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# P1.T2.20.7. Common univariate random variables (student's t, F-distribution, mixture)

#### Nicole Seaman

##### Director of FRM Operations
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Distinguish the key properties and identify the common occurrences of the following distributions: Student’s t and F-distributions ... Describe a mixture distribution and explain the creation and characteristics of mixture distributions.

Questions:

20.7.1. Let X = µ + σ*T be the generalized student's t distribution where X is a linear transformation of the classic student's t distribution, as represented by T ~ t(df) or sometimes T ~ t(v). Consequently, X is a generalized random variable denoted by X ~ Gen. t(v)(µ, σ^2). If the generalized student's t distribution happens to be standardized, then it could be represented as X ~ Gen. t(v)(0, 1). If X is a generalized student's t distribution with µ = 2.0 and σ = 3.0 with degrees of freedom, v = 8, then each of the following is true EXCEPT which is false?

a. The mean of X is 2.0
b., The variance of X is 12.0
c. The skew of X is 1.5; i.e., positive or right-skew
d. The kurtosis of X is 4.5; i.e., excess kurtosis is 1.5

20.7.2. Peter is comparing two algorithms that each generate a borrower's credit score based on a vector of values such as income, age, and other variables. Among a sample of 31 credit card borrowers (n = 31), the first algorithm's scores have a standard deviation of 20.0 while the second algorithm's scores have a standard deviation of 25.0. Peter is curious to know if the two variances (and standard deviations) are statistically different, as opposed to merely a function of sampling variation. He employs the F-distribution (aka, F-test) to conduct a comparison between the two variances. With 95.0% confidence, based on his F-test, are the variances truly different?

a. No, because the test statistic is 0.800
b. No, because the test statistic is ~1.563
c. Yes, because the test statistic is 37.50
d. Yes, because the test statistic is ~46.88

20.7.3. Let's assume two normal variables: X ~ N(0, 3.0^2) and Y ~ N(4.0, 9.0^2). Let V be a mixture distribution where X has a component weight of 80% and Y has a component weight of 20%. Let W = 0.80*X + 0.20*Y with the assumption that X and Y are independent; that is, this W is the sum of independent, random variables. Which is more likely (the mixture or the convolution) to realize a negative outcome; that is, is Pr(V<0) > Pr(W<0), or on the other hand, is Pr(W<0) > Pr(V<0)?

a. The mixture (V) has a greater probability of a negative outcome; i.e., Pr(V<0) > Pr(W<0)
b. The convolution (W) has a greater probability of a negative outcome; i.e., Pr(W<0) > Pr(V<0)
c. They have the same probability of realizing a negative outcome; i.e., Pr(W<0) = Pr(V<0)
d. It cannot be answered because it depends on correlation in the case of (W)