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Date Created: 11/02/15
Transforming a Random Variable Our purpose is to show how to nd the density function fy of the transformation Y gX of a random variable X with density function g Example We begin with a geometrical illustration in whichX UNF0 1 BETAl 1 and Y Xz We know thatf x 10x Here we use the indicatorfunction IAx 1 for x e A and IAx 0 otherwise The support of the random variable X is the unit interval It is crucial in transforming random variables to begin by finding the support of the transformed random variable here the support of Y is the same as the support ofX Now we approximate fy by seeing what happens under transformation to each of the intervals 0 01 01 02 00 10 shown in the xInterval column of the table Each of these intervals has length 01 and height 1 and thus area or probability 01 PO ltXlt 01 P01ltXlt 02 P09 ltXlt 1 01 The yInterval column of the table shows the transform of each interval Clearly the transformed random variable Y must have PO lt Ylt 001 P001lt Ylt 004 P081lt Ylt 1 01 Taking into account the unequal lengths of the y intervals we can see what the average height of the fY must be above each y interval in order to preserve the probability of the interval under transformation The results are shown in the column yHeigh and plotted in the rectangles of the graph on the next page The tops of the rectangles suggest the functional form of the density fy of Y Exercise IfX BETA2 1 that is fXx 2x1071x then use the same method to find the density function of Y 2X2 Notice that the entries in the Area column are not all equal for this problem and that the support of Y is not the same as the support of X 001004 03 333 025036 064081 CDF method A more formal approach to nd fy goes as follows The cdf of X is F Xx x for 0 lt x lt 1 Thus the cdf of Y can be found as FY01 PY Sy PX2 Sy PXSylzy1z for 0 lty lt 1 Dif ferentiating we have the density function of Y F Y39y fyy 1 2 y 121071y Thus we have shown that Y N BETA 12 1 Note Not all transformations Y X of a beta random variable X are beta The density function of Y is plotted in the figure on the next page This method of finding the distri bution of a transformed random variable called the cdf method is often applicable Exercises Use the cdf method to verify the functional form of the density function of Y 2X2 where X N BETA2 1 Also find the density function on Xl2 whereX UNF0 1 PDF method Notice that within the support of X the function y gx x2 is monotone increasing For any function y gx that is monotone increasing on the support of X we may carry out the cdf method in a general way FYy PY s y PgX s y PX s g 1y FXg 1y Whether or not we know the exact functional form of FX differentiation gives fyy dFXgquoty dy f2rg l 01 01g 1 y dy for y in the support of Y The factor with the derivative of g 1 results from applying the chain rule of differentiation For the example wherey gx xz we have 01 xlZ and olg39l y dx 12y39 z If g is monotone decreasing then My PltYsygt PltgltXgt sygtPltX2gquotltygtgt17PXltgquotltygtgt17 PltX5gquotygtgt17FXgquotygtgt 39Ihe nexttolast equality holds because the continuous random variable Xhas PX 100 0 Then differentiation gives fyy 7 dFXgquoty My fxgquoty dg lol My but here the sign ofdgquotydy is negative Thus for a function g that is monotone decreasing or increasing the density mction of Y is given My fxgquoty l 0116101de l1sltoy where SY designates the support of Y This is called the pdfmethod Exercises Use the pdfmethod to work previous two exercises Referring back to the geometrical illustration the expression ldg ly dy l ldxdyl can be viewed as the appropriate multiple to compensate for the change in the length of a small interval Ax as g transforms it to Ay in order to ensure thatPXe Ax PYe Ay Note It is o en best to use the cdf method if the transformation g is not monotone Exercise IfX is standard normal show that X 2 has a chisquared distribution with 1 degree of freedom Consider positive and negative values separately and combine Also prove this using generating mctions rnwwann nnmn FT h Convergence in Distribution and the Central Limit Theorem Convergence in distribution Let Y n be a sequence of random variables such that Y n has CDF F n02 Also let Y