唯學網(wǎng)小編提醒：根據(jù)金融分析師專家的專業(yè)分析后得出，定量投資是一種主動投資，因為它和指數(shù)化投資的理論基礎(chǔ)完全不同。指數(shù)化等被動投資建立在市場是完全有效的理論基礎(chǔ)上。而定量投資作為一種主動投資的方法，它的理論基礎(chǔ)是在于市場是無效的，或者是弱有效的。也就代表了投資人可以通過對于市場、行業(yè)基本面及個別公司的分析，能夠去建構(gòu)一個可以產(chǎn)生超額收益，戰(zhàn)勝市場的組合。

三、 Investment Tools: Quantitative Methods

1.A: Sampling and Estimation

a: Define simple random sampling.

Simple random sampling is a method of selecting a sample in such a way that each item or person in the population begin studied has the same (non-zero) likelihood of being included in the sample. This is the standard sampling design.

b: Define and interpret sampling error.

Sampling error is the difference between a sample statistic (the mean, variance, or standard deviation of the sample) and its corresponding population parameter (the mean, variance or standard deviation of the population).

The sampling error of the mean = sample mean - population mean = X bar -µ.

c: Define a sampling distribution

The sample statistic itself is a random variable, so it also has a probability distribution. The sampling distribution of the sample statistic is a probability distribution made up of all possible sample statistics computed from samples of the same size randomly drawn from the same population, along with their associated probabilities.

d: Distinguish between simple random and stratified random sampling.

Simple random sampling is where the observations are drawn randomly from the population. In a random sample each observation must have the same chance of being drawn from the population. This is the standard sampling design.

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Stratified random sampling first divides the population into subgroups, called strata, and then a sample is randomly selected from each stratum. The sample drawn can be either a proportional or a non-proportional sample. A proportional sample requires that the number of items drawn from each stratum be in the same proportion as that found in the population.

e: Distinguish between time-series and cross-sectional data.

A time-series is a sample of observations taken at a specific and equally spaced points in time. The monthly returns on Microsoft stock from January 1990 to January 2000 are an example of time-series data.

Cross-sectional data is a sample of observations taken at a single point in time. The sample of reported earnings per share of all Nasdaq companies as of December 31, 2000 is an example of cross-sectional data.

f: State the central limit theorem and describe its importance.

The central limit theorem tells us that for a population with a mean µ and a finite variance σ2, the sampling distribution of the sample means of all possible samples of size n will be approximately normally distributed with a mean equal to µ and a variance equal to σ2/n.

The central limit theorem is extremely useful because the normal distribution is relatively easy to work with when doing hypothesis testing and forming confidence intervals. We can make very specific inferences about the population mean, using the sample mean, no matter the distribution of the population, as long as the sample size is "large."

What you need to know for the exam:

1. If the sample size n is sufficiently large (greater than 30), the sampling distribution of the sample means will be approximately normal.

2. The mean of the population, µ, and the mean of all possible sample means, µx, are equal.

3. The variance of the distribution of sample means is σ2/n.

g: Calculate and interpret the standard error of the sample mean.

Standard error of the sample means is the standard deviation of the sampling distribution of the sample means. The standard error of the sample means when the standard deviation of the population is known is calculated by: σx = σ/√ n, where: σx = the standard error of the sample means, σ = the standard deviation of the population, and n = the size of the sample.

Example: The mean hourly wage for Iowa farm workers is $13.50 with a standard deviation of $2.90. Let x be the mean wage per hour for a random sample of Iowa farm workers. Find the mean and standard error of the sample means, x, for a sample size of 30.

The mean μx of the sampling distribution of x is μx = μ = $13.50. Since σ is known, the standard error of the sample means is: σx = σ/ √n = 2.90 / √30 = $.53. In conclusion, if you were to take all possible samples of size 30 from the Iowa farm worker population and prepare a sampling distribution of the sample means you will get a mean of $13.50 and standard error of $.53.

h: Distinguish between a point estimate and a confidence interval estimate of a population parameter.

Point estimates are single (sample) values used to estimate population parameters. The formula we use to compute the point estimate is called the estimator. For example, the sample mean X bar is an estimator of the population mean µ, and is computed using the following formula:

X bar = (Σ x / n)

The value we obtain from this calculation for a specific sample is called the point estimate of the mean.

A confidence interval is a range of estimated values within which the actual value of the parameter will lie with a given probability of 1 - α. The term α is also called the significance level of the test. It is also known as the confidence level.

i: Identify and describe the desirable properties of an estimate.

When we have a choice among several estimators, we want to select the one with the most desirable statistical properties: unbiasedness, efficiency, and consistency.

