The Delta Method for Estimating Functions of Parameters

The delta method is used to estimate quantities that are the functions of parameters in a distribution, where the parameters themselves need to be estimated through methods such as maximum likelihood estimation.

Let θ be a parameter of a distribution, let θ^ be an estimate of θ (for instance, a maximum likelihood estimate), and let g(θ) be a function of θ. Here, we are ultimately concerned with estimating g(θ) by using g(θ^).

If there is only one parameter under consideration, the following is true:

"Let θ^ be an estimator of θ that has an asymptotic normal distribution with mean θ and variance σ²/n. Then g(θ^) has an asymptotic normal distribution with mean g(θ) and asymptotic variance g'(θ)²σ²/n." (Klugman, Panjer, and Willmot 2008, p. 398).

Some of the problems in this section were designed to be similar to problems from past versions of Exam 4/C, offered jointly by the Casualty Actuarial Society and the Society of Actuaries. They use original exam questions as their inspiration - and the specific inspiration is cited to give students an opportunity to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Source:
Exam C Sample Questions and Solutions from the Society of Actuaries.

Loss Models: From Data to Decisions, (Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 15, pp. 397-398.

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

Problem S4C60-1. For an exponential distribution of random variable X, you are trying to find Pr(X > 50). However, the parameter θ is not known for the distribution and is estimated using the method of maximum likelihood. The sample used to estimate θ has size 34. Moreover, θ^, the maximum likelihood estimate of θ, follows an asymptotic normal distribution with mean θ and variance 9. Estimate the asymptotic variance of Pr(X > 50) using the delta method.

Solution S4C60-1. We are given that g(θ) = Pr(X > 50) = S(50) = e^-50/θ for an exponential distribution.

The asymptotic variance can be estimated via the formula Var^(g(θ)) = g'(θ)²σ²/n.

First we find g'(θ) = (50/θ²)(e^-50/θ). Thus, g'(θ)² = (2500/θ⁴)(e^-100/θ).

We are given that the variance of θ^ is σ²/n = 9. For an exponential distribution, the maximum likelihood estimate is the same as the sample mean, so Var(θ^) = Var(sample mean) = Var(X)/n. For an exponential distribution, Var(X) = θ², so Var(θ^) = θ²/n = 9. Since n = 34, θ² = 9*34 = 306 → θ = 17.49285568. Thus, our estimate Var^(g(θ)) is 9(2500/θ⁴)(e^-100/θ) =

9(2500/17.49285568⁴)(e^{-100/17.49285568}) = Var^(g(θ)) = 0.0007907575714*10^-4.

Problem S4C60-2. For an exponential distribution of random variable X, you are trying to find Pr(X > 50). However, the parameter θ is not known for the distribution and is estimated using the method of maximum likelihood. The sample used to estimate θ has size 34 and sample mean of 18. Moreover, θ^, the maximum likelihood estimate of θ, follows an asymptotic normal distribution with mean θ and variance 9. Use the delta method to develop a 90% confidence interval for Pr(X > 50).

Solution S4C60-2. Using our sample mean of 18 as the maximum likelihood estimate, we set as the center of the confidence interval the value S(50) ≈ e^-50/18 = 0.062176524.

We happen to know the true value of θ from Solution S4C60-1, and the variance of S(50) based on the true value of θ is 0.0007907575714*10^-4. Note that if we did not know the true value of θ, we would need to estimate Var(X) = θ^²/n, where θ^ would be the maximum likelihood estimate of 18.

The z-scores associated with a 90% confidence interval are found via the Excel inputs "=NORMSINV(0.95)" and "=NORMSINV(0.05)". These values are 1.644853627 and

-1.644853627, respectively.

Hence, Pr(X > 50) = 0.062176524 ± 1.644853627*√(0.0007907575714*10^-4) and our 90% confidence interval for Pr(X > 50) is (0.0159225623, 0.1084304857).

Problem S4C60-3. Similar to Question 180 of the Exam C Sample Questions from the Society of Actuaries. The mean of a sample of size 2 is 19. An exponential distribution is used to fit the data, and the parameter θ of the distribution is found via the method of maximum likelihood. Now a new sample of size 2 is taken. The mean of this new sample is denoted by random variable Z. Use the delta method to estimate the variance of F_Z(23), the probability that the mean of this new sample is 23 or less. Use the Exam 4 / C Tables where necessary.

Solution S4C60-3. The probability that the mean the new sample is less than or equal to 23 is the same as the probability that the sum of the 2 values in the new sample is less than or equal to 23*2 = 46. We can consider each value in the new sample as following an exponential distribution with parameter θ. Thus, the sum of the two values in the new sample follows a distribution that is the sum of these two exponential distributions - i.e. a gamma random variable with parameters α = 2 and θ. Since θ was estimated using the method of maximum likelihood, and the maximum likelihood estimate for an exponential distribution is just the sample mean, the estimate of θ is 19. We create a new random variable Q = 2Z, where Q is the sum of the values of the sample of size 2.

