Simulation. Two Rule For Inverting A Distribution Function

Transcription

1 Simulatio Two Rule For Ivertig A Distributio Fuctio Rule 1. If F(x) = u is costat o a iterval [x 1, x 2 ), the the uiform value u is mapped oto x 2 through the iversio process. Rule 2. If there is a jump i the graph of F at some poit x: F(x ) < F(x), the all values u i the iterval u [F(x ), F(x)) are mapped to x. Here is a example where we apply both: Example 1. Apply the iversio method for the followig distributio fuctio to determie simulated values from the uiform umbers i [0, 1]. 0.1x 0 x < 2 0.1x x < 4 F(x) = 0.2x x < x Solutio. 0 x < 2 0 u < 0.2 x=10u 2 x < 4 0. u < 0.5 x=10(u-0.1)=10u-1 4 x < u < 0.9 x=5(u+0.1)=5u x u=1 Now we fill the gaps for u: 1

2 0 x < 2 0 u < 0.2 x=10u 0.2 u < 0. x=2 2 x < 4 0. u < 0.5 x=10(u-0.1)=10u u < 0.7 x=4 4 x < u < 0.9 x=5(u+0.1)=5u u < 1 x=5 5 x u=1 F 1 (u) = 10u 0 u < u < 0. 10u 1 0. u < u < 0.7 5u u < u 1 Example 2. Apply the iversio method for the followig distributio fuctio to determie simulated values from the uiform umbers i [0, 1]. 0.1x 0 x < x < 4 F(x) = x < x Solutio. 0 x < 2 0 u < 0.2 x=10u 2 x < 4 u=0.4 4 x < 5 u=0.7 5 x u=1 2

3 Now we fill the gaps for u: 0 x < 2 0 u < 0.2 x=10u 0.2 u < 0.4 x=2 0.4 u < 0.7 x=4 0.7 u 1 x=5 F 1 (u) = 10u 0 u < u < u < u 1 Here is a example of completely discrete distributio: Example. Apply the iversio method for the followig distributio fuctio to determie simulated values from the uiform umbers i [0, 1] x < x < 4 F(x) = x < x Solutio. < x < 0 u=0 0 x < 2 u=0.1 2 x < 4 u=0.4 4 x < 5 u=0.7 5 x u=1 Now we fill i the gaps for u:

4 0 u < 0.1 x=0 0.1 u < 0.4 x=2 0.4 u < 0.7 x=4 0.7 u 1 x=5 The followig example provides the discrete distributio differetly: Example 4. The previous example ca be give to us like this: F(x) = 0.1 x = x = x = 4 1 x = 5 Solutio. Accordig to the solutio of the previous example, the u-itervals correspodig to x values are these: x F(x) u [0, 0.1) [0.1, 0.4) [0.4, 0.7) 5 1 [0.7, 1] F 1 (u) = 0 0 u < u < u < u 1 Example *. Isurace for a city s sow removal costs covers four witer moths. (i) There is a deductible of 10,000 per moth. (ii) The isurer assumes that the city s mothly costs are idepedet ad ormally distributed with mea 15,000 ad stadard deviatio 2,000. (iii) To simulate four moths of claim costs, the isurer uses the iversio method (where small radom umbers correspod to low costs). 4

5 (iv) The four umbers draw from the uiform distributio o [0, 1] are: 0.598, , 0.001, Calculate the isurer s simulated claim cost. Solutio. Φ(z 1 ) = z 1 = Φ(z 2 ) = z 2 = Φ(z ) = z =.0115 Φ(z 4 ) = z 4 = z i simulated claim cost Paymet x i = z i max(0, x i 10000) aggregate claim cost= =14,400. Example *. Four machies are i a shop. The umber eedig repair i each week has a biomial distributio with p = 0.5. For each machie, the repair time, i hours, is uiformly distributed o [0,10]. You are to estimate X, the total repair time (i hours) for a three-week period, usig simulatio. Use the followig umbers from the uiform distributio o [0, 1] to simulate the umber of machies eedig repair durig each of three weeks: 0., 0.6, 0.7 Use the followig umbers from the uiform distributio o [0,1] to simulate repair times: 0., 0.1, 0.7, 0.6, 0.5, 0.8, 0.1, 0. 5

