Characteristics of Estimators, Mean Square Error, and Assorted Exam Style Questions

This section of sample problems and solutions is a part of The Actuary's Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 60 of the Study Guide. See an index of all sections by following the link in this paragraph.

When you use an estimate Ŵ for a value W, then the mean square error associated with the estimate is defined as MSE= E[(Ŵ - W)²].

We now define what it means for an estimator to be consistent, following Larsen and Marx.

"An estimator Ŷ_n = h(W₁, W₂, ..., W_n) is said to be consistent for Y if it converges in probability to Y - that is, if for all ε > 0, lim_n→∞Pr(|Ŷ_n - Y| < ε) = 1" (406).

An estimator Ŷ of Y is an unbiased estimator if Ŷ does not "systematically err in any particular direction" (Larsen and Marx 2006, p. 381). More formally, an estimator Ŷ for Y is unbiased if E[Ŷ] = Y for all Y.

If you are given that values x₁,...., x_n follow an inverse exponential distribution with parameter θ, then this is the same as saying that the values 1/x₁,...., 1/x_n follow an exponential distribution with parameter 1/θ. Therefore, you should convert all relevant values to the values pertaining to an exponential distribution and apply known techniques to them.

Reminder: Both the maximum likelihood estimator and the method of moments estimator for the mean of an exponential distribution are equal to the mean of the sample drawn from that distribution.

Source: Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3L - Spring 2008.

Larsen, Richard J. and Morris L. Marx. An Introduction to Mathematical Statistics and Its Applications. Fourth Edition. Pearson Prentice Hall: 2006. pp. 381, 406.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L60-1. Similar to Question 2 from the Casualty Actuarial Society's Spring 2008 Exam 3L. The true value of a parameter Y is 3600. An estimator Ŷ_n of Y, for any positive integer sample size n, is as follows: Ŷ_n = 3600*n/(n+2). Which of the following statements about Ŷ_n are correct? More than one answer may be possible.

(a) Ŷ_n is an unbiased estimator of Y.

(b) Ŷ_n is a consistent estimator of Y.

(c) The mean square error of Ŷ₁₉ is less than 100000.

Solution S3L60-1.

We examine statement (a): Ŷ_n can be an unbiased estimator of Y only if E[Ŷ_n] = Y for all Y. In this case, each value of n corresponds to a single value of Ŷ_n, and for any value of n, 3600*n/(n+2) < 3600, so Ŷ_ncannot be equal to Y and thus cannot be an unbiased estimator of Y. Thus, (a) is false.

We examine statement (b): Ŷ_n can be a consistent estimator of Y if and only if for all ε > 0, lim_n→∞Pr(|Ŷ_n - Y| < ε) = 1.

|Ŷ_n - Y| = |3600*n/(n+2) - 3600|. lim_n→∞3600*n/(n+2) = 3600 by a single application of L'Hôpital's rule, so lim_n→∞Pr(|Ŷ_n - Y| < ε) = Pr(|3600 - 3600| < ε) = Pr(0 < ε) = 1, since ε > 0. Thus, the condition for consistency is met and Ŷ_n is a consistent estimator of Y. So (b) is true.

We examine statement (c): We use the formula E[(Ŵ - W)²] = E[(Ŷ₁₉ - Y)²] = (Ŷ₁₉ - Y)² = (3600*19/21 - 3600)² = MSE = 117551.0204 > 100000, so (c) is false.

Thus, only (b) is true.

Problem S3L60-2. Similar to Question 3 from the Casualty Actuarial Society's Spring 2008 Exam 3L. The random variable X follows an inverse exponential distribution with parameter θ. A sample of 5 values is drawn from this distribution: 45, 513, 95, 675, 134. What is the maximum likelihood estimator for θ?

Solution S3L60-2. Since X follows an inverse exponential distribution with parameter θ, it follows that 1/X follows an exponential distribution with parameter (mean) 1/θ. We can restate our sample values as values of the random variable 1/X: 1/45, 1/513, 1/95, 1/675, 1/134. Because 1/X follows an exponential distribution, the maximum likelihood estimator of 1/θ is just the sample mean: (1/45 + 1/513 + 1/95 + 1/675 + 1/134)/5 = about 0.0087284048. The maximum likelihood estimator for θ is just the inverse of the above result, or 1/0.0087284048 = about 114.5684724.

