4. Partial Sums and the Central Limit Theorem

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1 1 of 10 7/16/2009 6:05 AM Virtual Laboratories > 6. Radom Samples > Partial Sums ad the Cetral Limit Theorem The cetral limit theorem ad the law of large umbers are the two fudametal theorems of probability. Roughly, the cetral limit theorem states that the distributio of the sum of a large umber of idepedet, idetically distributed variables will be approximately ormal, regardless of the uderlyig distributio. The importace of the cetral limit theorem is hard to overstate; ideed it is the reaso that may statistical procedures work. Partial Sum Processes Defiitios Suppose that X = (X 1, X 2,...) is a sequece of idepedet, idetically distributed, real-valued radom variables with commo probability desity fuctio f, mea μ, ad variace σ 2. Let Y = i =1 X i, N Note that by covetio, Y 0 = 0, sice the sum is over a empty idex set. The radom process Y = (Y 0, Y 1, Y 2,...) is called the partial sum process associated with X. Special types of partial sum processes have bee studied i may places i this project; i particular see the biomial distributio i the settig of Beroulli trials the egative biomial distributio i the settig of Beroulli trials the gamma distributio i the Poisso process the the arrival times i a geeral reewal process Recall that i statistical terms, the sequece X correspods to samplig from the uderlyig distributio. I particular, (X 1, X 2,..., X ) is a radom sample of size from the distributio, ad the correspodig sample mea is M = Y = 1 i =1 X i By the law of large umbers, M μ as with probability 1. Statioary, Idepedet Icremets 1. Show that if m the Y Y m has the same distributio as Y m. Thus the process Y has statioary icremets.

2 2 of 10 7/16/2009 6:05 AM 2. Show that if the (Y 1, Y 2 Y 1, Y 3 Y 2,...) is a sequece of idepedet radom variables. Thus the process Y has idepedet icremets. 3. Coversely, suppose that V = (V 0, V 1,...) is a radom process with statioary, idepedet icremets, i the sese of Exercise 1 ad Exercise 2. Defie U i = V i V i 1 for i {1, 2,...}. Show that U is a sequece of idepedet, idetically distributed variables ad that V is the partial sum process associated with U. Thus, partial sum processes are the oly discrete-time radom processes that have statioary, idepedet icremets. A iterestig, ad much harder problem, is to characterize the cotiuous-time processes that have statioary idepedet icremets. The Poisso coutig process has statioary idepedet icremets, as does the Browia motio process. Momets 4. Suppose that N. Use basic properties of expected value ad variace to show that (Y ) = μ var(y ) = σ 2 5. Suppose that N + ad m N with m. Use basic properties of covariace ad the statioary ad idepedece properties to verify the followig results. Hit: Note that Y = Y m + (Y Y m ). c. cov(y m, Y ) = m σ 2 cor(y m, Y ) = m (Y m Y ) = m σ 2 + m μ 6. Suppose that X has momet geeratig fuctio G. Show that Y has momet geeratig fuctio G Distributios 7. Suppose that X has either a discrete distributio or a cotiuous distributio with probability desity fuctio f. Recall that the probability desity fuctio of Y is f * = f * f * * f, the covolutio power of f of order. M ore geerally, we ca use the statioary ad idepedece properties to fid the joit distributios of the partial sum process: 8. Suppose that 1 < 2 < < k. Show that (Y 1, Y 2,..., Y k ) has joit probability desity fuctio f 1, 2,..., k (y 1, y 2,..., y k) = f 1 (y 1) f 2 1 (y 2 y 1) f k k 1 (y k y k 1), (y 1, y 2,..., y k ) R k

3 3 of 10 7/16/2009 6:05 AM The Cetral Limit Theorem We will ow make the cetral limit theorem precise. From Exercise 4, we caot expect Y itself to have a limitig distributio. Note that var(y ) as if σ > 0 (Y ) as if μ > 0 ad (Y ) as if μ < 0 Thus, to obtai a limitig distributio that is ot degeerate, we eed to cosider, ot Y itself, but the stadard score of Y. Thus, let 9. Show that Z = Y μ σ (Z ) = 0 var(z ) = Show that Z is also the stadard score of the sample mea M : Z = M μ σ / The cetral limit theorem states that the distributio of the stadard score Z coverges to the stadard ormal distributio as. A special case of the cetral limit theorem (to Beroulli trials), dates to Abraham De M oivre. The term cetral limit theorem was coied by George Pólya i Proof of the Cetral Limit Theorem We eed to show that F (z) Φ(z) as for each z R, where F is the distributio fuctio of Z ad Φ the distributio fuctio of the stadard ormal distributio. Equivaletly we will show that χ (t) e 1 2 t 2 as for each t R where χ is the characteristic fuctio of Z, ad the expressio o the right is the characteristic fuctio of the stadard ormal distributio. The followig exercises sketch the proof of the cetral limit theorem. Ultimately, the proof higes o a geeralizatio of a famous limit from calculus. 11. Suppose that a a as. Show that 1 + a e a as