For a gamma distribution, F(x) = Γ(α; x/θ) = (1/Γ(α))₀^x/θ∫t^α-1e^-tdt. Here, x/θ = 46/19, so F_Q(46) =

Γ(2; 46/θ) = (1/Γ(2))₀^46/θ∫te^-tdt = ₀^46/θ∫te^-tdt.

To integrate te^-t, we use the Tabular Method:

Sign........u............dv

+...........t...............e^-t

-...........1..............-e^-t

+..........0..............e^-t

Thus, ∫te^-tdt = C -te^-t - e^-t, where C is a constant.

Hence, ₀^46/θ∫te^-tdt = (-te^-t - (e^-t)│ ₀^46/θ = 1 - exp(-46/θ) - (46/θ)exp(-46/θ) = F_Q(46) = F_Z(23) = g(θ).

Therefore, g'(θ) = (-46/θ²)exp(-46/θ) - (-46/θ²)exp(-46/θ) - (46²/θ³)exp(-46/θ) =

g'(θ) = -(46²/θ³)exp(-46/θ). Thus, g'(θ)² = (46⁴/θ⁶)exp(-92/θ). The maximum likelihood estimate for θ is 19, and the maximum likelihood estimate for the variance of the sample mean is the maximum likelihood estimate for the variance of θ (19²), divided by the sample size of 2. Thus, the estimate for the variance of the sample mean is 19²/2 = 180.5.

Likewise, g'(θ)² ≈ (46⁴/θ⁶)exp(-92/θ) = (46⁴/19⁶)exp(-92/19) = 0.0007509484486 with the maximum likelihood estimate 19 of θ substituted for θ.

Thus, the estimate of the variance of F_Z(23) is 0.0007509484486*180.5 = 0.135546195.

Problem S4C60-4. Similar to Question 231 of the Exam C Sample Questions from the Society of Actuaries. It is known that the 90% linear confidence interval for the hazard rate function H(x) is (0.45, 0.99). Use the delta method to find the 90% linear confidence interval for the survival function S(x).

Solution S4C60-4. First, we find the center of the confidence interval for H(x): (0.45+0.99)/2 = 0.72.

The z-scores associated with a 90% confidence interval are found via the Excel inputs "=NORMSINV(0.95)" and "=NORMSINV(0.05)". These values are 1.644853627 and

-1.644853627, respectively.

We know that 0.99 = 0.72 + 1.644853627*√(Var(H(x))). Thus,

0.27 = 1.644853627*√(Var(H(x))) →

0.1641483446 = √(Var(H(x))) →

Var(H(x)) = 0.026944679.

We can also find the center of the confidence interval for S(x) by noting that S(x) = exp(-H(x)), so the center of the interval for S(x) is exp(-0.72) = 0.486752256.

To find Var(S(x)), we use the delta method. Let g(θ) = S(x) = exp(-H(x)).

Then g'(θ) is the first derivative of S(x) with respect to H(x):

g'(θ) = -exp(-H(x)) and so g'(θ)² = exp(-2H(x)).

Hence,

Var^(S(x)) = exp(-2H(x))*Var(H(x)), where we use 0.72 as our estimate of H(x) in exp(-2H(x)).

Therefore, Var^(S(x)) = exp(-2*0.72)*0.026944679 = 0.0063839424.

Our 90% confidence interval for the survival function is thus

0.486752256 ± 1.644853627*√(0.0063839424) or (0.3553291469, 0.6181753651).

Problem S4C60-5. Similar to Question 277 of the Exam C Sample Questions from the Society of Actuaries. You have collected a sample consisting of the following values:

23, 34, 46, 68, 89, 120, 120, 150, 250.

This data is known to follow an exponential distribution, with the parameter θ being estimated by the method of maximum likelihood. Use the delta method to estimate the variance of the survival function S(200).

Solution S4C60-5. First, we note that, since the distribution in question is exponential the maximum likelihood estimate of θ is the sample mean or (23+34+46+68+89+120+120+150+250)/9 = 100.
Let S(200) = g(θ) = exp(-200/θ).

Then g'(θ) = (200/θ²)exp(-200/θ) and g'(θ)² = (200²/θ⁴)exp(-400/θ).

We substitute our maximum likelihood estimate of 100 for θ:

g'(100)² = (200²/100⁴)exp(-400/100) = 7.3262555555555*10^-6.

Now we find an estimate of Var(θ^) = Var(θ)/n = 100²/9 = 1111.11111111.

Thus, our estimate Var^(S(200)) = 1111.11111111*7.3262555555555*10^-6 =

Var^(S(200)) = 0.008140284.

See other sections of The Actuary's Free Study Guide for Exam 4 / Exam C.