**Q: Why does it take four times as much time to double the price of an at-the-money option? **

**A:** It is a well-known mathematical fact that it takes about __four times__ as much time in order to __double__ an at-the-money option’s price. This relationship is counterintuitive to new traders as is certainly seems logical that if you double the time remaining on the option that you should double its price.

Let’s move into the laboratory and take a closer look at this relationship by using the Black-Scholes Pricing Model with the following inputs:

Stock Price = $35

Strike Price = $35

Days to expiration = 30

Volatility = 25

Interest rate = 0%

The model tells us the theoretical price of this hypothetical $35 call is exactly $1 as shown in Figure 1-a below. However, keeping all other factors the same, if we quadruple the time to expiration to 120 days, the option’s price doubles to $2 as shown in the Figure 1 and Figure 2 below:

Figure 1:

Figure 2:

You can see that under the “laboratory” settings of a pricing model that the relationship holds true. But will we still observe this relationship by using real-world data?

Figure 2-a and 2-b below shows January and April Microsoft option quotes with 31 and 122 days to expiration respectively:

Notice that there is 122/31 = 3.9 times as much time remaining on the April option yet the price is only 2.20/1.02 = 2.1 times greater. The reasons these results are not coming out closer is that 1) we must contend with a bid-ask spread, 2) the $35 call is not quite at-the-money, and 3) there time factor is not exactly fourfold. Despite this, even with real-world data you can see that the counterintuitive relationship seems to hold. The April options with nearly four times as much time only sell for double the price.

What’s causing this relationship to occur? Why does it take four times as much time to double the price of an option and not twice the amount of time as most would suspect? For the answer, we need to turn our attention to a little option pricing theory.

**Option Pricing Theory
**An option’s price solely depends on the volatility (price uncertainty) of the underlying stock. An at-the-money $35 call is worthless if the stock is mandated by law to never move above $35. Under this condition, there is no chance for the option to ever gain intrinsic value and provides no incentive for anyone to pay a single cent for it.

However, if the stock were allowed to move to a maximum of, say $40, then it will have some value. Under this rule, we know the maximum price the option could be worth is the $5 potential intrinsic value (the difference between the $40 stock and $35 strike price). But until we know the probability for each stock price between $35 and $40, we cannot say for sure what the $35 call’s price will be.

If the stock’s price is allowed to move to as high as, say $200, then the $35 call will have a lot of value. These examples show that the price that someone is willing to pay for a call option solely depends on the price fluctuations “allowed” in the underlying stock.

When it comes to pricing options in the real world, we are not held by laws mandating the future potential price. Instead, we must rely on past stock price fluctuations – the volatility – of the stock to gauge how high the potential stock price may be within a given time. Once we know this, we can say what an option should be worth.

**Random Walks**The reason that price fluctuations play a significant role in option pricing is that even the smallest stock price fluctuations can cause a stock’s price to drift away – higher or lower – from its current price. One of the easiest ways to visualize this is to imagine a ball being dropped from a given height. Once it is released, it falls in a straight line to the ground. There are no “fluctuations” or forces to “jolt” it from its current path so it continues to fall in a straight line, just as Newton’s First Law of Motion says it will.

Now let’s assume that we drop the ball from the same height but this time we’re going to add some shocks – we’re going to add volatility. We freeze the ball in mid air every second and then flip a coin. If the coin lands heads (tails), we shift the ball one inch to the left (right) from its current location, release it, and then let it continue to fall for the next second. On average, we would expect the ball to end up directly underneath its starting point just as when no volatility was present. The reason is that we should expect an equal number of heads and tails to appear and the left and right random “jolts” should cancel each other out thus having no overall effect on the ball’s path.

But just because we *expect* and equal number of heads and tails does not mean that it will turn out that way. Anybody who has observed the opening football coin toss knows that some teams end up winning or losing more than might be expected. In 1992, for example, the Los Angeles Rams lost 11 straight opening coin tosses! A string of 11 heads is rare; it is expected to occur once in every 2,048 trials of tossing 11 coins. It is rare, but it does happen due to random fluctuations of heads and tails. Even though we would expect to see an equal number of heads and tails, randomness can play tricks on our perceptions.

Because of these random fluctuations, it is certainly possible for the ball to end up at a point that is not directly beneath its starting point. For example, if we happen to flip 10 heads in a row, the ball will land 10 inches to the left of its starting point. The random shocks would cause the ball to land in a very different spot from what Newton would predict a falling object with no outside forces acting on it.

