Method of Moments. which we usually denote by X or sometimes by X n to emphasize that there are n observations.

Transcription

1 Method of Moments Definition. If {X 1,..., X n } is a sample from a population, then the empirical k-th moment of this sample is defined to be X k Xk n n Example. For a sample {X 1, X, X 3 } the empirical second moment is X 1 +X +X 3 3. Example. The empirical first moment of a sample {X 1,..., X n } is X 1+ +X n n which we usually denote by X or sometimes by X n to emphasize that there are n observations. Assume a parametric distribution defined in terms of r parameters (θ 1,..., θ r ; for example the distribution Exponential(θ is define in term of only one parameter, while the Normal(µ, σ has two parameters. Suppose that we have observed n data points from the population under study. We want to use this data set to estimate the parameters of the model. Based on this sample we can calculate the empirical moments. Then to estimate the parameters of the model, we match the first r empirical moments with their theoretical counterparts: theoretical k-moment = (sample or empirical k-th moment, k=1,..., r So, here we have r equations in r unknowns (θ 1,..., θ r. By solving this system of equations we get the estimates for the parameters. Note. Some manuscripts use the notation E(M,..., E(M r to denote the sample moments. Example. Four losses are observed from a Gamma distribution. The observed losses are 00, 300, 350, and 450. Find the method of moments estimate for α. First Step: The Gamma distribution has two parameters α and θ. The theoretical 1

2 first moment is E(X = αθ and the theoretical second moment is E(X = α(α + 1θ. Now we calculate the empirical counterparts: Second Step: Calculate the sample moments: sample first moment = = 35 sample second moment = = 113, 750 Third Step: Form the equations: E(X = E(M αθ = 35 E(X = E(M α(α + 1θ = 113, 750 αθ = 35 (1 α(α + 1θ = 113, 750 ( = α(α + 1θ 113, 750 (αθ = (35 α + 1 = α = 13 α Example. You are given the following sample of five claims: 4, 5, 1, 99, 41 You fit a Pareto distribution using the method of moments. Determine the 95 th percentile of the fitted distribution. For the Pareto distribution we have E(X = other hand: E(M = = 110 E(M = = θ α 1 and E(X θ = (α 1(α. On the

3 Next Step: Form the equations: E(X = E(M θ α 1 = 110 E(X = E(M α(α + 1θ = θ α 1 = 110 (1 θ (α 1(α = ( θ / ( θ = (α 1(α α = (α 1 α = cross multiplying α = putting in (1 θ = ( θ α ( = F (x = 1 = 1 x = 369 x + θ x Example. For a sample of dental claims x 1, x,..., x 10, you are given: (i. x i = 3860 and x i = 4, 574, 80. (ii. Claims are assumed to follow a lognormal distribution with parameters µ and σ. (iii. µ and σ are estimated using the method of moments. Calculate E(X 500 for the fitted distribution. For the lognormal distribution we have E(X = exp(µ + 1 σ and E(X = exp(µ + σ. We equate these with their sample counterparts: exp(µ + 1 σ = = 386 exp(µ + σ = 4,574,80 10 = 457, 480. exp(µ + σ / ( exp(µ , 480. σ = (386 = 3.07 exp(σ =

4 σ = ln(3.07 = 1.1 exp(µ + (1.1 = 457, 480. µ = 5.39 Now (from the table: ( [ ( ] E(X r = e µ+ 1 ln r µ σ ln r µ σ Φ + r 1 Φ σ σ ( [ ( ] ln ln E(X 500 = 386 Φ Φ 1.1 σ [ ] = 386 Φ( Φ(0.77 = 59 Here is an example for mixture distribution: Example. You are given the following: The random variable X has the density function f(x = wf 1 (x + (l wf (x, 0 < x <, 0 w 1. A single observation of the random variable X yields the value 1. 0 xf 1 (xdx = 1 0 xf 1 (xdx = f (1 = f 1 (1 0 Determine the method of moments estimate of w. E(X = w E(X 1 + (1 we(x = w(1 + (1 w( = w But the single observation x 1 = 1, so the sample mean is also 1. By equating the theoretical mean with the sample mean, we get: 4