6 Determie X. Solutio. The distributioal values for for the Biomial(m = 4, p = 0.5) are P(N = k) = ( ) 4 (0.5) k (0.5) 4 k k k p k F(k) u 0 (0.5) 4 = [0, ) 1 4(0.5) 4 = [0.0625, 0.125) 2 6(0.5) 4 = [0.125, ) 4(0.5) 4 = [0.6875, 0.975) 0 (0.5) 4 = [0.975, 1] Accordig to the Two Rules, the iverted values for the u values u = 0., 0.6, 0.7 are N = 1 2,. So, the total umber of repairs required is = 6. So, we oly take the first 6 umber amog the umbers simulated, which the are: 0., 0.1, 0.7, 0.6, 0.5, 0.8 I order to ivert these umbers form [0, 1] to [0, 10] we juxct eed to multiply them by 10. So, the iverted values are, 1, 7, 6, 5, 8 Their sum is the total repair time for three weeks: = 0 Example *. You are to use simulatio to aalyze the medical claims, Y i, i = 1, 2,..., 9 of each perso i a group of ie. Isurace coverage pays for the excess, if ay, of each claim over $ 10,000. The distributio of Y i is give by: 6

7 y P(Y i = y) You are to simulate the Y i, i = 1, 2,..., 9 for the group usig the followig uiform radom umbers o [0, 1]: Determie the average amout per perso paid by the isurace. Solutio. Distributioal values ad the associated itervals of uiform (based o the Two-Rules): after imposig the deductible y F(y) U [0, 0.10) [0.10, 0.90) [0.90, 0.99) [0.99, 1.00] The from the ie probabilities simulated, oly the two values ad result i o-zero values, ad the correspodig o-zero values are ad Therefore, the average paymet is: seve zeros 9 = 14,444 Example *. A actuary is simulatig aual aggregate loss for a product liability product, where claims occur accordig to a biomial distributio with parameters m = 4 ad q = 0.5, ad severity is give by a expoetial distributio with parameter θ = 500,000. The umber of claims is simulated 7

8 usig the iverse trasform method (where small radom umbers correspod to small claim sizes) ad a radom value of 0.58 from the uiform distributio o [0, 1]. The claim severities are simulated usig the iverse trasform method (where small radom umbers correspod to small claim sizes) usig the followig values from the uiform distributio o [0, 1]: 0.5, 0.70, 0.61, 0.20 Calculate the simulated aual aggregate loss for the product liability policy. Solutio. For the biomial distributio we have: k p k F(k). u = p 0 [0, ) = p 0 + p 1 [0.0625, 0.125) = p 0 + p 1 + p 2 [0.125, ) The simulated value u = 0.58 is i the iterval [0.125, ), so k = 2. Therefore, we oly take the first two cases u = 0.5, u = By ivertig ay u for a expoetial distributio we have x = θ l(1 u), therfore the aggregate loss is: 50000l(1 0.5) 50000l(1 0.70) = Example *. Aual detal claims are modeled as a compoud Poisso process where the umber of claims has mea 2 ad the loss amouts have a two-parameter Pareto distributio with 8 = 500 ad a= 2. A isurace pays 80% of the first 750 of aual losses ad 100% of aual losses i excess of750. You simulate the umber of claims ad loss amouts usig the iversio method. The radom umber to simulate the umber of claims is 0.8. The radom umbers to simulate loss amouts are 0.60, 0.25, 0.70, 0.10, ad Calculate the simulated isurace claims for oe year. 8

9 Solutio. We have p k = e 2 2 k k!, therefore p 0 = e 2 = 0.15 p 1 = p 2 = p = k p k F(k). u = p 0 [0, 0.15) = p 0 + p 1 [0.15, ) = p 0 + p 1 + p 2 [0.4060, ) = p 0 + p 1 + p 2 + p [0.6767, ) Sice the simulated u = 0.8 is i the last iterval, so the umber of claims is N =, therefore we oly cosider the u umbers 0.60, 0.25, For Pareto distributio: ( ) u = F(x) = 1 x = x 1 u u = 0.60 x = u = 0.25 x = 77.5 u = 0.70 x = sum = Isurace paymet = (0.8)(750) + ( ) =

10 Determiig Risk Measures Via Simulatio If the uderlyig distributio is two complex, the we may use simulatio to estimate the risk measure VaR ad TVaR. Here is how: Suppose you have collected a sample y 1 y 2 y As we kow, VaR is a percetile value, p say, for which we let k = [p] + 1 ([p] beig the iteger part) ad the take the estimates: VaR p = y k TVaR p = 1 k + 1 j=k It is show i advaced topics that the variace of the estimator TVaR is ] 2 ] s Var [ TVaR 2 p + p[ TVaR VaR p = k + 1 where s 2 p is the sample variace (ote that there are k + 1 terms i the summatio): s 2 p = 1 k j=k ( ) 2 y j TVaR The estimator for the variace of TVaR p ca be used to form cofidece itervals. y j Example (from the textbook). Cosider a Pareto distributio with α = 2 ad θ = 100. Use 10,000 simulatios to estimate the risk measures with p = 0.95 Solutio. To start with, we simulate 10,000 umbers from Uiform(0, 1) ad the ivert them to get 10,000 draws of Pareto(α = 2, θ = 100). k = [p] + 1 = [(0.95)(10000)] + 1 = 9501 Puttig the umbers geerate i order, we see that y 9501 = The also: k + 1 = = 500 TVaR 0.95 = j=9501 y j =