The action we just described with this ball-drop experiment is one of many variations known as a “random walk.” A random walk has also been compared to a drunkard who tries to walk home in a straight line but staggers to the left and right with equal probability. If he tried to get home like this every night, we would expect him, on average, to find his front door. But just like the coin tosses, sometimes the random left and right staggering (jolts or volatility) will make him end up drifting away from his destination. There will certainly be times he’ll wake up in his neighbor’s front lawn.

**Galton’s Board**In the 19

In one of his experiments, he was curious to find out if random genetic fluctuations in nature created some sort of hidden pattern. Is there a way we could make predictions based on randomness? To find out, Galton devised an ingenious experiment by building a wooden board filled with an array of pegs in a repeating pattern. Directly from the top, he dropped balls which would, of course, be “jolted” left and right as they worked their way to the bottom. A version of “Galton’s Board” is shown in Figure 3:

**Figure 3:
**

Figure 3 shows the ball landing directly in the center bin which is what we’d expect on average. Of course, the sample path shows is only one of many combinations of paths the ball could take. Figure 4 shows another possible path:

**Figure 4:**

If we were to run hundreds of balls through the board we get a bell-shaped curve as shown in Figure 5:

**Figure 5:** The final result is a bell-shaped curve

What causes this bell-shaped pattern? When a ball is dropped, it lands on a peg and then, through pure chance, either bounces left or right where it then hits another peg and so on. Each time the ball strikes a peg, it has a “decision” to make as to whether it should bounce left or right. Because either direction is equally likely, it’s as if an invisible coin is tossed with “heads” moving the ball left and “tails” moving it right. It’s an actual working version of our hypothetical ball-dropping experiment we discussed earlier.

Most of the time we would expect to end up with roughly an equal number of heads and tails thus landing the ball in one of the bins near the center of the curve. Other times we may get an unexpectedly high number of heads or tails thus making the ball land to the far left or right ends of the curve. Naturally, these extreme events are rare so we shouldn’t expect to see too many balls in the ends of the curve. We should expect to see a large number in the center of the board and relatively fewer and fewer as we move toward the extreme ends. And that’s exactly what the experiment showed. It showed that there can be distinct patterns to randomness.

**Galton’s Board and Stock Prices**Galton’s Board has a lot of similarities with stock price movements. For example, let’s take a $100 stock and “inject” it with 20% volatility. Volatility is just a mathematical term for saying we are going to put some “pegs” in the stock’s price path to jolt it either higher or lower each trading day. The 20% figure is a way of determining just how big the “jolts” are for each day.

Now that we have our hypothetical $100 stock price loaded with 20% volatility, let’s turn it loose for 250 days, which is roughly the number of trading days in a year. Figure 6 shows five possible paths from a computer simulation:

**Figure 6: **Five sample stock price paths for a $100 stock with 20% volatility

Figure 6 shows that without volatility, the stock’s price would stay exactly at $100 for every trading day of the year. It would plot as a straight line at $100. However, because of volatility, we got some “jolts” in prices over time and that caused the final price to be somewhat different than the $100 starting price. In some cases, you can see that the final price is very close to $100. Others, however, drifted away either higher or lower. With Galton’s Board, it was the wooden pegs that caused shifts in the ball’s path. For stock prices, it is volatility. For stock prices, it is volatility that can cause the final stock price to be very different from its starting price.

Volatility has the same effect on stock prices as the wooden pegs do for Galton’s Board. Although volatility is caused by news and not wooden pegs standing in the way, the effect is the same. Some days the stock price closes higher and some days lower – and both directions are likely with about equal probability.

If we were to run this same experiment one thousand times and count the number of times we observed small “bins” of final prices, we would find the final stock prices formed a bell curve as shown in Figure 7. Think of it as a sideways version of Galton’s Board:

**Figure 7:** One thousand sample stock price paths for a $100 stock with 20% volatility

Figure 8 below shows actual data for the “percent daily returns” take for eBay for a one-year period. The “percent daily returns” just means that we are measuring the price “jolts” on a daily basis. For instance, if the stock closed at $100 yesterday and $101 today then the percent daily return is $101/$101 = 1.01, or one percent. By tracking the frequencies of the daily percent returns, Figure 8 shows that you will get a bell-shaped curve.