An unbiased estimator is one for which the expected value of the estimator is equal to the parameter you are trying to estimate.

An unbiased estimator is also efficient if the variance of its sampling distribution is smaller than all the other unbiased estimators of the parameter you are trying to estimate. The sample mean, for example, is an efficient estimator of the population mean.

A consistent estimator provided a more accurate estimate of the parameter as the sample size increases. As the sample size increases, the standard deviation (standard error) of the sample mean falls and the sampling distribution bunches more closely around the population mean.

j: Calculate and interpret a confidence interval for a population mean, given a normal distribution with a known population variance.

If the distribution of the population is normal and we know the population variance, we can construct the confidence interval for the population mean as follows:

?

X bar Zα/2 (σ /?√n)

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Example: Suppose we administer a practice exam to 100 CFA Level I candidates, and we discover the mean score on this practice exam for all 36 of the candidates in the sample who studied at least 10 hours a week in preparation for the exam is 80. Assume the population standard deviation is 15. Construct a 99% confidence interval for the mean score on the practice exam of candidates who study at least 10 hours a week.

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80 2.575 (15 / √36) = 80 6.4

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The 99% confidence has a lower limit of 73.6 and an upper limit of 86.4.

k: Describe the properties of Student's t-distribution.

The student's t-distribution is similar, but not identical to the normal distribution in shape. It is defined by a single parameter (the degrees of freedom), whereas the normal distribution is defined by two parameters (the mean and variance).

The student's t-distribution has the following properties:

· It is symmetrical.

· It is defined by a single parameter, the degrees of freedom (df), where the degrees of freedom are equal to the number of sample observations minus one. (n - 1).

· It is less peaked than a normal distribution, with more probability in the tails.

· As the degrees of freedom (the sample size) gets larger, the shape of the t-distribution approaches a standard normal distribution.

l: Calculate and interpret a confidence interval for a population mean, given a normal distribution with an unknown population variance.

Example: Suppose you take a sample of the past 30 monthly returns for McCreary Inc. The mean return is 2%, and the sample standard deviation is 20%. The standard error of the sample was found to be 3.6%. Construct a 95% confidence interval for the mean monthly return.

Because there are 30 observations, the degrees of freedom are 30 - 1 = 29. Remember, because this is a two-tailed test, we want the total probability in the tails to be α = 5%; because the t-distribution is symmetrical, the probability in each tail will be 2.5% when df = 29. From the t-table, we can determine that the reliability factor for tα/2, or t2 5, is 2.045. Then the confidence interval is:

2 2.045 (20 / √30) = 2% 7.4%

The 95% confidence interval has a lower limit of -5.4% and an upper limit of 9.4%.

m: Discuss the issues surrounding selection of the appropriate sample size.

When the distribution is non-normal, the size of the sample influences whether or not we can construct the appropriate confidence interval for the sample mean. If the distribution is non-normal, but the variance is known, we can still use the Z-statistic as long as the sample size is large (n > 30). We can do this because the central limit theorem assures us that the distribution of the sample mean is approximately normal when the sample is large.

If the distribution is non-normal and the variance is unknown, we can use the t-statistic as long as the sample size is large (n > 30).

This means that if we are sampling from a non-normal distribution (which is sometimes the case in finance), we cannot create a confidence interval if the sample size is less than 30. So, all else equal, make sure you have a sample larger than 30, and the larger, the better.

n: Define and discuss data-snooping/data-mining bias.

Data-snooping bias occurs when the researcher bases his research on the previously reported empirical evidence of others, rather than on the testable predictions of well-developed economic theories.

Data snooping often leads to data mining, when the analyst continually uses the same database to search for patterns or trading rules until he finds one that "works." For example, some researchers argue that the value anomaly (in which value stocks appear to outperform growth stocks) is actually the product of data mining. Because our data set of historical stock returns is quite limited, we don't know for sure whether the differences between value and growth stock returns is a true economic phenomenon, or simply a chance pattern that we stumbled on after repeatedly looking for any pattern in the data.

o: Define and discuss sample selection bias, survivorship bias, look-ahead bias, and time-period bias.

Sample selection bias occurs when some data is systematically excluded from the analysis, usually because of the lack of variability. Then the sample isn't random, and any conclusions we draw from the observed sample can't be applied to the population because the observed sample and the rest of the population we didn't observe are different.

The most common form of sample selection bias is survivorship bias. A good example of the problems associated with survivorship bias in investments is the study of mutual fund performance. Most mutual fund databases (like Morningstar's) only include funds currently in existence (the "survivors"), and do not include funds that have been closed down or merged.