5 w = 1 w = 1 Here is an example for non-classical distributions: Example. You are given the following: The random variable X has the density function f(x = θ (θ x, 0 < x < θ. A random sample of two observations of X yields values 0.50 and Determine θ, the method of moments estimator of θ. E(X = θ θ 0 x(θ xdx = θ θ (θ θ x dx x dx = θ ( θ θ3 = θ 3 3 On the other hand, the sample first moment is: = 0.7 Matching the two values gives us: θ = 0.7 θ =.1 3 Here is an example for dealing with discrete distributions: Example. We want to estimate the parameters β and r in the negative binomial distribution. The first and second empirical moments are 6 and 60. Find the method of moment estimate of P (N. 5

6 E(N = rβ E(N = Var(N + E(N = rβ(1 + β + (rβ rβ = 6 rβ(1 + β + (rβ = 60 By putting rβ = 6 in the second equation, we get 6(1 + β + 36 = 60 β = 3 r = P (N = 1 P (N = 0 P (N = 1 = 1 1 (1 + β r rβ (1 + β r+1 = = Below is an example for dealing with negative moments: Example. We have observed the following claim sizes 10, 13, 16, 0, 3 We want to fit an inverse exponential model to this data. Calculate the method of moments estimate for the probability of claim being higher than 1. For this distribution only the negative moments exist. If θ is the parameter of this distribution, then we have E(X 1 = θ 1 On the other hand, the sample negative moment is: Matching the two values gives us: = Harmonic average =

7 E(X 1 = E(M 1 θ 1 = θ = Using the table, we will have: F (x = exp( θx P (X > 1 = 1 F (1 = 1 exp( = Below is an example dealing with censoring. But before getting to the example, note that in presence of right censoring the expected value of claim is actually the limited expected value E(X u where u is the policy limit. The moments for this distribution are E[(X u k ]. These moments will be used for the purpose of method of moments estimation. So, the model distribution and the sample distribution are both censored. Example. We have observed the following 10 values of claim sizes: 100, 100, 150, 170, 170, 00, 0, 300 +, 300 +, where the last three are censored observations. We want to fit the distribution Pareto(α =, θ to this data. What is the method of moments estimation for θ?. E(M 300 = = 01 Using the table, we will have: E(X 300 = [ θ ( ] θ α 1 1 = 300θ α θ θ E(X 300 = E(M θ θ = θ = 01(300 + θ t = (01( =

8 Note. If we had α = 3, then we would need to solve the equation θ which would require a numerical method to solve it. [ ( ] 1 θ 300+θ = 01 Below is an example dealing with left truncation. For this type of situations, the model distribution and the sample distribution must be truncated. Note that in this process, an estimation for the underlying distribution is found, but not for the truncated one. Example. The following values of a random variable have been observed 0.55, 0.60, 0.73, 0.8, 0.95 We want to fit the distribution with density f(x = 1 + c + cx 0 x 1 where 0 c 1. What is the MME for c if the distribution is assumed to be truncated at x = E(M M > = = E(X X > = x f(x dx = f(x dx x (1 + c + cx dx = (1 + c + cx dx (x + cx + cx dx (1 + c + cx dx = [ ( 1+c x + 3 cx3 ] 1 [(1 + cx + cx ] 1 = c c E(X X > = E(M M > c c = c = Below is an example dealing with (complete grouped data. So in general assume that we have k intervals formed by the boundary points: c 0 < c 1 < < c k 8

9 and that n is the total number of observations, and that n j is the number of observations falling in j-th interval (c j 1, c j ]. So we have n = k j=1. For a grouped data no adjustment is needed for the model being fitted, but on the other hand since the exact values of the sample points are not known, we need to approximate the sample moments. If in each interval (c j 1, c j ] no point has an advantage over another point, then we can assume that the the sample points have uniform distribution over this interval. Then the average value in each interval is just the midpoint of the interval. Then the weighted average k j=1 is taken for the sample mean x. c j 1 + c j n j n = 1 n k j=1 c j 1 + c j n j Example. Consider the following grouped data we had sometime ago.. Interval Number of observations (0, ] 5 (, 10] 10 (10, 100] 10 (100, 1000] 5 We want to fit an Exponential(θ to this data. Find the MME for θ. E(M = 1 n = 1 50 k j=1 [( 0 + c j 1 + c j (5 + n j ( + 10 E(X = E(M θ = (10 + ( (10 + ( ] (5 =

10 Finally we see here in an example how to calculate the limited expected value E(X u for a grouped data. Suppose you are given the following grouped data: Interval Number of observations (0, ] 5 (, 10] 10 (10, 100] 10 at This data is right-censored at 100. Then the limited expected value E(X 100 is calculated through: E(X 100 = 1 90 [( 0 + (5 + ( + 10 (10 + ( ] (10 + (100(45 =