# Relation between volatility and exercise timing of American Options

Hopefully someone can help me with intuition. Suppose that we have a stock whose value evolves per the geometric brownian motion $$dX_t=X_t\mu dt+X_t\sigma dW_t$$, for $$\sigma>0$$, $$\mu\in\mathbb{R}$$ and $$W_t$$ a standard Brownian Motion, and with $$X_0>0$$. I am trying to understand how an increase in $$\sigma$$ affects the optimal exercise timing. Suppose the options are perpetual. What we know:

1) A call option is exercised when $$X_t>x^*_c>0$$ for some boundary value $$x^*_c>0$$; likewise, a put option is exercised when $$0, for some boundary value $$x^*_p$$. Suppose that the initial condition is $$X_0\in(x^*_p, x^*_c)$$;

2) From Shiryaev's Essentials of Stochastic Finance (1999), chapters VIII a and b, one can see that $$x^*_c$$ increases with $$\sigma$$ and $$x^*_p$$ decreases with $$\sigma$$. This is consistent with the well known idea that volatility increases the value of the options: for every given $$X_0$$, the value of holding the option is higher, so it requires an even higher $$X_t$$ to exercise a call option or a lower $$X_t$$ to exercise a put option.

3) From this paper, one can arrive to the fact that the optimal exercise timing for an american call option is $$\mathbb{E}(\tau)=\frac{\log(x^*_c/X_0)}{\mu-\frac{1}{2}\sigma^2}$$, which is consistent with the expected hitting time of an upper boundary by geometric brownian motion when $$X_0 and $$\mu>\frac{1}{2}\sigma^2$$. It is easy to see then that if $$\sigma$$ grows, $$\mathbb{E}(\tau)$$ grows as well: the denominator is smaller and the numerator is bigger. Intuitively, this makes sense: if having the option is more valuable as $$\sigma$$ grows, I can expect the investor to hold the option for longer;

4) When I make the same calculations for $$\mathbb{E}(\tau)$$ in the case of a put option, I find that $$\mathbb{E}(\tau)=\frac{\log(X_0/x^*_p)}{\frac{1}{2}\sigma^2-\mu}$$, which is symmetric to the case of the call option. However, now when $$\sigma$$ grows, the denominator is bigger and the decrease in $$x^*_p$$ is not enough to compensate, especially in logs, so I found (numerically) that $$\mathbb{E}(\tau)$$ decreases. In fact, from Shiryaev's Essentials of Stochastic Finance (1999), chapter VIII b, one can see that the probability of ever hitting $$x^*_p$$ is increasing in $$\sigma$$, somehow consistent with this. My question is the following: Can someone explain me or give me some intuition on why when $$\sigma$$ grows, even though the value of a put option increases, I would want to exercise it earlier?

• Intuitively: The bigger $\sigma$ is the bigger the chance of hitting the "exercise boundary" which is (for a Put) usually in my experience quite a bit below the current stock price. And yes, at the time you exercise a Put it has a large value, that is why you exercise it. An early exercise involves a big fall in the stock price and hence the put has a big intrinsic value, which you want to collect. Mar 11 '20 at 22:07
• @noob2 thanks for your reply! that definitely helps me building intuition. However, under the same idea, shouldn't I expect to exercise a call also earlier? however the expected exercise time is later for call and earlier for put... I'm wondering if this could be a feature of the geometric brownian motion structure: perhaps the put boundary decreases but not as much as needed to compensate for the increase in variance, while an uper boundary does... Mar 11 '20 at 23:01
• The asymmetry is a feature of the timing of the cash flows: when you are long a put, you receive the exercise price which is good to do early (especially in a high interest rate environment). when you are long a put you pay the exercise price, which is best postponed in a high i.r. environment. Spend money late, get your cash early. Mar 12 '20 at 0:04