The context is this theoretical result from Black-Scholes-Merton differential equation that the effects of theta and gamma cancel each other.

Equation (5.23) from the book titled "Option Trading" by Euan Sinclair: $\cfrac{1}{2}\sigma^2 S^2 \Gamma+\theta = 0$.

I understand that this is for a hedged position, and that this holds in an average sense etc. But let us take the following real life situation (numbers are accurate as of August 16th 2022).

GOOGL stock is trading at 122. The 21st Oct 2022 expiry, 150 strike call option price is 0.21. As you might know, the all time high is around 150. So a trader takes a bet that there is no way it will hit 150 in three months. He holds 1000 shares so he sells a covered call of 10 lots. There are two ways to get out of this position.

One is if the stock goes down by a few dollars. If the stock falls even by 10 dollars, the call prices fall down dramatically for the 150 call. Trader can buy it back and keep the difference as profit.

The other way is to simply wait. Theta will kill the call price with time. In case the trader is right and if the stock is nowhere close to 150 closer to the expiry, theta would have killed the call price and he can square off his position.

In all of this, the only big hypothesis is that GOOGL will not hit 150 or even get close to that price in three months. This seems reasonable as 150 is kind of the all time high for GOOGL stock. Isn't covered call in a way hedged as the short calls are covered? Doing what this trader plans to do seems like a solid way to make a risk free profit and he is essentially trading theta. Effect of gamma does not seem to cancel effect of theta here. This can be repeated every time he gets to square off his position based on where GOOGL stock price is at that time, and the call price premiums at a ridiculously high strike price.

I understand that the the profit is really low for selling a 3 month out call. I also understand that theta and gamma cancel each other under certain circumstances and may not hold in this covered call scenario. However, it does seem like we can trade theta. Is there an error in this thought process?


1 Answer 1


You are neglecting the PnL from the stock position. Let us say you hold 1,000 shares at \$122 per unit. You’ve sold calls at \$0.21 per unit of stock, thus receiving \$210 in premiums. If the stock price decreases by \$10, you’re suffering a loss of \$10 x 1,000 = \$10,000 on the stock position and the calls expire worthless, for a full PnL equal to \$210-\$10,000=-\$9,790. This is without accounting for a potential funding cost of the stock position.

Let $v(t,T)$ be the value at time $t$ of a call with expiry $T$. Recall that $\theta:=\partial v/\partial t$ yet by the chain rule: \begin{align} \frac{\partial v}{\partial t} =\frac{\partial v}{\partial \tau}\frac{\partial \tau}{\partial t} =-\frac{\partial v}{\partial \tau} \end{align} where $\tau:=T-t$ is the time to expiry. Now let us consider a trading strategy where we hold a call with expiry $T^\prime$ and short a call with expiry $T<T^\prime$, for a total of $1/(T^\prime-T)$ units - this is known as a calendar spread. Then its value is equal to: \begin{align} \frac{v(t,T^\prime)-v(t,T)}{T^\prime-T} =\frac{v(t,T^\prime)-v(t,T)}{\tau^\prime-\tau} \end{align} Note that $\tau$ is in bijection with $T$ therefore $v(t,T)$ is the call value for time to expiry equal to $\tau$, thus: \begin{align} \frac{v(t,T^\prime)-v(t,T)}{\tau^\prime-\tau} \approx\frac{\partial v}{\partial \tau} \approx-\frac{\partial v}{\partial t} \end{align} Hence a calendar spread allows you to trade theta $-$ approximately, the wider the maturity spread is the worse it approximates theta.

  • 2
    $\begingroup$ I was not aware that the calendar spread allows us to trade theta. You are right about me not accounting for the contribution of change in stock price to PnL. Covered calls can be sold by long term investors where instead of letting the stocks sit untraded, selling calls lets the investor make some chump change. The catch is to choose the strike so high that the call strike is never reached and the trader gets to keep his stocks. Thanks a bunch for answering my question! $\endgroup$ Aug 17, 2022 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.