be a random variable with CDF F 2 We say that dlimYn Y if and only lim Fny F 2 for all continuity points y of F Examples 1 Let Y be a RV that puts probability ln on each of the n points ln 211 n ln 1 Then dlim Y Y UNIFO l The CDFs F of the RVs Y n are stairstep functions that get ever closer to the CDF of UNIFO l 2 Let Y be a RV that puts probability 12 on each of the points 0 and ln And let Ybe the degenerate RV with PY O 1 Then dlim Y Y lim Fny l fory gt 0 lim Fn0 l2 and lim Fn0 O fory lt O The disagreement between lim Fn0 12 and F O l is quotexcusedquot by the condition in the de nition Comments If the RVs Y n have MGFs mnt and Y hasMGF ml then convergence in distribution is equivalent to lim mnl ml Earlier we used MGFs to show that BINOMn Mn converges to POISO That was an example of convergence in distribution If discrete RVs Y have lim PYn k pk 2 0 where kak 1 then Ywith PY k pk is aRV with dlim Y Y So the binomial convergence to Poisson also illustrates this case If plim Y c then also dlim Y 0 So the second Example above is a case where we have both convergence in probability and convergence in distribution Convergence in distribution is often shown using MGFs In particular the CLT can be proved using MGFs Central Limit Theorem Let X1 X2 be a sequence of random variables iid with LL EX and 62 VX De ne 3 X1 X2 Xnn the sample mean of the rst 11 observations Then E n and V 6211 De ne Z n12n LLG Then dlim Z Z NORMO 1 Comments We already know that if the RVs X are iid NORNKH 6 then 3 NORMnon12 and so 2 NORMO 1 so this is a trivial example of the CLT The interesting thing is that averages A7 based on any distribution tend to standard normal In practice we use the fact that if Z is approximately standard normal then 3 is approximately NORNKH onlZ Simulations We will not prove the CLT here but will illustrate it Via some simple simulation experiments This is feasible because for many distributions of interest Z is approximately NORMO l or equivalently A7 is approximately NORMQJ 611 for n of only moderate size m lt 10000 n lt 12 x lt runifmn DTA lt matrixx nrowm z lt rowSumsDTA 6 cutp lt seqfloorminz ceilingmaxz by5 histz breakscutp ylimc0 4 probT 22 lt seq3 3 by 01 1ineszz dnormzz meanz varz gt meanz varz 1 000783211 1 09933388 Densz Histogram ofz m lt 10000 n lt 12 alp lt 12 bet lt 12 x lt rbetamn alp bet DTA lt matrixx nrowm xbar lt rowMeansDTA mu lt a1pa1pbet va lt a1pbeta1pbet2a1pbet1 z lt xbar musqrtvan cutp lt seqfloorminz ceilingmaxz by5 histz breakscutp ylimc0 4 probT 22 lt seq 3 3 by 01 1ineszz dnormzz meanz varz gt meanz varz 1 001301487 1 0974018 Dem Histogram of z m lt 10000 n lt 50 Skewed distributions require larger n x lt rexpmn rate1 DTA lt matrixx nrowm xtot lt rowSumsDTA Plot unstandardized sum lo lt floorminxtot up lt ceilingmaxxtot cutp lt seqlo up1 by2 histxtot breakscutp probT ylimc0dnorm5050sqrt50 22 lt seqlo up by 01 1ineszz dnormzz 50 sqrt50 CLT Approximation 1ineszz dgammazz 50rate1 colquotdarkgreenquot Exact meanxtot varxtot gt meanxtot varxtot 1 500202 1 5029844 Density 001 0039 003 004 000 Histogram of mm x tot Summary A Let X1 X2 X be a random sample from a population with mean u and standard deviation 6 Then X is a random variable with E0 u and SDX onlZ This statement by itself is not of much use in actual probability computations The information is enough to get a Chebyshev bound B LetX1X2 X be a random sample from a normal population with mean u and standard deviation 6 Then X is a normal random variable with E0 LL and SDX GnlZ that is X NORNKH onlZ If we know the population is normal this statement is useful in probability computations about X See below C CLT LetXl X2 X be a random sample from a population with mean u and standard deviation 6 Then X is variable with E0 u and SDX GnlZ and its distribution is approximately NORNIQJ onlZ This approximation gets better as 11 increases If the population distribution is roughly symmetrical then the approximation is computationally useful for moderate 11 Example Suppose a population of male swimmers has weights in kg distributed as NORM70 7 If X is the weight of one man chosen at random from this population then P63 ltXlt 77 P63 707 lt X uo lt 77 707 P 1 lt Z lt 1 06826 If A7 is the sample mean weight of 9 men chosen at random from