5
$\begingroup$

I am trying to predict the future probability of stock returns based on the return distribution. Therefore I calculate the returns as $\frac{P(t)}{P(t-1)}$ for the whole daily data and fit a probability density function $f(x)$ to the data. Now the probability that the future return lies in the interval $[a,b]$ should be given as $\int_a^b f(x) dx$.

enter image description here

1.) What are some caveats about this approach?

I am particular concerned about how changing volatility could bias my results. As far as I understand it, my fitted pdf will take an average volatility value and yield the return probabilities based on that.

2.) How could I account for changing volatility? Pre selecting data on the most recent data?(but: fit will not be well with less data) Maybe usage of of volatility models?

3.)Are there different approaches to calculate the probability of future returns?

$\endgroup$
1
  • $\begingroup$ You can imply the future distribution of a stock from the second derivative of the u discounted option prices. Try comparing this to the distributions you're generating (which won't account for the market's expectation of the outcomes of any upcoming events...). $\endgroup$
    – will
    Commented Jun 17, 2017 at 16:01

1 Answer 1

6
$\begingroup$

I have written an entire paper on this approach at https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2828744

As to your specifics

1) "Volatility" as defined by variance does not exist, which is why it is changing. The first moment is undefined so the second cannot exist. See the paper as to why. Your fitted pdf will treat the outcomes as having a single, joint scale parameter. You won't want to do fitting like this, though, because you won't be able to use a sufficient statistic to create it and so must lose information as to the true location. A Bayesian method will correct for this.

2) You do not need to, it is supposed to change. You will have to use a Bayesian method because there does not exist an admissible non-Bayesian method. The Bayesian likelihood function always is minimally sufficient and a statistic created using Bayesian methods is always admissible.

3) Yes, it is called the Bayesian Predictive distribution. It is defined as $$\Pr(\tilde{x}|\mathbf{x})=\int_{\theta\in\Theta}\Pr(\tilde{x}|\theta)\Pr(\theta|\mathbf{x})\mathrm{d}\theta,\forall\theta\in\Theta,$$ where $\Theta$ is the parameter space. What is important to note is that the prediction, $\tilde{x}$, only depends upon $\mathbf{x}$, the data. The integration process marginalizes the parameters, removing their effect. Your prediction does not depend upon a parameter estimate so $\hat{\theta}$ isn't used. This differs from Frequentist methods which depend upon a parameter estimate. The question you asked is "can I produce a distribution?" You did not ask "what are the parameter estimates?" Those are two different questions.

Do note that Bayesian methods are neither biased nor unbiased. They do not care about bias, it's not important in Bayesian methods. You should treat them as more accurate, but biased as both statements are generally true. This is because of the fact that you cannot stochastically dominate a Bayesian estimator.

$\endgroup$
2
  • $\begingroup$ Why is there no volatility or (standard deviation/variance) in the observed data? Can you explain that or refer to the chapter in your paper? In case of a fitted Cauchy distribution ( I am also fitting a sum of a normal and cauchy) I understand that there are no moments. But if I just look at the data without any model and apply the definition of the variance, I observe a changing variance. The Bayesian approach seems powerful. I will try to apply it $\endgroup$
    – Mh47
    Commented Jun 19, 2017 at 21:28
  • 1
    $\begingroup$ @Mh47 The Frequentist algorithm to calculate standard deviation is a fixed calculation that does not automatically measure anything other than the sample standard deviation. Because there is no population variance the algorithm has no basin of attraction to fix to. This makes the algorithm a random number generator. It just floats around. It is why you see volatility clusters. $\endgroup$ Commented Jun 21, 2017 at 18:15

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.