# Option pricing: Relationship between Theta and early exercise

I am confused about the following:

For a European put option, the parameter $$\Theta$$ is given by $$\Theta= \frac{d V}{dt} = -\frac{SN'(d_1) \sigma}{2 \sqrt{T-t}} + rK e^{-r(T-t)}N(-d_2).$$

My textbook claims the following:

For deep-in-the-money European puts, $$S<. Hence, $$d_1, d_2 \approx - \infty$$, which implies that $$N'(d_1) \approx 0$$ and $$N(-d_2) \approx 1$$. Hence, $$\Theta >0$$. This shows that it can be optimal to exercise a deep-in-the-money American put before maturity.

I have two questions regarding this statement:

1. I have a deep confusion. $$\Theta$$ measures the rate of change of $$V$$ with respect to $$t$$ by its definition. Hence, a positive value of $$\Theta$$ should imply that the put option increases in value over time. This means that we should wait further for the future increase in the value of the option and therefore wait and not exercise at the moment.

2. The formula of $$\Theta$$ comes from the Black-Scholes formula of $$V$$, which is only valid for European options, from what I know. Therefore, I am puzzled about the conclusion regarding the American counterpart in my textbook.

Any ideas? Thanks!

Would you rather have \$100 today or an option that is currently worth \$95 but is expected to be worth \$100 in one year? You would probably choose the former despite the expected increase in the latter's value. • I am still confused: why is it even possible to have the option of getting \$100 in cash when the value of the option is only \\$95??? In other words, your argument is only valid if the amount of cash that one can retrieve at this moment is greater than/equal to the final value of the option at maturity after the increase. – Richard Oct 1 '19 at 3:39