a discrete random variable, the mode is the value associated with the highest probability. As with population and sample data sets, the mode of a discrete random variable need not be unique.
Sample Problem
Question:
At the start of the year, a bond portfolio consists of two bonds, each worth $100. At the end of the year, if a bond defaults, it will be worth $20. If it does not default, the bond will be worth $100. The probability that both bonds default is 20 percent. The probability that neither bond defaults is 45 percent. What are the mean, median, and mode of the year-end portfolio value?
Answer:
We are given the probability for two outcomes:
At year-end, the value of the portfolio, V, can only have one of three values, and the sum of all the probabilities must sum to 100 percent. This allows us to calculate the final probability:
The mean of V is then $140:
The mode of the distribution is $200; this is the most likely single outcome. The median of the distribution is $120; half of the outcomes are less than or equal to $120.
CONTINUOUS RANDOM VARIABLES
We can also define the mean, median, and mode for a continuous random variable. To find the mean of a continuous random variable, we simply integrate the product of the variable and its probability density function (PDF). In the limit, this is equivalent to our approach to calculating the mean of a discrete random variable. For a continuous random variable, X, with a PDF, f(x), the mean, μ, is then:
(3.26)
The median of a continuous random variable is defined exactly as it is for a discrete random variable, such that there is a 50 percent probability that values are less than or equal to, or greater than or equal to, the median. If we define the median as m, then:
(3.27)
Alternatively, we can define the median in terms of the cumulative distribution function. Given the cumulative distribution function, F(x), and the median, m, we have:
(3.28)
The mode of a continuous random variable corresponds to the maximum of the density function. As before, the mode need not be unique.
Sample Problem
Question:
Using the now-familiar probability density function discussed previously:
What are the mean, median, and mode of x?
Answer:
As we saw in a previous example, this probability density function is a triangle, between x = 0 and x = 10, and zero everywhere else.
Probability Density Function
For a continuous distribution, the mode corresponds to the maximum of the PDF. By inspection of the graph, we can see that the mode of f(x) is equal to 10.
To calculate the median, we need to find m, such that the integral of f(x) from the lower bound of f(x), zero, to m is equal to 0.50. That is, we need to find:
First we solve the left-hand side of the equation:
Setting this result equal to 0.50 and solving for m, we obtain our final answer:
In the last step we can ignore the negative root. If we hadn't calculated the median, looking at the graph it might be tempting to guess that the median is 5, the midpoint of the range of the distribution. This is a common mistake. Because lower values have less weight, the median ends up being greater than 5.
The mean is approximately 6.67:
As with the median, it is a common mistake, based on inspection of the PDF, to guess that the mean is 5. However, what the PDF is telling us is that outcomes between 5 and 10 are much more likely than values between 0 and 5 (the PDF is higher between 5 and 10 than between 0 and 5). This is why the mean is greater than 5.
Expectations
On January 15, 2005, the Huygens space probe landed on the surface of Titan, the largest moon of Saturn. This was the culmination of a seven-year-long mission. During its descent and for over an hour after touching down on the surface, Huygens sent back detailed images, scientific readings, and even sounds from a strange world. There are liquid oceans on Titan, the landing site was littered with “rocks” composed of water ice, and weather on the moon includes methane rain. The Huygens probe was named after Christiaan Huygens, a Dutch polymath who first discovered Titan in 1655. In addition to astronomy and physics, Huygens had more prosaic interests, including probability theory. Originally published in Latin in 1657, De Ratiociniis in Ludo Aleae, or The Value of All Chances in Games of Fortune, was one of the first texts to formally explore one of the most important concepts in probability theory, namely expectations.
Like many of his contemporaries, Huygens was interested in games of chance. As he described it, if a game has a 50 percent probability of paying $3 and a 50 percent probability of paying $7, then this is, in a way, equivalent to having $5 with certainty. This is because we expect, on average, to win $5 in this game:
(3.29)
As one can already see, the concepts of expectations and averages are very closely linked. In the current example, if we play the game only once, there is no chance of winning exactly $5; we can win only $3 or $7. Still, even if we play the game only once, we say that the expected value of the game is $5. That we are talking about the mean of all the potential payouts is understood.
We can express the concept of expectation more formally using the expectations operator. We could state that the random variable, X, has an expected value of $5 as follows:
(3.30)
where E[ ·] is the expectation operator.1
In this example, the mean and the expected value have the same numeric value, $5. The same is true for discrete and continuous random variables. The expected value of a random variable is equal to the mean of the random variable.
While the value of the mean and the expected value may be the same in many situations, the two concepts are not exactly the same. In many situations in finance and risk management the terms can be used interchangeably. The difference is often subtle.
As the name suggests, expectations are often thought of as being forward-looking. Pretend we have a financial asset for which the mean annual return is equal to 15 percent. This is not an estimate; in this case, we know that the mean is 15 percent. We say that the expected value of the return next year is 15 percent. We expect the return to be 15 percent, because the probability-weighted mean of all the possible outcomes is 15 percent.
Now pretend that we don't actually know what the mean return of the asset is, but we have