this population then P63 ltlt 77 P363 707 lt 11120 uo lt 377 707 P 3 lt Zlt 3 09973 Also if n 9 6 7 and LL is unknown then a 95 CI for LL is Xi 19661112 or Xi l9673 Furthermore by the CLT if the population is roughly symmetrical but not exactly normal and n 9 u 70 o 7 then P63 ltlt 77 z 997 Notes 1 For general 71 an approximate 95 CI for LL based on the CLT is Xi 196 671 2 or more simply ii 2 671 2 S to estimate 6 and get the doubly approximate CI X i 2 5771 Here S2 2 X Ean 1 For normal data ESZ 62 and for most data with large n ES z 6 3 Strictly speaking for 2 If 6 is unknown and n is large enough say ngt 30 then we can use normal data T 77120 uS has Student39s tdistribution with df n l and the 95 CI becomes Xi l SnlZ where PT gt l 0025 4 For simulations based on very large numbers iterations the CI in 2 often gives an idea how close the simulated result is likely to be to the theoretical one xx lt7 seq50 so by01 1m lt7 Normal Densltles of Pop and Blue Mean of 119 Obs plotxx dnormxx 70 y yllmc0 2 malnlbl xlab kg llnesxx dnom2o 70 73 Col blue llnesc63 77 cm 0 col red lwd4 Normal Densmu upon Ind Blue Main nrna onsv n 1 1mm 70 n w L05 a n5 xx lt seq50 95 by01 1b1 lt quotGamma Densities of Pop and Blue Mean of n9 Obsquot p1oxx dgammaxx 1oo rate107 type quot1quot y11mc0 2 main1b1 xlabquotkgquot 900 rat 0 7 usquot 11nesc63 77 Co o Colquotredquot 1wd4 H 1 m m gtlt n u 3 gtlt u n o gt pgalllma77900rate907 pgamllla63900rate907 1 09972461 Gamma Densities of Pop and Blue Mean of n9 Obs 015 dgammawx 100 rate 107 0 10 Note For population H 0W 70 and OZ 0W Z 49 imply 7 107 and 0L 100 Not quite symmetn39cal For the mean ofn 9 0L 900 and 7 907 give E0 70 and V0 499 Ver nearly symmetn39cal Markov and Cheybshev Inequalities and the Law of Large Numbers Proofs and Illustrations Markov39s Inequality Suppose a random variable X takes only nonnegative values so that PX2 0 1 How much probability is there in the tail of the distribution of X More speci cally for a given value a gt 0 what can we say about the value of PX 2 a Markov39s inequality takes LL EX into account and provides an upper bound on PX 2 a that depends on the values of a and LL We give the derivation for a continuous random variable X with density function fx u EltXgt Low xfx dx f0a xfx dx face xfx dx 2 IM xfx obc 2 Lam afx antsx dx MD a from which we obtain Markov39s Inequality PX2 a S ua In the above I The rst inequality holds because the integral ignored is nonnegative I The second inequality holds because a S x for x in 61 00 The proof for a discrete random variable is similar with summations replacing integrals Markov39s Inequality gives an upper bound on PX 2 a that applies to any distribution with positive support Practical consequences For most distributions of practical interest the probability in the tail beyond a is noticeably smaller than LLa for all values of 6 Below for several continuous and discrete distributions each with u l we use R to show that the nonincreasing quotreliability functionquot Ra l Fa PXgt a is bounded above by LLa la For continuous distributions there is no difference between PXS a and PXlt a and for discrete distributions the discrepancy is not noticeable in our graphs The Markov bound 16 is not useful for a lt 1 that is 16 gt 1 because no probability exceeds 1 Problems a Verify the functional form of Ra in each example b Make a plot of R and the Markov bound for the degenerate random variable X with PX l l Exponential distribution with mean 1 a lt seq0 4 lengthleOO aa lt seql 4 length 500 plota 1 pexpa ratel typequotlquot ylimc0l Ylab R mainquotMarkov Bound for EXP1quot linesaa laa colquotredquot Markov Bound for EXP1 R 1 O x x x 00 02 O4 06 08 Uniform distribution on 0 2 a lt seqO 4 1ength1000 aa lt seq1 4 length 500 plota 1punifa02 typequot1quot ylimc01 ylab R main Markov Bound for UNIF02quot 1inesaa 1aa colquotredquot Markov Bound For UNI FWJ 1H J g A f f DO 02 05 08 Uniform distribution on 0911 a lt seqO 4 1ength1000 aa lt seq1 4 length 500 plota 1 punifa 911 typequot1quot ylimc 0 1 ylab Rquot main Markov Bound for UNIF911quot 1inesaa 1aa colquotredquot Markov Bound for UMP311 At a l the bound nearly touches R A distribution more tightly concentrated about LL 1 would come even closer to touching So as a general statement