How does $\Theta$ change for deep out-of-the money options? Looking at the below graph, it seems the time decay is highest for ATM options and increases rapidly as we approach maturity of the option. From the graph, it seems the deep OTM options have flat $\Theta$ throughout the entire term strucuture. Shouldn't the OTM options experience the most decay?
No because they are worthless in the first place. Theta is in dollar space and therefore, if something is worthless, it is hard for it to lose much more value.
Think about it this way. When you are buying an option, you are really buying gamma from BS PDE. The cost of gamma is theta. Where is gamma highest? ATM
$\Theta$ measures the rate of change of the option value $V$ with time $t$ if the underlying asset $S$ doesn't move. since deep OTM options are almost worthless this change will be small if the asset will not move - they still will be worthless: at least they cannot change much in price since are almost worth 0.
write Black-Scholes equaton as:
since $\Gamma$ for OTM call option is close to 0 theta will be higher. and $V$ and $\Delta$ don't change(vary) much, so as the $\Theta$
ofcourse this is just the rule of thumb since formula for $\Theta$ is not so easy to understand at the first glance, or even at 100th
I made a picture which might help to understand this: notice relative stability of hadged portfolio $(V-\Delta S)$, negative (in this case) value of this doesn't vary much with respect to changes in spot when OTM, and vary more when close to ATM (ATM spot strike is 1.5178). this is the change in the second term of equation for $\Theta$ that introduces much to the variation of it, it is $-\frac12\sigma^2S^2\Gamma$. and as said before, since $\Gamma$ converges to 0 for OTM options the shape of this term is as we can see in the picture.