**Figure 8:**

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The important point to understand is that even though we have volatility in this $100 stock there are only so many ending stock prices we should expect. Using properties of bell curves, we could certainly calculate the probability that the stock’s price will fall within certain “bins” such as between $105 and $105.50. As long as we specify a range, no matter how small, we can calculate the probability of the stock’s price ending in that range if we know two conditions. First, we must know the volatility. Second, the volatility must remain constant throughout the time period; in this case, one year.

**Stock Prices and Volatility**We said previously that option prices depend solely on the volatility of the underlying stock. Now you should have a better understanding why. When you enter the “stock price” into the Black-Scholes Model, you’re telling the calculator where the “center” of the bell curve lies. When you input the volatility, you are telling the model how “wide” the curve is.

The time to expiration also helps to scale the volatility. The exercise price tells us at which point we want to calculate the probability. For example, using Figure 7, if we input an exercise price of $120 into the Black-Scholes Model, we’re really telling it to calculate the probability that the stock price is $120 or greater at expiration. Once we know the probabilities for all ranges of stock prices that exceed the strike price, the Black-Scholes model can calculate the theoretical price (fair value) of the option.

Notice that of all five factors in the Black-Scholes Model – stock price, exercise price, time to expiration, risk-free rate of interest, and volatility – the only *unknown* factor is volatility. In other words, if many people were to calculate the theoretical option value, they must all use the same current stock price, exercise price, time to expiration, and risk-free rate. The only variable left for interpretation is volatility. It is the only true unknown factor in the model and that’s why option prices depend solely on your projection of the future volatility of the underlying stock. If you knew for certain what the volatility of the underlying stock will be during the life of the option, you could calculate a true fair value for the option. Remember: Option prices depend solely on volatility.

**The Solution**Now that you know that option prices depend on proper assessment of the future volatility of the underlying stock, let’s go back to our original question: Why does it take four times as much time to double the price of an option? The answer has to do with the distribution of stock prices under the bell curve (or, equally valid, the probabilities under a bell curve).

Probabilities can be calculated by listing all combinations that can occur and then counting how many of those fit a specific definition. For instance, if we toss a coin four times, what is the probability we will observe two heads? All we have to do is find all possible combinations of heads and tails that could occur from four coins and then count how many of those have exactly two heads. Once we have that number, we divide by the total number of possible combinations and that gives us the probability. Table 1 shows every possible pairing of heads and tails for four coin tosses. The second column shows the number of heads found in that particular combination:

Table 1 shows there are sixteen possible combinations of heads and tails that could occur by tossing a coin four times. This can be verified mathematically by realizing that the two sides of the first coin can be matched with two sides of the second coin and so on. Using this logic, we can calculate there must be 2 * 2 * 2 * 2 = 16 combinations of heads and tails for four coins. To find the probability that we observe exactly two heads, we just need to count the number of “2s” occurring in the second column and find there are six. Therefore, the probability of getting exactly two heads in four tosses of a coin is 6/16, or 3/8. Once we convert that to a decimal, we find the probability is 0.375 or just under 38%.

We can find the distribution for the number of possible heads by counting the number of ones, twos, threes, and fours that appear in the second column of Table 1. We find there is only one way to get zero heads, four ways to get one or three heads, and six ways to get two heads as shown in Table 2:

**Table 2:**

As suggested by Galton’s Board, the middle value (two out of four) should be the most frequently occurring area and we see it is true here. If we plot the date in Table 2, we get the following bar chart (technically called a histogram) that is starting to resemble a bell curve:

**Figure 9: **Bar chart showing frequency for “number of heads” for four coin flips

Notice in Table 1 that we have more combinations than individual elements. In other words, we have only four coins but sixteen combinations. *There will always be more combinations of the elements than there are elements in the set.*

As another example, if there are 10 people in a room we can find 90 unique pairings, which are called permutation. There are 10 ways we could select the first person and then 9 ways to select the second person so 10 * 9 = 90 different pairings (taking order into account). Again, the pairings (90) is greater than the number of elements (10 people).

To understand why it takes four times as much time in order to double the price of an option, we must now find out more about the behavior of these unique pairings.