Look-ahead bias occurs when the analyst uses historical data that was not publicly available at the time being studied. For example, financial information is not usually available until 30 to 60 days after the end of the fiscal year. A study that uses market-to-book value ratios to test trading strategies might estimate the book value as reported at fiscal year end and the market value two months later in order to account for this bias.

Time-period bias can result if the time period over which the data is gathered is either too short or too long. If the time period is too short, the results may reflect phenomenon specific to that time period, or perhaps data mining. If the period is too long, the fundamental economic relationships that underlie the results may have changed.

1.B: Hypothesis Testing

a: Define a hypothesis and describe the steps of hypothesis testing.

A hypothesis is a statement about the value a of population parameter developed for the purpose of testing a theory or belief. Hypothesis testing is a procedure based on evidence from samples and probability theory used to determine whether a hypothesis is a reasonable statement and should not be rejected, or is an unreasonable statement and should be rejected. The process of hypothesis testing consists of:

Stating the hypothesis.

Selecting the appropriate test statistic.

Specifying the level of significance.

Stating the decision rule regarding the hypothesis.

Collecting the sample and calculating the sample statistics.

Making a decision regarding the hypothesis.

Making a decision based on the results of the test.

b: Define and interpret the null hypothesis and alternative hypothesis.

The null hypothesis is the hypothesis that the researcher wants to reject. This is the hypothesis that is actually tested and is the basis for the selection of the test statistics. The null is generally stated as a simple hypothesis, such as:

H0:?μ = μ0

Where μ is the population mean and μ0 is the hypothesized value of the population parameter. In our option return example, μ is the return on options (the true return for the population) and μ0 is zero.

c: Discuss the choice of the null and alternative hypotheses.

The null hypothesis is what the researcher wants to disprove. The alternative hypothesis is what is concluded if the null hypothesis is rejected by the test. It is usually the alternative hypothesis that you are really trying to assess. Why? Since you can never really prove anything with statistics, when you discredit the null hypothesis you are implying that the alternative is valid.

d: Distinguish between one-tailed and a two-tailed hypothesis tests.

The alternative hypothesis can be one-sided or two-sided (a one-sided test is referred to as a "one-tailed" test and a two-sided test is referred to as a "two-tailed" test). Whether the test is one or two-sided depends on the theory. If a researcher wants to test whether the return on options is greater than zero, she may use a one-tailed test. However, if the research question is whether the return on options is different from zero, the researcher may use a two-tailed test (which allows for deviation on both sides of the null value). In practice, most tests are constructed as two-tailed tests.

e: Define and interpret a test statistic.

Hypothesis testing involves two statistics: the test statistic calculated from the sample data, and critical value of the test statistic (i.e., the critical value). The comparison of the calculated test statistic to the critical value is a key step in assessing the validity of the hypothesis.

We calculate a test statistic by comparing the estimated parameter (that is, a parameter such as mean return that is calculated from the sample) with the hypothesized value of the parameter (that is, what is specified in the null hypothesis).

Test statistic = (sample statistic - hypothesized value) / (standard error of the sample statistic).

f: Define and interpret a significance level and explain how significance levels are used in hypothesis testing.

Recall the assertion that you can't prove anything with statistics (you can only support or reject an idea or hypothesis). This is because we are using a sample to make inferences about a population. Hopefully, a well-selected sample will yield insight into the true parameters of the population, but we must be careful to note that there is a chance that our sample is somehow unrepresentative of the population.

g: Define and interpret a Type I and a Type II error.

With hypothesis testing, there are basically two possible errors: 1) rejection of a hypothesis when it is actually true (a type I error) and 2) acceptance of a hypothesis with it is actually false (a type II error). The significance level is the risk of making a type I error (rejecting the null when it is true). For instance, a significance level of 5% means that, under the assumed distribution, there is a 5% chance of rejecting a true null hypothesis (this is also called an alpha of .05).

 

	Null Hypothesis is True	Null Hypothesis is False
Fail to Reject Null Hypothesis	Correct decision	Incorrect decision (Type II error)
Reject Null Hypothesis	Incorrect decision (Type I error)	Correct decision

h: Define the power of a test.

The power of a test is the probability of correctly rejecting the null hypothesis when the null hypothesis is indeed false. The power of a test statistic is important, because we wish to use the test statistic that provides the most powerful test among all the possible tests.

i: Define and interpret a decision rule.

The decision rule for rejecting or failing to reject the null hypothesis is based on the test statistic's distribution. For example, if a z-statistic is used, the decision rule is based on critical values determined from the normal distribution.

