Comparing Staggered and Periodic Short-Call Strategies

# Shorting an option every day vs shorting only at maturity


# Shorting an option every day vs shorting only at maturity












Suppose we have 2 strategies :

- strategy A : every $N$ days, we short a call option with a time-to-maturity of $N$ days;

- strategy B : every day, we short $\frac{1}{N}$ of a call option with a time-to-maturity of $N$ days.

When would strategy B be considered better than strategy A ?

## Answer by RandyF (score 4, accepted)

https://quant.stackexchange.com/a/24937

Draw a picture. For each scenario, there are obvious circumstances that the payoff for each would be better.

For the N day option, the payoff would be better if there was a slow gradual decline in price and a slow gradual increase over the same period, such that the final difference in the price of the underlying was largely unchanged. For multiple options issued over that period, at expiration, there would be several options that would need to pay off, so that would be worse.

In the event of quick downward and upward movements, I'm unable to say which would be better. The multiple options would benefit from increased premium from higher implied volatilities, but will be further out of the money.

In the event of a slow decreasing or increasing underlying price, the multiple options would benefit from strike prices that are struck at different levels and therefore, the overall volatility of the strategy would be lower than just selling one option.

You will find that if you look at these strategies using historical prices, the multiple options will have a lower return, but the decrease in volatility will push the information ratio of the strategy higher than the individual sale. Given you are selling options, and depending on your cash coverage requirements, this could mean that, for the same volatility, you can put on a larger position in the second scenario and have a larger return than the first.

## Answer by user59 (score 3)

https://quant.stackexchange.com/a/24856

This really isn't worth the bounty, but it's too long for a comment.

Quoting https://www.tradeking.com/education/options/option-greeks-explained#theta

> At-the-money options move at the square root of time. This means if a one-month ATM option is trading for \$1, then a two-month ATM option would be trading for 1 x sqrt of 2 or \$1.41. A three-month ATM option would be trading for 1 x sqrt of 3 or \$1.73.

As you can see from this example, selling 3 1 month options over 3 months would be worth \$3, whereas a single 3 month option would be worth only \$1.73.

Formula-wise, this means the price of an at money option expiring in $t$ days is $k \sqrt{t}$ (for some value of $k$ that depends on the volatility). So, the money you make per day is $\frac{k \sqrt{t}}{t}$ or $\frac{k}{\sqrt{t}}$. As t becomes smaller, this number becomes larger.

Thus, to maximize your per-day income, sell options as frequently as possible.

Of course, this assumes the underlying's price doesn't change. As I noted in the comments, per the rule of arbitrage, there is no guaranteed way to make money: this method only works on the assumption the underlying's price is relatively stable (ie, more stable than the volatility would indicate).

Another source re theta decay as a square root:

http://www.optionseducation.org/strategies_advanced_concepts/advanced_concepts/understanding_option_greeks/theta.html

(you can also derive this directly from the Black-Scholes formula or from first principles [the sum of two normal distributions with standard deviation $s$ is a normal distribution with standard deviation $s\sqrt{2}$, not $2s$])

## Answer by Yugmorf (score 2)

https://quant.stackexchange.com/a/24902

Strategy A has fatter tails and should outperform when the volatility surface is convex.

Note that both strategies have the same average option maturity = N/2 days. However, for Strategy A option maturity fluctuates between 0 and N days, while for Strategy B the average maturity is always N/2 days (after the initial N day run up to full investment). For any given vol curve, the expected cost of the two strategies should be the same, however expected returns to Strategy A will have fatter tails while returns to Strategy B will have higher kurtosis.

So long as the volatility curve is smiling convexly (at most maturities) then Strategy A gives exposure to these tails at the same cost as strategy B.

## Answer by onlyvix.blogspot.com (score 1)

https://quant.stackexchange.com/a/24967

Given the information provided in the question, expected value of A and B is the same. Simple example: consider this binomial model with stock paths on the left, and call option prices on the right.

What you can easily calculate is that expected value of strategy A and strategy B is the same. If we label paths as UU, UD, DU, DD (up & up, up & down, etc ) the payoffs are:

A: -15,5,5,5

B: -12.5,7.5,2.5,2.5

Like @Yugmorf noted "For any given [arbitrage-free] vol curve, the expected cost of the two strategies should be the same"; with A having higher variance and higher kurtosis than B. This statement is correct in general, not only for a two-period model, and for convex or flat vol surface.

In practical cases however, if you expect change in volatility, or have an opinion on direction, A or B may be preferable.