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 $0<X_t<x^*_p$, 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 $X_0<x^*$ 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?

  • $\begingroup$ 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. $\endgroup$
    – nbbo2
    Mar 11, 2020 at 22:07
  • $\begingroup$ @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... $\endgroup$ Mar 11, 2020 at 23:01
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    $\begingroup$ 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. $\endgroup$
    – nbbo2
    Mar 12, 2020 at 0:04


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