Artificially Intelligent

Any mimicry distinguishable from the original is insufficiently advanced.

A Math Student's Guide to Options

| 2709 words

Disclaimer: This represents slightly more than everything I knew about options when I put my entire net worth into SPY puts in March.1 I learned most of this in about 30 minutes of reading. This introduction should be accessible to someone that knows statistics and calculus, but little finance.

The Basics

Assets are things like houses, gold, and stocks. Assets can be bought or sold for money. Options on top of assets are the ability to buy or sell those assets at a certain price on a certain date. Options can also be bought or sold for money. For example, if I own $1000 of gold and am worried that the price of gold will drop, I might be willing to pay money for the ability to sell my gold for$900 in a year. In effect, this will “insure” me against large drops in the price of gold over the next year.

An option that allows you to sell an asset is called a put option. An option that lets you buy an asset is called a call option.

In the above gold example, the option was a put option. Buying call options is “insuring” yourself against the price of the asset going up. For instance, an airline company might want to buy call options on fuel to protect themselves against increasing costs. If a ton of fuel currently costs $1000, then the airline might buy a call option that lets them buy a ton of fuel for$1100 in one year. This means the maximum the airline will have to pay for a ton of fuel in one year is $1100, insuring them against a rapid increase in price that might put them out of business. Options have two parameters: the strike price and the expiration date. Recall that options are the ability to buy or sell an asset at a certain price on a certain date. The strike price is that price and the expiration data is that date. In the above gold put option example, the strike price is$900 and the expiration date is one year from now. In the above fuel call option example, the strike price is $1100 and the expiration date is also one year from now. The price of an option is determined by the distribution of price of the underlying asset on the expiration date. For example, suppose that you knew in one year, an ounce of gold would be worth$500 with 1/2 chance and $1500 with 1/2 chance. The value of a call option with strike price$1000 and an expiration date one year from now can be calculated by considering both cases. If gold was worth $500 an ounce in a year, the option to buy gold for$1000 an ounce is worthless - you can simply buy it at the current market price. However, if gold is worth $1500 and ounce, the option to buy gold for$1000 an ounce is worth $500 - you can buy for$1000 and sell for $1500, making$500 in profit. Since each case happens with 1/2 chance, the call option is worth $250. The value of a put option with strike price$1000 and an expiration date one year from now is calculated similarly, except the option to sell at $1000 is worthless if gold is worth$1500 and worth $500 if gold is worth$1000.

In general, let your option have strike price $$K$$ and expiration date $$T$$. Let $$p_T(x)$$ be the probability that your underlying asset has price $$x$$ on date $$T$$. If it is a call option, then you make $$\(x - K)$$ if $$x > K$$ and zero otherwise. Thus the expected value of the call option can be computed as $$\int_K^\infty p_T(x)(x - K)dx$$. If the option is a put option, you make $$\(K - x)$$ if $$x < K$$ and zero otherwise. Thus the expected value of the a put option can be computed as $$\int_{-\infty}^Kp_T(x)(K-x)dx$$.[^2]

The above is a calculation of the expected value. Since people’s utility with respect to money deviates predictably from linear, the price of an option will be systematically different from the expectations given by the above formula. For instance, put options are often used as insurance. Insurance is used to lower risk, which is worth money to people who are not risk neutral. Thus people who buy options as insurance expect to lose money, but they also think that this is worth it for the risk reduction. This means the price of a put option should be systematically higher than its expected value.