The critical values are determined based on the distribution and the significance level chosen by the researcher. The significance level may be 1%, 5%, 10%, or any other level. The most common is 5%.

If we reject the null hypothesis, the result is statistically significant; if we fail to reject the null hypothesis, the result is not statistically significant.

j: Explain the relationship between confidence intervals and tests of significance.

A confidence interval is a range of values within which the researcher believes the true population parameter may lie. The range of values is determined as:

{sample statistic - (critical value)(standard error) < population parameter < sample statistic + (critical value)(standard error)}

The strict interpretation of a confidence interval is that for a level of confidence of, say 95%, we expect 95% of the intervals formed in this manner to have captured the true population parameter.

k: Distinguish between a statistical decision and an economic decision

A statistical decision is based solely on the sample information, the test statistic, and the hypotheses. That is, the decision to reject the null hypothesis is a statistical decision. An economic decision considers the relevance of that decision after transaction costs, taxes, risk, and other factors - things that don't enter into the statistical decision. For example, it is possible for an investment strategy to produce returns that are statistically significant, but which are not economically significant once one considers transactions costs.

l: Discuss the p-value approach to hypothesis testing.

The p-value is the probability that lies outside the calculated test statistic and (in the case of a two-tailed test) its negative or positive value. In other words, the p-value is the probability, assuming the null hypothesis is true, of getting a test statistic value at least as extreme as the one just calculated. You can make decisions using the p-value instead of the critical value of the test-statistic.

If the p-value is less than the significance level of the hypothesis test, H0 is rejected.

If the p-value is greater than the significance level, then H0 is not rejected.

m: Identify the test statistic for a hypothesis test about the population mean of a normal distribution with (1) known or (2) unknown variance.

The choice between the t-distribution and the normal (or z) distribution is dependent on sample size and whether the variance of the underlying population is known (in the real world, the underlying variance of the population is rarely known). Tests of a hypothesis with a known population variance can use the z-statistic. In contrast, tests of hypothesis regarding the mean when the population variance is unknown require the use of the t-distributed test statistic.

The test statistic for testing a population mean is the following t-distributed test statistic (often referred to as the t-test):

tn – 1 = (sample mean – μ0)/(s/√n)

Where:

μ0 is the hypothesized sample mean (i.e., the null)

s is the standard deviation of the sample

n is the sample size

This t-distributed test statistic has n-1 degrees of freedom.

Test of hypothesis regarding the mean when the population variance is known require the use of a z-distributed test statistic. The test statistic for testing a population mean is the following z-distributed test statistic (often referred to as the z-test):

z = (sample mean - μ) / (σ / √n)

The test of a hypothesis regarding the mean when the population variance is unknown and the sample size is large allows the use of the z-distributed test statistic. The test statistic for testing a population mean is the following test statistic (often referred to as the t-test):

z = (sample mean - μ0) / (s / √n)

Example: An investor believes that options should have an average daily return greater than zero. To empirically access his belief, he has gathered data on the daily return of a large portfolio of options. The average daily return in his sample is .001, and the sample standard deviation of returns is .0025. The sample size is 250. Explain which type of test statistic should be used and the difference in the likelihood of rejecting the null with each distribution.

The population variance for our sample of returns is unknown (we had to estimate it from the sample). Hence, the t-distribution is appropriate. However, the sample is also large (250), so the z-distribution would also work (it is a trick question - either method is appropriate). What about the difference in likelihood of rejecting the null? Since our sample is so large, the critical values for the t and z are almost identical. Hence, there is almost no difference in the likelihood of rejecting the null.

n: Explain the use of the z-test in relation to the central limit theorem.

The standard deviation of the standard normal distribution is 1. The standard deviation of the t-distribution is df / (df - 2), where df is the degrees of freedom.

If the degrees of freedom are 10, the standard deviation is 1.25. If the degrees of freedom are 30, the standard deviation of the t-distribution is 1.07. For large df, the t and z distributions are similar.

The central limit theorem (CLT) is a theorem that states that for any given distribution with a mean of µ and a variance of σ2 / N, the sampling distribution of the mean approaches a normal distribution with a mean µ and a variance of σ2 / N, the sample size increases.

o: Formulate a null and an alternative hypothesis about a population mean and determine whether the null hypothesis is rejected at a given level of significance.

Example: When your company's gizmo machine is working properly, the mean length of gizmos is 2.5 inches. However, from time to time the machine gets out of alignment and produces gizmos that are either too long or too short. When this happens, production is stopped and the machine is adjusted. To check the machine, the quality control department takes a gizmo sample each day. Today a random sample of 49 gizmos showed a mean length of 2.49 inches. The population standard deviation is 0.021 inches. Using a 5% significance level, should the machine be shut down and adjusted?