11 We get from the simulatio that: [ ] Var TVaR p = The a 95% cofidece iterval for the true value of TVaR is : ± ± The true value is TVaR=47.21 which happes to be i the cofidece iterval. 11

12 Bootstrap Approximatio We recall that if ˆθ is a estimator for a parameter θ, the the mea square error is [ ] MSE(θ) = E ( ˆθ θ) 2 θ I calculatig this expectatio we use the desity whe the true parameter is assumed to be θ. But whe the uderlyig distributio is ot kow, we calculate the Bootstrap Approximatio for the mea square error defied by: [ ] MSE F (θ) = E ( ˆθ θ(f )) 2 θ i which the expectatio is calculated uder the empirical distributio, so all probabilities are equal to 1 for a sample of size. Here is how the bootstrap procedure is doe: (i) Draw a sample of size (or it might be give to us). Calculate the parameter of iterest as if this sample is the populatio. So, if the parameter of iterest is the mea of the populatio, the the sample mea is take for ˆθ = θ(f ). (ii) Assumig that the sample i part (i) is the populatio, we resample with replacemet from it. These ew samples are called bootstrap samples. There are of them. I some sese, we are dealig with a i.i.d. {X 1..., X } from this ew populatio. (iii) For each i = 1,2,..., bootstrap sample, calculate the parameter of iterest ad deote it by ˆθ i. (iv) The bootstrap estimate for MSE is the simple average: i=1 MSE F (θ) = 1 ( ˆθ i ˆθ) 2 Note. The bootstrap approximatio is a radom variable a value of which is observed through samplig. It ca be show (but ot i this course) that the bootstrap approximatio has this ice property that it coverges asymptotically (i.e. whe the sample size gets large) to the true value of the parameter. 12

13 Example (from the textbook). A sample of size from a populatio produced 2,, ad 7. Determie the bootstrap estimate of MSE of the sample mea as a estimate for the populatio mea. Solutio. The sample mea of this particular sample is ˆθ = = 4 Now we resample from the ew populatio {2,, 7}. There are = 27 possibilities. There is oly oe case of {2, 2, 2}, so we assig probability of 1 27 to this for which ˆθ i = = 2. There are possibilities of havig oe ad two 2 s. The mea of each of them is 7 ad we asig the probability of 27 to this case. Ad so o. The the bootstrap estimate of MSE is: ( ) ( ) 1 MSE( X) = (2 4) 2 + ( )2 + = 14 9 Shortcut. It ca be show that with the aid of some algebra work that whe estimatig the populatio mea with the sample mea, the bootstrap of MSE( X) is equal to MSE( X) = 1 2 (X i X) 2 i=1 Example. Aswer the previous questio usig this formula. Solutio. With the aid of the ew formula we calculate: MSE( X) = 1 9 { (2 4) 2 + ( 4) 2 + (7 4) 2} = 14 9 Example. I a study o three lives, oe life dies at time, oe leaves the study at time 4, ad oe dies at time 5. The product-limit estimator is used to estimate S(). Calculate the bootstrap approximatio of the mea square error of the product-limit estimator. Solutio. The product-limit estimate is 2. Now we resample from this sample. I each draw, either a comes up or ot. The chace of drawig a three i each draw is 1. The umber of s draw i three attempts is biomial with parameters m = ad q = 1. So, the probability of havig k = 1,2, of s is 1

14 ( k )( 1 ) k ( ) 2 k So, here are the situatios we have: Case Probability Estimate of S() three s two s 6 27 oe o s MSE = 1 ( 0 2 ) ( ) ( ) ( 1 2 ) 2 = Here the examples 92.2 ad 92. of the Fia s study guide were solved i class 14