Suppose that you had $1000 and knew in one year, an ounce of gold would be worth$500 with 1/4 chance and $1500 with 3/4 chance. However, this you have inside knowledge - the rest of the world thinks an ounce of gold will be worth$500 with 1/2 chance and $1500 with 1/2 chance and thus an ounce of gold can currently be bought for$1000. One way to make money (in expectation) is to buy an ounce of gold for $1000, which has expected value $$1/4 * \500 + 3/4 * \1500 = \1250$$, making you$250 (in expectation). However, recall from our previous example that a $1000 strike price call option with one year out expiration date is valued at$250. Given your inside knowledge, however, you value this call option at $$\500 * 3/4 = \375$$; buying a call option will make you $125 (in expectation). You have$1000, so you can buy four of this option, making an expected $500, double the expected profits of simply buying gold. However, if the price of gold turns out to be$500, you will have zero money if you bought options compared to $500 if you bought an ounce of gold. In general, buying options instead of assets gives you more leverage, it lets you expose yourself to higher risk and higher returns with the same amount of capital. Buying assets is investing; buying options is betting. The Greeks Note: this section contains more detailed information that I understand less well and is less relevant unless you’re doing serious trading. However, if you’re doing serious trading then you should really not be getting your information from me. I do not know enough about “the greeks” to know how their values should affect trading strategies, except for the obvious inferences one can make based on their definitions. Traders often think about options in terms of “the greeks”, which are parameters for a given option that give information. These parameters are calculated using some model that I do not understand. For the following, let $$O$$ be the option price, $$K$$ be the option strike price and $$A$$ be the asset price. The delta of an option is how much the price of the option changes when the price of the asset changes. More specifically, $$\Delta = \frac{dO}{dA}$$. Delta is positive for calls and negative for puts (the option to buy at a fixed price is worth more the more the asset is worth, the opposite is true for puts). For puts, if $$A << K$$, i.e. the asset will almost certainly be worth less than the strike price of the option when the option expires, then $$\Delta \approx -1$$. Similarly for calls, if $$A >> K$$, i.e the asset will almost certainly be worth more than the strike price of the option, then $$\Delta \approx 1$$.2 It is always the case $$-1 \leq \Delta \leq 1$$. The logic underlying all of the previous facts is that delta is the probability that the price of the asset on the expiration date is below the strike price (opposite for puts), i.e. that the option expires “in the money”. For example, suppose that you know that in one year, the price of an ounce of gold will either be +$500 or -$500 from its current price. Suppose also that its current price is$1000. From above, we know that a call option with a strike price of $1000 and expiration date of a year from now is worth$250. However, if the price of the asset goes up to $1001, the option is now worth an extra dollar in worlds where the price goes up, but still worthless in the world where the price goes down. Thus the value of the option is now$250.50, implying a delta of 0.5.

To be more specific, under the assumption that a change in asset price only shifts the distribution, a change in the asset price by $$-\alpha$$ is equivalent to a change in the strike price by $$+\alpha$$. If the strike price of a put option increases by $$\alpha$$, there are two sources of additional money. First, in all situations where the option expires with an asset price lower than the strike price, you will make an extra $$\\alpha$$. Second, there is some probability that the asset is valued between $$K$$ and $$K + \alpha$$ when the option expires, so you might profit in some additional worlds. However, as $$\alpha \to 0$$, this probability goes to zero, so the change in the option price with respect to $$\alpha$$ is exactly the chance the asset is valued lower than the strike price when the option expires. Since we were subtracting $$\alpha$$, we thus have (for put options) $$\Delta = -\int_{-\infty}^Kp_T(x)dx$$. For a call option, we would have $$\Delta = \int_{K}^\infty p_T(x)dx$$.

Computing explicitly for a put option, we have $$O = \int_{-\infty}^Kp_T(x)(K-x)dx$$. To reasonable approximation, a change in asset to $$A - \alpha$$ implies our option value goes to $$\int_{-\infty}^Kp_T(x + \alpha)(K-x)dx$$. Substituting, this is equal to $$\int_{-\infty}^{K + \alpha}p_T(x)(K-x+\alpha)dx = \int_{-\infty}^{K}p_T(x)(K-x+\alpha)dx + \int_{K}^{K + \alpha}p_T(x)(K-x+\alpha)dx$$.3

Thus we have: $$\frac{dO}{dA} = \lim_{\alpha \to 0} -\frac{dO}{d\alpha} = \lim_{\alpha \to 0} -\frac{d\int_{-\infty}^{K + \alpha}p_T(x)(K-x+\alpha)dx}{d\alpha}$$. Substituting, this equals $$\lim_{\alpha \to 0} -\frac{d\int_{-\infty}^{K}p_T(x)(K-x+\alpha)dx}{d\alpha} + \frac{d\int_{K}^{K + \alpha}p_T(x)(K-x+\alpha)dx}{d\alpha}$$. Evaluating, we have $$\lim_{\alpha \to 0} -\int_{-\infty}^Kp_T(x)dx - (p_T(K + \alpha) - p_T(K))\alpha = -\int_{-\infty}^Kp_T(x)dx$$.

The gamma of an option is how much the delta changes with the price of the asset changes. More specifically, $$\Gamma = \frac{d^2O}{dA^2}$$. Since the value of the option should be approximately linear with respect to the price of asset except for changes in the probability that the asset is worth less than the strike price when the option expires (for put options), the value of gamma at $$A$$ is proportional to the change in $$\int_{K}^\infty p_T(x)dx$$ when the price of the underlying asset changes. Since the future price distribution of an asset has more bulk closer to the asset price, the value of gamma will thus be larger when $$K \approx A$$. If $$A << K$$, then the asset will almost certainly be lower than the strike price when the option expires, so $$\Gamma \approx 0$$.

