-1$ as it changes nothing and is just a little bit of extra work.

Now the question is what is the distribution of $R_t$ as $S_t,S_{t+1}$ are actual data while $R_t$ is not data, but rather a statistic; that is to say, it is a function of the data.

As is well known, from Curtis at:

Curtiss, J.H. (1941) On the Distribution of the Quotient of Two Chance Variables. Annals of Mathematical Statistics, 12, 409-421,

the solution to any ratio of continuous random variables, where $Z=\frac{Y}{X}$ is $$p(z)=\int_{-\infty}^\infty|x|f(x,zx)\mathrm{d}x$$ For the normally distributed variables that are in equilibrium, the solution is very well known and goes back in various forms to Fermat and Cardano as $$\frac{1}{\pi}\frac{\sigma}{\sigma^2+(z-\mu)^2}.$$

There is an assumption of allowing infinitely negative returns. If you restrict the domain then the constant of integration changes from $$\pi^{-1}$$ to $$\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}.$$

For our purposes, the constant of integration does not matter, though it will create a serious error in estimation if you drop it in the real world.

The above distribution is famous for a variety of reasons. When Laplace first sent his proof of what we now call the "central limit theorem" to his former student Poisson, Poisson returned the proof to him with an exception to when the rule holds. It fails to hold when the distribution is as above. From that observation, when that distribution is present, then you can no longer use things such as t-tests, F-tests and so forth, subject to the qualification that as the sample size exceeds 100, the t-test will work if you hold it to one degree of freedom.

You can find a discussion of this at:

Fama, E. F. and Roll, R. (1968). Some properties of symmetric stable distributions. Journal of the American Statistical Association, 63(323): pp. 817–836.

However, the Fama and Roll discussion does not apply to the case of limiting liability to $-100%$. I am building a separate discussion for those in another paper.

The next appearance of this distribution is in a battle between Augustin Cauchy and Irénée-Jules Bienaymé. Augustin Cauchy had just produced a method of regression in a journal article. Bienaymé produced an article that showed that ordinary least squares was the "best" way to do regression. Cauchy took this as a personal attack and then went to work to determine when OLS will ALWAYS fail with probability 1.

Whenever the above distribution is present, OLS will produce purely spurious results. The reason is that the above distribution, which has acquired the moniker "the Cauchy distribution," has no mean and so cannot have a variance.

While the Cauchy principal value is $\mu$, higher moments do not exist, even about the Cauchy principal value. The second raw moment is infinity or does not exist depending on how you define the integral.

As to estimating the scale parameter of returns, you cannot use a non-Bayesian method. There does not exist an unbiased admissible Frequentist estimator for real data. I have estimated the scale parameter for all disaggregated equity securities in another paper, but for one security, what you should do is solve: $$\Pr(\sigma|\mathbf{R},\mu)=\int_{-\infty}^\infty\frac{\prod_{i=1}^n{\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}}\frac{\sigma}{\sigma^2+(R_i-\mu)^2}\Pr(\mu;\sigma)}{\int_0^\infty\int_{-\infty}^\infty{\prod_{i=1}^n{\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}}\frac{\sigma}{\sigma^2+(R_i-\mu)^2}\Pr(\mu;\sigma)}\mathrm{d}\sigma\mathrm{d}\mu}\mathrm{d}\mu.$$

The posterior density of $\sigma$ is well behaved. If you need a point estimate, you can minimize a cost function over the density. You should even be able to use quadratic loss because, for a sufficiently flat prior, the posterior density should converge to the ratio distribution of two standard deviation distributions. I haven't taken the time to prove that, however. It is possible that it is not true, but it should be as $\sigma$ is the ratio of the standard deviation of $S_{t+1}$ and the standard deviation of $S_t$.

You will want to restrict your prior probabilities, $\Pr(\mu,\sigma)$ to proper priors as I have not found generalized Bayes rules in the literature for the truncated case and there is no reason to believe that the posterior is well behaved under the joint distribution of $(\mu,\sigma)$ with a uniform or other improper prior.

From this, it should be sufficient to argue that mean-variance finance cannot exist. Consequently, any $\beta$ style model in raw data is invalid. In log-transformed data, it is suspect as the likelihood function is the hyperbolic secant distribution and it admits nothing resembling a covariance matrix. Since nothing can covary, what are you measuring?

This is not to say they cannot co-move. When looking at multiple firms returns of the firms cannot be independent, though asymptotically none of them can covary. This is part of what makes this distribution famous. The variables are not independent, but they do not covary as the sample size goes to infinity.

Finally, risk-neutral behavior cannot exist at the margin. I know I am making your day.

There are two arguments for this. The first isn't a true argument, but should warrant a pause. If you assume risk-aversion, then from deFinetti coherence principle and assumption of a willingness to accept all finite bets at stated prices, then Kolmogorov's axioms fall out as theorems. If you do not assume risk-aversion, then this does not happen. You then have to add the assumptions that: $$\Pr(A)\ge{0},$$ $$\Pr(\Omega)=1,$$ and for any countable sequence of disjoint sets $$\Pr(\cup_{i=1}^\infty{A_i})=\sum_{i=1}^\infty\Pr(A_i).$$ One should give a moment of pause when nature provides a solution that minimizes both the assumptions and matches reality sufficiently often.

The second argument is from rationality. If the marginal actor was risk-loving then they would pay a premium to take a risk. This is the same as saying that $K_{t+1}=RK_t+\epsilon_{t+1},R<1,\forall{t}$. Given sufficient time the capital stock of the planet would go to zero and all humans would die.

This does not mean that risk-loving actors do not exist, nor does it mean that they are never the marginal actor. It implies that they can be the marginal actor only a minority of the time.

The assumption of risk-neutrality was only ever a mathematical convenience created by using the normal distribution. The probability of risk-neutrality must be zero from this second argument.

It goes like this, risk-neutrality exists at exactly one point. A single point over a continuum of possible points has measure zero and hence a probability of zero. Even if it were true, it could never be measured and risk-loving behavior is impossible. Hence, by being required to use Bayesian statistics, risk-neutral behavior is functionally excluded as a possibility.

For an extended discussion of the Cauchy distribution see: Why the Cauchy Distribution Has No Mean