# Expected stock price range using implied volatility calculated by Black-Scholes

What's the correct way to calculate the expected stock price range using implied volatility, without the simplifying assumption that the stock price follows a normal distribution?

• Please be more specific: what is the range, and the expectation under what measure? Jul 26 at 10:37
• @FridoRolloos I'm trying to figure out an unsimplified version of this. Jul 26 at 10:43
• @emot can you expand this a little and post it as an answer so that I can mark it as accepted? Please also add where I can read more about it. Jul 26 at 13:33

Consider stock price process (Geometric Brownian Motion): $$S_t=S_0exp((\mu-0.5\sigma^2)t+\sigma W_t) \tag{1}$$

where $$W_t$$ is a Wiener process and $$\mu$$ is a drift - or average return. If you are not familiar with Wiener process you can see this equation as: $$S_t=S_0exp((\mu-0.5\sigma^2)t+\sigma \sqrt t Z) \tag{2}$$ where $$Z$$ is standard normal random variable. Then $$S_t$$ follows log-normal distribution.

For this process we have: $$E^P[S_t]=S_0exp(\mu t)$$ The parameter $$\mu$$ means that the stock earns $$\mu t$$ on average for the period $$t$$. This is just a continuous compounding formula.

When we price options in Black-Scholes setting we assume initially that the stock prices follows this process. But then we change the drift parameter $$\mu$$ to risk-free rate $$r$$ because this allows us to calculate replicating portfolio value. We call this step - changing real world measure to risk neutral measure. This comes from the fact that we assume that the market is arbitrage-free and we can replicate the option by proper trading strategy (replicating portfolio). If it is possible, then the option price is just a price of replicating portfolio that replicates the option payoff and to calculate the price of replicating portfolio (and the option itself) we have to change the real drift parameter to risk free rate and then evaluate expectation of the discounted payoff.

Therefore when we change measure (i.e. change the drift) we have: $$S_t=S_0exp((r-0.5\sigma^2)t+\sigma W_t^Q) \tag {3}$$

Therefore in this "new" process we have: $$E^Q[S_t]=S_0exp(rt)$$

Which just means that in that risk neutral measure assets earn risk free rate on average. This measure is just a way to calculate the option value.

You should bear in mind that in reality we assume that $$\mu > r$$ i.e. that on average the stock market beats the risk free rate, because in stock market there is additional risk that should be compensated by higher $$\mu$$.

But then we can ask ourselves what is the price range of the stock price? Should we calculate it w.r.t $$r$$ or $$\mu$$? If you want to know real world probability (also called physical), which I assume you do, then you should use $$\mu$$. But knowing real $$\mu$$ is hard. You can try to estimate it with historical data, but the problem with $$\mu$$ is that in reality it is not a constant, it can change day to day. But for short time horizon knowing $$\mu$$ is not very important to know the bands, because in our equation it is really small (and scaled linearly) compared to volatility (which scales with square root of time). Therefore we can assume that $$\mu = 0$$.

Then if we assume that $$\mu = 0$$, we have:

$$S_t=S_0exp((-0.5\sigma^2)t+\sigma \sqrt t Z) \tag{4}$$

and we can easily calculate the bands:

up: $$S_{up}=S_0exp((-0.5\sigma^2)t+\sigma \sqrt t * \Phi^{-1}(0.5+q/2))$$ down: $$S_{down}=S_0exp((-0.5\sigma^2)t+\sigma \sqrt t * \Phi^{-1}(0.5-q/2))$$

where $$q$$ is the band (for example 68%) and $$\Phi^{-1}$$ is inverse of standard normal CDF.

Therefore to calculate 68% band, for Stock $$S_0=100$$ and volatility $$\sigma = 0.30$$ in 1 month time $$t= 1/12$$ we get: $$S_{up}=100*exp((-0.5*0.3^2) * 1/12 + 0.3 * \sqrt {1/12} * \Phi^{-1}(0.5+0.68/2))$$ $$=100*exp((-0.5*0.3^2) * 1/12 + 0.3 * \sqrt {1/12} * 1)=108.59$$ and $$S_{down}=91.41$$

But if true $$\mu = 0.1$$ and we didn't know it and assumed $$\mu=0$$ then the correct band is $$S_{up}=109.49$$ and $$S_{down}=92.17$$ and our calculation is just slightly off.