Let’s start with the numbers 2, 4, 8, 16, and 32 and find out how many unique pairings there are for each. A unique pair is found by multiplying the number by one less than the number. For example, if there are four items in a group then there are 4 *3 = 12 unique pairings. For instance, if we take the letters a, b, c, d, then there are 12 unique pairs of letters:

ab, ac, ad, bc, bd, cd

ba, ca, da, cb, db, dc

Again, notice that we are taking order into account. The second row of letters above is just the reverse of the first row. Using the same reasoning, there must be 56 unique ways to arrange eight letters since there are eight ways to choose the first and seven ways to choose the second, or 8 * 7 = 56 ways.

Table 3 shows the number of unique pairs for our numbers 2, 4, 8, 16, and 32:

**Table 3:**

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Notice that we intentionally chose the sequence of numbers in the first column to always be twice that of the previous number as shown by the arrows. However, notice that the unique pairings in the second column do not double at each step. Instead, they increase by a factor of 6 when we jump from 2 to 12 but then gradually appear to be getting closer to a factor of four.

If we continue our sequence in the first column, will the second column ever increase by less than a factor of four? Let n be any number in the first column. We can then say the next number in the same column is 2n since it is twice as big. To find the number of unique pairs for any n, we take n * (n-1). The following number in the second column must therefore have 2n * (2n-1) unique pairs.

For example, if n=4, we know the number of unique pairs must be n * (n-1), or 4 * 3 = 12. The next n in the first column must be 2n, or 4 and its unique number of pairs must be 2n * (2n-1), or 56 so the formulas check out. We know that the ratio for the first column will always be 2n/n, which always equals two. But what can we say about the numbers is the second column?

The ratio for any two numbers in the second column can be found by:

__2n * (2n-1)
__ n * (n-1)

The n in the denominator cancels out the n in the numerator since we are multiplying thus leaving:

__2 * (2n-1)
__ n-1

After factoring the numerator, we end up with:__
4n-2
__ n-1

Now take a close look at this resulting formula. It shows that as n gets bigger and bigger that the numerator must be getting closer to four. In other words, for any n, if you multiply by four and subtract off two then the effect of subtracting two must have less and less of an impact on the resulting value. Therefore, as n gets bigger, subtracting off two is effectively subtracting zero. The numerator is getting closer to 4n.

For similar reasons, the denominator is getting closer and closer to n. The bigger that n gets, the less effect that subtracting of one has on the resulting value. Therefore, this formula is really getting closer and closer to 4n/n, which equals four. This shows that no matter how big of a number you choose for n, the ratio of the unique pairings cannot fall below four.

Let’s put the theory to the test and see if it works. If n=5,000 then the number of unique pairs is 5,000 * 4,999 = 24,995,000. If we double n to 10,000 then there are 10,000 * 9,999 = 99,990,000 unique pairs. The ratio between these unique pairs is 99,990,000/24,995,000 = 4.0004. The bigger you make n, the closer and closer you’ll get to four.

To relate this relationship back to Galton’s Board, it means that you must quadruple the height of the ball before you will get twice the spread in the curve. The ball must fall four times as long in order to double the probabilities under the bell curve. If the spread of the curve is doubled then so too is the option’s price. You can verify this by using a Black-Scholes Model. If you double volatility, you’ll double the price of the at-the-money option. However, you must quadruple the time to expiration to get the same increase.

This is often called the “square root” rule of option pricing. That is, an option’s price is related to the square root of time. If time is increased by a factor of four then the option’s price is increased by a factor equal to the square root of four, which is two. If you double the time remaining, you’ll increase the option’s price by a factor equal to the square root of two, which is 1.41. If a one-month option is priced at $1 then a two month option is worth $1.41 and a four-month option is worth $2.

We can show this relationship another way. The probability of getting a 50-50 split between “heads” and “tails” for four coin tosses is 0.375. If we double the number of coins to eight, the probability of an equal split is 0.2734. The relationship between these two probabilities is 0375/0.2734 = 1.37, which is pretty close to 1.41, which is the square root of two. Now let’s quadruple the number of coins from four to sixteen. The probability of an equal split between “heads” and “tails” for sixteen tosses is 0.1964 thus making the relationship 0.375/0.1964 = 1.91, which is close to two. It takes four times the number of tosses in order to cut the probability in half. The probability at the center is cut in half thus doubling the spread of the curve.It must be emphasized that this relationship only applies to at-the-money options. If we were to run the same experiment for far out-of-the-money options it will not even come close. We’ll save that answer for another class.