The Investment Advisor Body of Knowledge + Test Bank. IMCA

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СКАЧАТЬ 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 10 years' worth of historical data, for which the sample mean is 15 percent. In this case the expected value may or may not be 15 percent. In most cases if we say that the expected value is equal to 15 percent, we are making two assumptions: first, we are assuming that the returns in our sample were generated by the same random process over the entire sample period; second, we are assuming that the returns will continue to be generated by this same process in the future. These are very strong assumptions. In finance and risk management, we often assume that the data we are interested in are being generated by a consistent, unchanging process. Testing the validity of this assumption can be an important part of risk management in practice.

      The concept of expectations is also a much more general concept than the concept of the mean. Using the expectations operator, we can derive the expected value of functions of random variables. As we will see in subsequent sections, the concept of expectations underpins the definitions of other population statistics (variance, skew, kurtosis), and is important in understanding regression analysis and time series analysis. In these cases, even when we could use the mean to describe a calculation, in practice we tend to talk exclusively in terms of expectations.

      Sample Problem

      Question:

      At the start of the year, you are asked to price a newly issued zero-coupon bond. The bond has a notional value of $100. You believe there is a 20 percent chance that the bond will default, in which case it will be worth $40 at the end of the year. There is also a 30 percent chance that the bond will be downgraded, in which case it will be worth $90 in a year's time. If the bond does not default and is not downgraded, it will be worth $100. Use a continuous interest rate of 5 percent to determine the current price of the bond.

      Answer:

      We first need to determine the expected future value of the bond, that is, the expected СКАЧАТЬ