Problem S3L60-3. Similar to Question 4 from the Casualty Actuarial Society's Spring 2008 Exam 3L. This question is a review of the concepts in Section 55. It is hypothesized that the number of rabid rhinoceros attacks experienced by a member of a particular tribe in a year follows a Poisson distribution with mean 0.2. For 1000 tribe members, this is the distribution of rabid rhinoceros attacks during the year.

Number of members suffering 0 attacks:879

Number of members suffering 1 attack: 98

Number of members suffering 2 or more attacks: 23.

Find the chi-square statistic associated with the hypothesis above.

Solution S3L60-3. We use the formula D = _i=1^t∑[(X_i - np_i)²/np_i], where n = 1000, the X_i's are the given sample values, and the p_i's are calculated using probabilities of the Poisson distribution. We find p₀ = e^-0.2, so np₀ = 1000e^-0.2 = 818.7307531.

We find p₁ = 0.2e^-0.2, so np₁ = 200e^-0.2 = 163.7461506.

We find p_{2 or more} = 1 - p₀ - p₁ = 1 - 1.2e^-0.2, so np_{2 or more} = 1000(1 - 1.2e^-0.2) = 17.52309631.

Thus, we can find D = (879 - 818.7307531)²/818.7307531 + (98 - 163.7461506)²/163.7461506 +

(23 - 17.52309631)²/17.52309631 = D = about 32.40033295.

Problem S3L60-4. Similar to Question 6 from the Casualty Actuarial Society's Spring 2008 Exam 3L. This question is a review of concepts in Section 56. You have a sample of 51 values. The sample mean is 94500 and the sample standard deviation is 20000. You perform a two-sided hypothesis test for the mean of the distribution in question, with H₀: µ = 98000 and H₁: µ ≠ 98000. Which of the following decisions would you legitimately make with regard to accepting or rejecting the null hypothesis H₀ at significance levels α.

(a) Do not reject at α = 0.1.
(b) Do not reject at α = 0.05, but reject at α = 0.1.
(c) Do not reject at α = 0.02, but reject at α = 0.05.

(d) Do not reject at α = 0.01, but reject at α = 0.02.

(e) Reject at α = 0.01.

Use the Table of Percentage Points of the t Distribution where appropriate.

Solution S3L60-4. Since we have sample size n = 51, sample standard deviation 20000, and sample mean µ_actual = 94500, we can use the Student t distribution with test statistic

t = (µ_actual - µ_hypothesized)/(S/√(n)). The t distribution used here will have 51 - 1 = 50 degrees of freedom. We find t = (94500 - 98000)/(20000/√(50)) = about -1.237436867. Since we have a two-sided test, it does not really matter whether our t value is positive or negative. We do to a t value of -1.237436867 precisely what we would do to a t value of 1.237436867.

We use the table linked above to find the row of values of t associated with 50 degrees of freedom. We refer to the two-tail probabilities associated with each entry in that row. We find that for 50 degrees of freedom, the t-statistic associated with α = 0.1 is 1.299 and 1.237436867 < 1.299. Thus, our t statistic is small enough for us to not reject the null hypothesis at the 0.1 significance level. Therefore, (a) is true.

Problem S3L60-5. Similar to Question 7 from the Casualty Actuarial Society's Spring 2008 Exam 3L. You have a random sample of four values x, and you know that the values are uniformly distributed over the interval 1000 < x < 2400. Find the probability that no more than three of these values will be greater than 1900. (Hint: This is the kind of problem that you can solve without any special knowledge, except basic knowledge of what the uniform distribution is. Think about how you might be able to do this.)

Solution S3L60-5. The only possible way that more than three of these values can be greater than 1900 is if all four of the values are greater than 1900. The probability of any given value being greater than 1900 is (2400 - 1900)/(2400 - 1000) = 5/14. So the probability of all four values being greater than 1900 is (5/14)⁴. We desire the complement of that probability or

1 - (5/14)⁴ = about 0.9837307372.

See other sections of The Actuary's Free Study Guide for Exam 3L.