# Probability of a stock price using implied volatility

I have attempted to use the fact of having implied volatility, but have not been able to come up with a viable way to calculate the probability, any ideas? Suppose that a stock $$S_t$$ follows a lognormal model and that on May 29,2019 the closing price of the stock was $$S_0$$ was 61.5 and the implied volatility of the options with maturity T=0.4 was 120%. If we assume that the annual return required by the investors is 30% what is the probability $$P(40\le S_t \le 55)$$

• On top of hypothesizing a (system of) equation(s) that describe the stock behavior, you could even follow a data-driven approach by plugging your implied volatility into some classification model to train and get a probability as output. But that would just be a toy model, don't expect to earn money with that. – Lisa Ann Jun 11 at 11:45
• @lamdaepsilo: if the answers below are satisfactory, could you pls accept one of them (by clicking on the tick mark) so that your answer can be marked as "answered"? – Jan Stuller Jun 11 at 19:35

I assume you want to real-world probability, because the risk-neutral probability is not a probability in the 'likelihood' sense.

Under the real-world measure, we model the stock under the B-S model as:

$$X(t)=X(0)+\int^{t}_{0}\mu X(h)dh+\int^{t}_{0}\sigma X(h)dW(h)$$

If the market demands a 30% annual return, I will take that as the real-world rate $$\mu$$. Strictly speaking, we should also take the volatility estimated from a historical time series if we deal with the real-world measure, but I will just take your implied vol here for simplicity:

$$X(t)=61.5+\int^{t=0.4}_{0}0.3 X(h)dh+\int^{t=0.4}_{0} 1.2 X(h)dW(h) = \\ = 61.5exp \left( \left[ 0.3 - 0.5* 1.2^2 \right] 0.4 + 1.2 * \sqrt(0.4) Z \right) = \\ = 61.5exp\left( -0.456+0.759Z\right)$$

Therefore:

$$\mathbb{P}\left( 40

Now:

$$\mathbb{P}\left(X_t<55\right)=\mathbb{P}\left(61.5exp\left( -0.456+0.759Z\right)<55\right) = \\= \mathbb{P}\left(ln(61.5) +\left( -0.456+0.759Z\right)

You can do the same for $$\mathbb{P}\left( X_t<40\right)$$, work out the numbers yourself and you should get the answer.

Important Note: The above was just to demonstrate how real world probability could be calculated by blindly plugging numbers into the B-S model. However, pls note that if you want an actual real probability of a stock ending up within a specific range, the B-S model framework is not really suitable for that. Every market agent will have his or her (Bayesian) view of the state of the world and every market agent will view the probabilities differently. Even the choice of model that you will use to compute the probability is a Bayesian choice in itself. It's a really interesting problem, but its more of an "existentialistic" problem, rather than "practical" problem. High-frequency algo traders try to estimate probabilities all the time. They all use different models, different input data, etc...

• Hi there was a typo in my question I meant to say $P(40\le S_.4 \le55)$ it seems that you understood this. Just double checking that you accounted for this in the solution correct? – lambdaepsilon Jun 11 at 23:30
• Also thank you sincerely I greatly appreciate it. – lambdaepsilon Jun 11 at 23:31
• Yeh, I did the computation with $t=0.4$ in mind. Once you work out the probabilities, let me know what you get!!! :) – Jan Stuller Jun 12 at 7:07

Asset prices follow a random walk, so assuming probabilities and forecasting stock prices are not that accurate. Hence, investors try to project volatility rather than asset prices (i.e. implied vol) using GARCH, EWMA, or other vol forecasting models.

The optimal portfolio is to invest long term in a globally diversified portfolio with a focus on uncorrelated asset styles (growth assets, real assets, and hedge assets) or asset classes.

If you figure out a way to project asset prices, with somewhat strong accuracy. Please let me know so that we can set up an investment management firm 😗. Joking aside, I wish you the best of luck on figuring this out and I look forward for your findings.