For example, suppose I know the price of an ounce of gold in one year will be its current value plus a random number drawn uniformly from $$[-50, 50]$$. Suppose also that the price of gold is currently $1000. By symmetry, a call option with a strike price$1000 and an expiration date of a year from now is worth $1000. Since there is a 1/2 chance that the asset is worth less than the strike price on the expiration date, the delta is thus 1/2. Suppose, however, that the asset increased in value to$1001. Now, as long as the random number is in [-1, 50], the asset will expire “in the money”. Since this is 51 of the 100 possible values, the delta of the option becomes 0.51. This implies that gamma is equal to 0.01.

To be more specific, under the assumption that a change in asset price only shifts the distribution, a change in the asset price by $$-\alpha$$ is equivalent to a change in the strike price by $$+\alpha$$. We know that the delta for a put option is the probability that the asset is below the strike price when the option expires. The change in the delta is thus the amount this probability shifts, which is exactly the probability that the asset is valued at between $$K$$ and $$K + \alpha$$ when the option expires. As $$\alpha \to 0$$, this is equal to the probability that the asset is valued at exactly $$K$$, thus $$\Gamma = p_T(K)$$. Note that since the probability that an asset is valued between $$K - \alpha$$ and $$K$$ is equal to the probability that the asset is valued between $$K$$ and $$K + \alpha$$, the expression for gamma is the same for both put and call options.

Computing explicitly for a put option, $$\Gamma = \frac{d^2O}{dA^2} = \frac{d\Delta}{dA} = \lim_{\alpha \to 0} -\frac{d \Delta}{d \alpha} =\lim_{\alpha \to 0} \frac{d\int_{-\infty}^{K + \alpha}p_T(x)dx}{d\alpha} =\lim_{\alpha \to 0} p_T(K + \alpha) = p_T(K)$$.

The theta of an option measures how much the price of the option changes with respect to time. More specifically, if $$t$$ is time then $$\Theta = \frac{dO}{dt}$$. The way to think about this is that the uncertainty over the asset price will decrease as you get closer to the expiration date (less time for the price to move). Generally, the option becoming worth more when the asset moves in a certain way is balanced by an equal and opposite probability of the asset moving the other direction. However, since an option cannot be worth less than zero, sufficiently large asset movements do not have equal and opposite movements in the other direction. Higher variance means more probability of such an unbalanced asset movement occurring, thus the value of the option will be high when variance of the distribution of the asset price on the expiration date is high. Since variance decreases with time, theta is always thus always negative.

For example, suppose I know the price of an ounce of gold will be either $730 or$0 in a year, with a 1/2 chance for each. Suppose I expect my uncertainty to decrease linearly with respect to time, ending in a final asset price of $365 in one year. This means that every day, I expect to become more certain by$1, i.e. tomorrow I expect the price of an ounce of gold is $729 or$1 with 1/2 chance each. Today, a put option with a strike price of $365 and expiration date a year from now is worth $$\365/2 = \182.50$$. However, tomorrow, it will be worth $$\364/2 = \182$$. Thus, we have that $$\Theta = \0.50/\text{day}$$. There are two other greeks called vega and rho. Since I do not understand them that well, I will refrain from trying to explain. From what little I understand of them, they are not very important. 1. To first order approximation, you should never do this. 2. If it is not clear, here is the full explanation. From above, $$O = \int_{-\infty}^Kp_T(x)(K-x)dx$$. Since the efficient market hypothesis is approximately true, the future price of an option should be a distribution centered around the current price of the option. This suggests the effect of an asset dropping by$1 should mean that the put option is now worth $$\int_{-\infty}^Kp_T(x)(K-(x-1))dx$$. We have $$\int_{-\infty}^Kp_T(x)(K-(x-1))dx = \int_{-\infty}^Kp_T(x)(K-x)dx + \int_{-\infty}^Kp_T(x)dx = O + 1 - \int_{K}^\infty p_T(x)dx$$. If $$A << K$$, then the asset will almost certainly be worth less than the strike price of the option when the option expires, so $$\int_{K}^\infty p_T(x)dx \approx 0$$, thus $$\Delta \approx -1$$.

3. In following the above interpretation of this equation, $$\int_{-\infty}^{K}p_T(x)(k-x-\alpha)dx$$ is the price of the option assuming the asset price shifted by $$\alpha$$ if we only profit when the asset is valued less than $$K$$ when the option expires. $$\int_{K}^{K + \alpha}p_T(x)(k-x-\alpha)dx$$ is an adjustment for profit earned when the asset is valued between $$K$$ and $$K + \alpha$$ when the option expires. However, as $$\alpha \to 0$$, the probability mass on the asset being valued between $$K$$ and $$K + \alpha$$ goes to zero, hence the derivative only depends on the first piece.