the Markov bound cannot be improved ma d nmmmnw hn2mmpl a lt seq01 4 by001 aa lt seq1 4 length 500 plota 1pbinoma25 typequot1quot ylimc01 ylab Rquot mainquotMarkOV39Bound for BINOM25 1inesaa 1aa colquotredquot Markov Bound li39nr BIMOM25 1 I I14 DE IZIB H 02 Poisson distribution with 7 1 a lt seq 01 4 by001 aa lt seql 4 length 500 plota 1 ppoisal typequotlquot ylimc0l YlabquotRquot mainquotMarkov Bound for POIS1 quot lines aa laa colquotredquot Markov Bound for P0ls1 10 R 00 02 O4 06 08 Chebyshev39s Inequality Suppose a random variable Y has EY u and VY oz lt oo Then setting X Y L02 we have uX EX VY and PX2 O1 Now apply Markov39s Inequality to X with b gt O to obtain PX2 b2 PY ml 2 b2 PY m 2 b 3 52192 Letting b k6 we obtain Chebyshev39s Inequality PY u 2 k6 S lkz which by the complement rule is sometimes written as PY m lt k6 2 1 118 In words not more than lk2 of the probability in a distribution can lie beyond k standard deviations away from its mean The following table compares the some of the information from Chebyshev39s Inequality with exact information about a normal distribution upon which the Empirical Rule is based Interval Exact Chebyshev LL 6 LL 6 0681 Uninformative LL 156 LL 156 0866 2 1 49 0556 LL 26 LL 26 0950 2 1 14 0750 LL 36 LL 36 0997 2 1 19 0889 Below for several continuous and discrete distributions each with LL 1 and 61 we use R to show that the nondecreasing function Qk PY LL S k6 PY 1 S k is bounded above by l lkz Problems 0 d In the normal example below verify the values in the table above as closely as you can by reading the graph In the remaining examples verify that the variance is l and nd the functional forms of Q Draw a similar sketch of Q compared with the Chebyshev bound for the distribution that places probability US at each of the points 0 and 2 and probability 34 at l Normal distribution with H 1 and 5 1 k lt seq 01 4 by001 kk lt seql 4 length 500 plotk pnormlkll pnorm1 kll typequotlquot ylimc0l ylabquotQquot mainquotChebyshev Bound for NORM11quot Chebyshev Bound for NORM11 10 O x 00 02 O4 06 08 lineskk 1 lkkAZ colquotredquot Exponential distribution with mean 1 k lt seq01 4 by001 kk lt seq1 4 length 500 plotk pexp1k11pexp1k11 type 1 ylimc01 ylab Q mainquotChebyshev Bound for EXP1quot Ch byshev Bnund rm EXP Q P F 39 r I fr d r m f 4 o x Q 3 C X C r q xquot g a 9 rquot 1 J D r39 139 3 P I 3 I 5 I l l 1ineskk 1 1kk2 colquotredquot Uhmwmd ummmnon Vamp1V k lt seq01 4 by001 kk lt seql 4 length 500 plotk punif1k1sqrt31sqrt3punif1 k1sqrt31sqrt3 typequot1quot ylimc01 ylabquotQquot mainquotChebyshev Bound for UNIF07322732 quot 1ineskk 1 1kk2 colquotredquot cnebyshev Bound For LI NIFIJ322732n 10 CI GHQ I3 5 08 CHI 12 Poisson distribution with 7 1 k lt seq 01 4 by001 kk lt seql 4 length 500 plotk ppoislklPp0isl k 0011 typequotlquot ylimc0l YlabquotQquot mainquotChebyshev Bound for POIS2quot chebyshev Bound for P0ls2 10 O 00 02 O4 06 08 lineskk 1 lkkAZ colquotredquot For relatively large values of k the Chebyshev bound can be used to get a reasonable if rough approximation to the probability in the tails of a distribution Of course when the exact distribution is known and its probability distribution is available as a function in R it is preferable to get the exact value Law of Large Numbers for Coin Tossing An important theoretical use of Chebyshev39s Inequality is to prove the Law of Large Numbers for Coin Tossing If a coin with PHeads p is tossed n times then the heads ratio Z Headsn has mean H EZn 19 and 62 VZn p1 pn Thus for arbitrarily small 8 k6 gt O Chebyshev39s inequality gives 12PZ p lt8 2 1 p1 pn82 gt 1 asn gtoo Thus PZ p lt e gt 1 Note Here 8 k6 kpl pn12 so lk2 pl pn82 prob We say that Z converges to p in probability and write Z gt p In words as the number of tosses increases to a suf ciently large number we see that the heads ratio is within 8 of p with probability as near 1 as we please As a practical matter Z is nearly normally distributed for large n Thus the normal distribution is a better way than the Chebyshev bound to assess the accuracy of Zn as an estimate of p Roughly speaking this amounts to using the Empirical Rule As a speci c example in 10000 tosses ofa fair coin 2 SDV10000 2pq10000 2 240 000 2 2200 001 So we can expect the heads ratio to be within 001 of 12 