4 4 of 10 7/16/2009 6:05 AM Now let χ deote the characteristic fuctio of the stadard score of a sample variable X, ad let χ deote the characteristic fuctio of the stadard score Z : χ(t) = exp i t X μ ( ( σ )), χ (t) = (exp(i t Z )), t R 12. Show that c. χ(0) = 1 χ (0) = 0 χ (0) = Show that Z = 1 X i μ i =1 σ 14. Use properties of characteristic fuctios to show that t χ (t) = χ ( ), t R 15. Use Taylor's theorem (amed after Brook Taylor) to show that t χ ( ) = χ (s ) t where s t 16. I the cotext of previous exercise, show that s 0 ad hece χ (s ) 1 as Fially, show that χ (t) = ( χ (s ) t2 ) e 1 2 t 2 as Normal Approximatios The cetral limit theorem implies that if the sample size is large the the distributio of the partial sum Y is approximately ormal with mea μ ad variace σ 2. Equivaletly the sample mea M is approximately ormal with mea μ ad variace σ 2. The cetral limit theorem is of fudametal importace, because it meas that we ca approximate the distributio of certai statistics, eve if we kow very little about the uderlyig samplig distributio. Of course, the term large is relative. Roughly, the more abormal the basic distributio, the larger must be for ormal approximatios to work well. The rule of thumb is that a sample size of at least 30 will usually suffice; although for may distributios smaller will do.

5 5 of 10 7/16/2009 6:05 AM 18. Let Y deote the sum of the variables i a radom sample of size 30 from the uiform distributio o [ 0, 1]. Fid ormal approximatios to each of the followig: P(13 < Y < 18) The 90th percetile of Y. 19. Let M deote the sample mea of a radom sample of size 50 from the distributio with probability desity fuctio f (x) = 3 x 4, x 1. This is a Pareto distributio, amed for Vilfredo Pareto. Fid ormal approximatios to each of the followig: P(M > 1.6) The 60th percetile of M. The Cotiuity Correctio A slight techical problem arises whe the samplig distributio is discrete. I this case, the partial sum also has a discrete distributio, ad hece we are approximatig a discrete distributio with a cotiuous oe. 20. Suppose that X takes iteger values ad hece so does the partial sum Y. Show that for ay h ( 0, 1] ad k Z, the evet {k h < Y < k + h} is equivalet to the evet {Y = k}. I the cotext of the previous exercise, differet values of h lead to differet ormal approximatios, eve though the evets are equivalet. The smallest approximatio would be 0 whe h = 0, ad the approximatios icrease as h icreases. It is customary to split the differece by usig h = 0.5 for the ormal approximatio. This is sometimes called the cotiuity correctio. The cotiuity correctio is exteded to other evets i the atural way, usig the additivity of probability. 21. Let Y deote the sum of the scores of 20 fair dice. Compute the ormal approximatio to P(60 Y 75). 22. I the dice experimet, set the die distributio to fair, select the sum radom variable Y, ad set = 20. Ru the simulatio 1000 times, updatig every 10 rus. Compute the followig ad compare with the result i the previous exercise: P(60 Y 75) The relative frequecy of the evet {60 Y 75} Normal Approximatio to the Gamma Distributio

6 6 of 10 7/16/2009 6:05 AM If Y k has the gamma distributio with shape parameter k N + ad scale parameter b ( 0, ) the k Y k = i =1 X i where (X 1, X 2,..., X k ) is a sequece of idepedet variables, each havig the expoetial distributio with scale parameter Sice (X i ) = b ad var(x i ) = b 2, it follows that if k is large, the gamma distributio ca be approximated by the ormal distributio with mea k b ad variace k b 2. The same statemet actually holds whe k is ot a iteger; more precisely, the distributio of the stadardized variable below coverges to the stadard ormal distributio as k. Z k = Y k k b k b 23. I the gamma experimet, vary k ad b ad ote the shape of the probability desity fuctio. With k = 10 ad b = 2, ru the experimet 1000 times with a update frequecy of 10 ad ote the apparet covergece of the empirical desity fuctio to the true desity fuctio. 24. Suppose that Y has the gamma distributio with shape parameter k = 10 ad scale parameter b = 2. Fid ormal approximatios to each of the followig: P(18 Y 23) The 80th percetile of Y. Normal Approximatio to the Chi-S quare Distributio The chi-square distributio with degrees of freedom N + is the gamma distributio with shape parameter k = 2 ad scale parameter b = 2. From the previous subsectio, it follows that if is large the chi-square distributio ca be approximated by the ormal distributio with mea ad variace 2. More precisely, if Y has the chi-square distributio with degrees of freedom, the the distributio of the stadardized variable below coverges to the stadard ormal distributio as Z = Y I the chi-square experimet, vary ad ote the shape of the desity fuctio. With = 20, ru the experimet 1000 times with a update frequecy of 10 ad ote the apparet covergece of the empirical desity fuctio to the probability desity fuctio. 26. Suppose that Y has the chi-square distributio with = 20. Fid ormal approximatios to each of the followig: P(18 < Y < 25)

