The Investment Advisor Body of Knowledge + Test Bank. IMCA
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Название: The Investment Advisor Body of Knowledge + Test Bank
Автор: IMCA
Издательство: John Wiley & Sons Limited
Жанр: Зарубежная образовательная литература
isbn: 9781118912348
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A related problem has to do with back-testing. Good risk managers should regularly back-test their models. Back-testing entails checking the predicted outcome of a model against actual data. Any model parameter can be back-tested.
In the case of VaR, back-testing is easy. Each period can be viewed as a Bernoulli trial. In the case of one-day 95 percent VaR, there is a 5 percent chance of an exceedance event each day, and a 95 percent chance that there is no exceedance. Because exceedance events are independent, over the course of n days, the distribution of exceedances follows a binomial distribution:
(3.83)
In this case, n is the number of periods that we are using to back-test, k is the number of exceedances, and (1 – p) is our confidence level.
Sample Problem
Question:
As a risk manager, you are tasked with calculating a daily 95 percent VaR statistic for a large fixed income portfolio. Over the past 100 days, there have been four exceedances. How many exceedances should you have expected? What was the probability of exactly four exceedances during this time? Four or less? Four or more?
Answer:
The probability of exactly four exceedances is 17.81 percent:
Remember, by convention, for a 95 percent VaR the probability of an exceedance is 5 percent, not 95 percent.
The probability of four or fewer exceedances is 43.60 percent. Here we simply do the same calculation as in the first part of the problem, but for zero, one, two, three, and four exceedances. It's important not to forget zero:
For the final result, we could use the brute force approach and calculate the probability for k = 4, 5, 6, … , 99, 100, a total of 97 calculations. Instead we realize that the sum of all probabilities from 0 to 100 must be 100 percent; therefore, if the probability of K ⩽ 4 is 43.60 percent, then the probability of K > 4 must be 100 percent – 43.60 percent = 56.40 percent. Be careful, though, as what we want is the probability for K ⩾ 4. To get this, we simply add the probability that K = 4, from the first part of our question, to get the final answer, 74.21 percent:
EXPECTED SHORTFALL
Another criticism of VaR is that it does not tell us anything about the tail of the distribution. Two portfolios could have the exact same 95 percent VaR, but very different distributions beyond the 95 percent confidence level.
More than VaR, then, what we really want to know is how big the loss will be when we have an exceedance event. Using the concept of conditional probability, we can define the expected value of a loss, given an exceedance, as follows:
(3.84)
we refer to this conditional expected loss, S, as the expected shortfall.
If the profit function has a probability density function given by f(x), and VaR is the VaR at the α confidence level, we can find the expected shortfall as:
In most cases the VaR for a portfolio will correspond to a loss, and Equation 3.85 will produce a negative value. As with VaR, it is common to reverse the sign when speaking about the expected shortfall.
Expected shortfall does answer an important question. What's more, expected shortfall turns out to be subadditive, thereby avoiding one of the major criticisms of VaR. As our discussion on back-testing suggests, though, the reliability of our expected shortfall measure may be difficult to gauge.
Sample Problem
Question:
In a previous example, the probability density function of Triangle Asset Management's daily profits could be described by the following function:
We calculated Triangle's one-day 95 percent VaR as a loss of
Answer:
Because the VaR occurs in the region where π < 0, we need to utilize only the first half of the function. Using Equation 3.85, we have:
Thus, the expected shortfall is a loss of 7.89. Intuitively this should make sense. The expected shortfall must be greater than the VaR, 6.84, but less than the minimum loss of 10. Because extreme events are less likely (the height of the PDF decreases away from the center), it also makes sense that the expected shortfall is closer to the VaR than it is to the maximum loss.
Part VI Linear Regression Analysis
Linear Regression (One Regressor)
One of the most popular models in statistics is the linear regression model. Given two constants, α and β, and a random error term, ϵ, in its simplest form the model posits a relationship between two variables, X and Y:
As specified, X is known as the regressor or independent variable. Similarly, Y is known as the regressand or dependent variable. As dependent implies, traditionally we think of X as causing Y. This relationship is not necessary, and in practice, especially in finance, this cause-and-effect relationship is either ambiguous or entirely absent. In finance, it is often the case that both X and Y are being driven by a common underlying factor.
The linear regression relationship is often represented graphically as a plot of Y against X, as shown in Figure 3.9. The solid line in the chart represents the deterministic portion of the linear regression equation, Y = α + βX. For any particular point, the distance above or below the line is the error, ϵ, for that point.
FIGURE 3.9 Linear Regression Example
Because there is only one regressor, this model is often referred to as a univariate regression. Mainly, this is to differentiate it from the multivariate model, with more than one regressor, which we will explore later in this chapter. While everybody agrees that a model with two or more regressors is multivariate, not everybody agrees that a model with one regressor is univariate. Even though the univariate model has one regressor, X, it has two variables, X and Y, which has led some people to refer to Equation 3.86 as a bivariate model. From here on out, however, we will СКАЧАТЬ