Step 1: Let μ be the mean length of all gizmos made by this machine and X bar the corresponding mean for the sample.

Step 2: State the hypothesis.

H0: μ = 2.5 (machine does not need an adjustment)

HA: μ?≠ 2.5 (machine needs an adjustment)

This is a two-tailed test.

Step 3: Select the appropriate test statistic.

z = (X bar - μ0) / (σ /?√n)

Step 4: Specify the level of significance. You are willing to make a Type I error 5% of the time so the level of significance is 0.05.

Step 5: State the decision rule regarding the hypothesis. The?≠ sign in the alternative hypothesis indicates that the test is two-tailed with two rejection regions, one in each tail of the normal distribution curve. Because the total area of both rejection regions combined is 0.05 (the significance level), the area of the rejection region in each tail is 0.025. The table value is +/- 1.96. This means that the null hypothesis should not be rejected if the computed z value lies between -1.96 and +1.96, and rejected if it lies outside of these critical values.

Step 6: Collect the sample and calculate the test statistic. The value of X bar from the sample is 2.49. Since population standard deviation is given as 0.021, we calculate the z test statistic using σ as follows:

z = (2.49 - 2.5) / (0.021 /?√49) = (-.01 / 0.003) = -3.33

Step 7: Make a decision regarding the hypotheses. The calculated value of z = -3.33 is called the calculated, computed, or observed test statistic. This z-value indicates the location of the observed sample mean relative to the population mean (3.33 size-adjusted standard deviations to the left of the mean). Now, compare the calculated z to the critical z value. The calculated z value of -3.33 is less than the critical value of -1.96, and it falls in the rejection region in the left tail. Reject H0.

p: Identify the test statistic for a hypothesis test about the equality of two population means of two normally distributed populations based on independent samples.

A test of differences in means requires using a test statistic chosen depending on two factors: 1) whether the population variances are equal, and 2) whether the population variances are known. This test can be used to test the null hypothesis:

H0: μ1 - μ2 = 0 vs. the alternative of HA: μ1 - μ2?≠ 0

The test when population means are normally distributed and the population variances are unknown but assumed equal uses a pooled variance. Use the t-distributed test statistic:

t = {(X bar1 - X bar2) - (μ1 - μ2)} / {(sp2 / n1) + (sp2 / n2)}1/2

The test when the population means are normally distributed and population variances unknown and cannot be assumed to be equal uses the t-distributed test statistic that uses both samples' variances:

t = {(X bar1 - X bar2) - (μ1 - μ2)} / {(s12 / n1) + (s22 / n2)}1/2

q: Formulate a null and an alternative hypothesis about the equality of two population means (normally distributed populations, independent samples) and determine whether the null hypothesis is rejected at a given level of significance.

Sue Smith is investigating whether the announcement period abnormal returns to acquiring firms differ for horizontal and vertical mergers. She estimated the abnormal announcement period returns for a sample of acquiring firms associated with horizontal mergers and a sample of acquiring firms involved in vertical mergers. She estimated the abnormal announcement period returns for a sample of acquiring firms associated with horizontal mergers and a sample of acquiring firms involved in vertical mergers.

Abnormal returns for horizontal mergers: mean = 1%, standard deviation = 1%, sample size = 64.

Abnormal returns for vertical mergers: mean = 2.5%, standard deviation 2.0%, sample size = 81.

Assume that the population means are normally distributed, and that the population variances are equal. Is there a statistically significant difference in the announcement period abnormal returns for these two types of mergers?

Step 1: State the hypothesis.

H0: µ1 - µ2 = 0

HA: µ1 - µ2 ≠ 0

Where µ1 is the mean of the abnormal returns for the horizontal mergers and µ2 is the mean of the abnormal returns for the vertical mergers.

Step 2: Select the appropriate test statistic. Use the test statistic formula that assumes equal variances (from LOS 1.B.p).

Step 3: Specify the level of significance. There are n1 + n2 - 2 degrees of freedom. Therefore, there are 64 + 81 - 2 = 143 degrees of freedom. We will use the common significance level of 5%.

Step 4: State the decision rule regarding the hypothesis. Using the t-distribution table and the closest degrees of freedom, the critical value for a 5% level of significance is 1.980.

Step 5: Collect the sample and calculate the sample statistics. Inputting our data into the formula:

sp2 = (63)(0.0001) + (80)(0.0004) / 143 = 0.000268

t = -.015 / 0.00274 = -5.474