with probability 95 Fifty thousand tosses would allow approximation of p with 2place accuracy Problem e What does ChebysheV39s inequality say about P212500 p lt 1150 What does the normal approximation say about this probability The graph on the next page illustrates ChebysheV and CLT bounds for tossing a fair coin III lt 10000 p lt 12 q lt 1 p n lt 1m sd lt sqrtpqn h lt rbinomm1 p s lt cumsum pha lt sn kch lt sqrt1alpha kc1 lt qnorm1 alphaZ kch t p1on pha c y11mcmax0pZ m1n1p2 mainquotcoin Tosses with CLT red and Chebyshev Boundsquot 11nesn U 11nesn x x x x x 0 2000 4000 6000 8000 10000 The Gamma Function and Gamma Distribution The gamma function Ft is de ned for t gt 0 as TU I xHe x dx Integrating by parts one can show that Ht 87311167 ltFt 1 0 1tFt 1 Hence Ft 1 ITO Thus if we know values of the function in 0 1 we can nd corresponding values in 1 2 then in 2 3 and so on The function is undefined at t 0 Although we will not have use for Ht at negative values of t we note that this same relationship can be used to extend the function to negative values except for negative integer values where it is unde ned See the plot below Consider the values of the F function at positive integer values of t It is easy to see that F1 1 0 and so F2 F1 1 1 F3 2F2 2 2 H4 3H3 3 and so on Thus for any positive integer t Ht t 7 1 In this sense the F function can be regarded as a generalization of factorials to noninteger values Setting 1 222 in the integral over the positive half line of the standard normal density function pz we see that F12 TtlZ 177245 from which we can get quothalfintegerquot values of the F function F32 088623 etc For tgt 0 R gives 1 2 1461632 as the point where the locally rather at function attains its minimum value 2 08856032 For positive values of the parameters X and B the gamma family of probability distributions has the density function y K yoquot1 e 7W for y 2 039 y 0 elsewhere The constant K that causes this function to integrate to 1 over the positive half line is K BOTX 1 For B 1 this is obvious from looking at the de nition of FX and it can easily be seen for other values of B by making the change of variable x yB The parameter X governs the shape of the gamma density and B is a scale parameter Some books and computer languages use the quotratequot parameter 7 1B For example in R pgamma 12 2 5 returns PXS 12 07127025 whereXhas a gamma distribution with X 2 and 7 1B 5 We get the same result with pgamma 12 2 scale 2 The moment generating function of a gamma distribution is mt 1 7 Br From the mgf it is easy to see that the sum of r independent exponential random variables each with mean B or rate 7 1B has a gamma density with shape parameter r X and scale parameter If X N GAMMAX B then EX XB and VX XB2 There are two notable subfamilies of the gamma family An exponential distribution with mean B is GAMMA1 A chisquared distribution with quotdegrees of freedomquot parameter V is GAlllIAV2 2 Gamma Function gammax 0 Copyright 2004 by Bruce E Trumbo All rights reserved Department of Statistics CSU Hayward Correctionssuggestions btrumbocsuhaywardedu Transforming a Random Variable Our purpose is to show how to nd the density function fy of the transformation Y gX of a random variable X with density function fX LetX have probability density function PDF fo and let Y gX We want to nd the PDF fY0 of the random variable Y Geometrical Example Suppose X UNF0 1 BETA1 1 and that Y X2 gX We know that fXx 101x Here we use the indicator function IAx 1 forx e A and IAx 0 otherwise The support of the random variable X is the unit interval 0 1 It is crucial in transforming random variables to begin by finding the support of the transformed random variable Here the support of Y is the same as the support of X Now we approximate fy by seeing what the transformation does to each of the intervals 0 01 01 02 00 10 In the table below these intervals are in the xlnterval column Each of these intervals has length 01 The height of fXx is 1in each interval so area or probability in within each interval is 01 P0 ltXlt 01 P01ltXlt 02 P09 ltXlt 1 01 The ylnterva column of the table shows the transform of each interval Clearly the transformed random variable Y must have P0 lt Ylt 001 P001lt Ylt 004 P081lt Ylt 1 01 Taking into account