15 Determiig the sample size Example. To estimate E[X], you have simulated X 1, X 2, X, X 4, ad X 5 with the followig results: 1, 2,, 4, 5 You wat the stadard deviatio of the estimator of E[X] to be less tha Estimate the total umber of simulatios eeded. Solutio. The estimator of E(X) is X = X 1+ +X whose stadard deviatio is σ. But sice the true value σ is ot reachable, we use the estimate s (sample stadard deviatio) istead of it. This s ca be calculated from ay sample; so we use the sample give to us. s 2 = (1 )2 + (2 ) 2 + ( ) 2 + (4 ) 2 + (5 ) 2 4 = 2.5 s < < 0.05 > (0.05) 2 Example. We wat to estimate a probability P( < X 70]. For this, we collect a sample of 100 from the populatio where X represets ad the result is: iterval umber of observatios (0, ] 12 (, 70] 66 (70, 95] 22 Estimate the umber of simulatios eeded for beig 95% cofidet that the estimated probability is withi 1% of the expected. Solutio. Let p = P( < X 70]. Let the simulated observatios are associated with the i.i.d. {X 1,..., X }. Let {Y 1,..., Y } be the collectio of Beroulli radom variables such Y i takes o the value 1 if the evet < X i 70 occurs. Let P be the sum Y Y which deotes the umber of observatios that happe to fall i the iterval (, 70]. Accordig to the Cetral Limit Theorem, the 15

16 radom variable P = Ȳ is approximately ormal with mea P( < X 70] ad variace σ 2 (Y 1 ) = p(1 p). We use Ȳ to estimate P( < X 70]. We wat so large to have ( ) P( Ȳ p 0.01p) 0.95 P Ȳ E(Ȳ) 0.01E(Ȳ) 0.95 ( P N(0,1) 0.01E(Ȳ) Std(Ȳ) ) E(Ȳ) 1.96 Std(Ȳ) 0.01p p(1 p) p p But sice the exact value of p is ot reachable, we use its approximate value foud based o the data set: ˆp = But sice Ȳ is ormal, this is equivalet to havig p(1 p) p(1 p) p p (by the term expected they refer to the true value). But sice p is ot reachable, we estimate it with the value = 0.66 that we read from the sample above (0.66)(0.4) (1.96)2 (0.66)(0.4) (0.0066) Note. The same techique ca be used to solve example 89.7 of the Fia s study guide which is actually oe of the examples of the textbook. For this topic, the examples 89.6 ad 89.7 of the Fia s study guide were solved i class. Some details will be added to example 89.7 later. 16

17 Simulatig from discrete mixtures Suppose you wat to simulate from a discrete mixture distributio such as F X (x) = a 1 F X1 (x) + + a k F Xk (x) You would do this i two steps: Step 1. Simulate a value from the discrete radom variable Y with P(Y = j) = a j j = 1,...,k. Suppose that the simulated value is j. Note that to simulate a value from Y you eed to first divide the iterval (0,1) accordig to the probabilities {a 1, a 2,..., a k } (the first subiterval would have the legth a 1, the secod iterval would have legth a 2 ad so o, from the left to the right.) The simulate a umber for Uiform(0,1). If that umber falls i the j -th subiterval,say, the we proceed with the secod step: Step 2. Simulate a observatio from F j (x). Note. This techique is called Radomized Samplig. Be careful. If you wat to simulate 5 observatios, for example, you eed to do both steps for each of the 5 observatios. Example (questio 81 of the SOA sample questios). You wish to simulate a value, Y from a two poit mixture. With probability 0., Y is expoetially distributed with mea 0.5. With probability 0.7, Y is uiformly distributed o [-, ]. You simulate the mixig variable where low values correspod to the expoetial distributio. The you simulate the value of Y, where low radom umbers correspod to low values of Y. Your uiform radom umbers from [0, 1] are 0.25 ad 0.69 i that order. Calculate the simulated value of Y. 17

18 Solutio. Sice the first uiform umber is 0.25 ad this is i the iterval (0, 0.) for the expoetial distributio, so Y must be chose from the expoetial distributio. Now we look for y satisfyig: F(y) = exp( y ) = 0.69 y =

19 Simulatig p-values Example. The 6 observatios 2,, 6, 7, 8, 9 are fitted to a uiform distributio o [0,10]. To test the fit, a chi-square test is performed. The data are split ito two groups, [0,5) ad [5,10]. Because there are so few observatios, the chi-square statistic is simulated. You perform four simulatio rus with the followig sets of uiform radom umbers o [0,1]: Ru 1: Ru 2: Ru : Ru 4: Determie the simulated p-value of the fit. Solutio. Step 1. Calculate the chi square statistic for the observed data set: χ 2 = (2 )2 + (4 )2 = 2 Step 2. We ivert the values i each ru to get values i [0,10]. For that we just eed to multiply by 10. The the iverted values of ru 1 become: Iverted ru 1 values: for which we have a chi square value of 2. Similarly, we get these values: chi-square for ru 1 : 2 chi-square for ru 2 : 2 chi-square for ru 1 : 8 chi-square for ru 1 : 0 19

20 Now, three out of four simulated chi-square values are larger tha or equal to the observed oe. So p-value =