7 7 of 10 7/16/2009 6:05 AM The 75th percetile of Y. Normal Approximatio to the Biomial Distributio If Y has the biomial distributio with trial parameter N + ad success parameter p ( 0, 1), the Y = i =1 X i where (X 1, X 2,..., X ) is a Beroulli trails sequece with success parameter p, that is, a sequece of idepedet idicator variables with P(X i = 1) = p for each i. It follows that if is large, the biomial distributio with parameters ad p ca be approximated by the ormal distributio with mea p ad variace p (1 p). The rule of thumb is that should be large eough for p 5 ad (1 p) 5. More precisely, the distributio of the stadardized variable Z give below coverges to the stadard ormal distributio as : Z = Y p p (1 p) 27. I the biomial timelie experimet, vary ad p ad ote the shape of the probability desity fuctio. With = 50 ad p = 0.3, ru the simulatio 1000 times, updatig every 10 rus. Compute the followig: P(12 Y 16) The relative frequecy of the evet {12 Y 16} 28. Suppose that X has the biomial distributio with parameters = 50 ad p = 0.3. Compute the ormal approximatio to P(12 Y 16) (do't forget the cotiuity correctio) ad compare with the results of the previous exercise. Normal Approximatio to the Poisso Distributio If Y has the Poisso distributio with parameter N +, the Y = i =1 X i where (X 1, X 2,..., X ) is a sequece of idepedet variables, each with the Poisso distributio with parameter 1. Sice (X i ) = var(x i ) = 1, it follows from the cetral limit theorem that if is large, the Poisso distributio with parameter ca be approximated by the ormal distributio with mea ad variace. The same statemet holds whe is ot a iteger; more precisely, the distributio of the stadardized

8 8 of 10 7/16/2009 6:05 AM variable below coverges to the stadard ormal distributio as. Z = Y 29. Suppose that Y has the Poisso distributio with mea 20. Compute the true value of P(16 Y 23). Compute the ormal approximatio to P(16 Y 23). 30. I the Poisso experimet, vary the time ad rate parameters t ad r (the parameter of the Poisso distributio i the experimet is the product r t). Note the shape of the desity fuctio. With r = 5 ad t = 4, ru the experimet 1000 times with a update frequecy of 10 ad ote the apparet covergece of the empirical desity fuctio to the probability desity fuctio. Normal Approximatio to the Negative Biomial Distributio If Y k has the egative biomial distributio with trial parameter k N + ad success parameter p ( 0, 1) the k Y k = i =1 X i where (X 1, X 2,..., X k ) is a sequece of idepedet variables, each havig the geometric distributio o N + with success parameter p. Sice (X i ) = 1 p ad var(x i) = 1 p, it follows that if k is large, the egative 2 biomial distributio ca be approximated by the ormal distributio with mea k 1 p ad variace k p. More p 2 precisely, the distributio of the stadardized variable below coverges to the stadard ormal distributio as k. p Z k = p Y k k k (1 p) 31. I the egative biomial experimet, vary k ad p ad ote the shape of the probability desity fuctio. With k = 5 ad p = 0.4, ru the experimet 1000 times with a update frequecy of 10 ad ote the apparet covergece of the empirical desity fuctio to the true desity fuctio. 32. Suppose that Y has the egative biomial distributio with trial parameter k = 10 ad success parameter p = 0.4. Fid ormal approximatios to each of the followig: P(20 Y 30) The 80th percetile of Y.

9 9 of 10 7/16/2009 6:05 AM Partial Sums with a Radom Number of Terms Suppose ow that N is a radom variable takig values i N, with fiite mea ad variace. The N Y N = i =1 X i is a radom sum of the idepedet, idetically distributed variables. That is, the terms are radom of course, but so also is the umber of terms N. We are primarily iterested i the momets of Y N. Idepedet Number of Terms Suppose first that N, the umber of terms, is idepedet of X, the sequece of terms. Computig the momets of Y N is a good exercise i coditioal expectatio. 33. Show that (Y N N) = N μ (Y N ) = (N) μ 34. Show that var(y N N) = N σ 2 var(y N ) = (N) σ 2 + var(n) μ Let H deote the probability geeratig fuctio of N. Show that the momet geeratig fuctio of Y N is H G (e t Y N N) = G(t) N (e t YN ) = H(G(t)) Wald's Equatio Some of these results geeralize to the case where the radom umber of terms N is a stoppig time for the sequece X. This meas that the evet {N = } depeds oly o (techically, is measurable with respect to) (X 1, X 2,..., X ) for each N. 36. Prove Wald's equatio amed after Abraham Wald: (Y N ) = (N) μ. c. Show that Y N = i =1 X i 1(i N) Show that X i ad {i N} are idepedet for each i. Coclude that (X i 1(i N)) = μ P(N i)

10 10 of 10 7/16/2009 6:05 AM d. e. Suppose that X i 0 for each i. Take expected values term by term i (a) to establish Wald's equatio i this special case. The iterchage of sum ad expected value is justified by the mootoe covergece theorem. Now establish Wald's equatio i geeral by usig the domiated covergece theorem Virtual Laboratories > 6. Radom Samples > Cotets Applets Data Sets Biographies Exteral Resources Key words Feedback

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