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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?

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The graph would be more instructive if you normalize it by some unit of convexity risk (either gamma or vega). Then you'd actually see what is your theta for similar risk position and be able to judge if it's "high" or "low". –  Strange Mar 15 '13 at 17:54

2 Answers 2

up vote 6 down vote accepted

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

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what about your promise? : p –  0d0a Mar 15 '13 at 18:14
    
sorry.. haven't had time yet.. at work –  Andrew Mar 15 '13 at 18:17
    
"The cost of gamma is theta." -- this is by far the best explanation of what it really means to go long an option. Is this the reason why gamma and theta look the opposites of one another? Gammma peaks ATM and decays lower with time, just like the graph above but transposed on the y-axis. –  jessica Mar 15 '13 at 18:52
    
@jessica, that is not an explanation of what it means to go long an option. It simply means that theta is the cost of buying gamma. Hence the inverse albeit still non linear relationship between the two. Andrew provided a very good explanation imho. Upvoted. –  Matt Wolf Mar 15 '13 at 19:17
    
@Andrew, for correctness sake, may I suggest you update your comment re "worthless" to "almost worthless" or similar terminology. Any option that has not yet expired has time value left, even those without intrinsic value. Just a suggestion. –  Matt Wolf Mar 18 '13 at 4:32

$\Theta$ 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:

$\Theta+\frac12\sigma^2S^2\Gamma+rS\Delta-rV=0$

$\Theta=rV-\frac12\sigma^2S^2\Gamma-rS\Delta=r(V-S\Delta)-\frac12\sigma^2S^2\Gamma$

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.

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Comments have been purged. –  chrisaycock Mar 18 '13 at 11:11

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