the unequal lengths of the y intervals we can see what the average height of the must be above each yinterval in order to preserve the probability of the interval under transformation The results are shown in the column yHeight and plotted in the rectangles of the graph on the next page The tops ofthe rectangles suggestthe functional form of the dansxtyY of y 1 IfXBEFA21d1atxsXx 2x1mxthen use the same methodto fmdthe denstty funetmn of Y X2 Nouee that the entnes m th problem and that the su e ea column are not 111 Equal for this ppon of Yxs not Lhe same as the support ofX CDF method A more formal approach to finding goes as follows The cdfoins FXx x for 0 ltx lt1 Thus the cdf of Y can be found as my PY s y PX2 s y PXSy12 y12 for 0 lt y lt 1 Differentiating we have the density function of Y FY39V JV 12y 121o1y Thus we have shown that Y BETAl2 1 Not all transforms Y Xk of a beta random variable X are beta The density function of Y is plotted in the figure This method of finding the distribution of a transformed random variable is called the cdfmethod It is very widely applicable Exercises 2 Use the cdf method to verify the functional form of the density function of Y 2X2 where X BETA2 1 3 Find the density function of Y X12 where X UNF0 l PDF method For monotone increasing or decreasing functions g the CDF method can be carried out in general allowing one to deal only with PDFs In our example within the support of X the function y gx x2 is monotone increasing For any function y gx that is monotone increasing on the support of X we may carry out the CDF method in a general way FM PY S y PgX S y PX S 100 Fxg 1y Whether or not we know the exact functional form of F X differentiation gives My 0 Fxg 1y dy fxg 1y 0 g 1 V dy for y in the support of Y provided that g is a monotone increasing function The factor with the derivative of g71 results from applying the chain rule of differentiation For the example where y gx x2 we have g lo x 2 and d g lo alx 12y 2 If g is monotone decreasing then my M s y mm s y PX2 g lo 1 PXlt g lo 1 PX g lo 1 an lo The nexttolast equality holds because the continuous random variableX has PX g 1y 0 Then differentiation gives CV 01565749 dy fxg 1y 015510 dy but here the sign of d 102 dy is negative Thus for a function g that is monotone decreasing or increasing the density function of Y is given by CV fxg 1y 015510 dy l Isa9 Where SY designates the support of Y This is called the PDF method Exercise 4 Use the PDF method to work previous two exercises Geamenimlimerpmmhn Refemhg back to the geomemcal illustration the expression tam My MM eah be mewed as the appropnate mulhple to eompehsate for the change m the length ofa small mtem as gh39ansforms xtto Ay m order to ensure that PXe Ax PYe Ay m w w M W w Note If the transforming function is not monotone on the support of X it is often best to use the cdf method Advanced Exercises 5 If X is standard normal show that X 2 has a chisquared distribution with 1 degree of freedom Hint What is the density function of V X Then use the PDF method to find the density function of V2 6 Do Problem 5 using generating functions Hint Find EeXpIX2 using the standard normal density Complete the square Recognize the integral of a density that must equal 1 Compare the simplified result with the MGF of CHISQ1 Smlations For square of standard uniform density inction derived above is superimposed superimposing a density is possible if the histogram i drawn on a density scale parameter freqF or probT m lt 10 0 0 0 0 x lt runif m 2 his Y freqF yy lt seq0 l by001 linesyy dbetayy 5 1 Histogram ofy Square of standard uniform with larger sample also with more histogram bins as speci ed by breaks parameter m lt 500000 x lt runifm lt xAZ histY breaks50 freqF yy lt seq0 ll by001 linesyy dbetayy 5 1 Histogram ofy Square of standard normal with density of CHISQ1 superimposed see exercise above m lt 10 0 0 0 0 x lt mom m xAZ his y freqF xlimc0 10 yy lt seq0 10 by0001 lines yy dchisqyy 1 Histogram ofy Density 0 3 o 5 o e 02 Advanced Exercise 7 Use simulation to illustrate that the sum ofthe squares oftwo independent standard normal random variables is exponential with mean 2 rate 12 CVPYId39M 211mm mmby me 2 mm All rams mma Bernoulli Trials and Related Distributions A single Bernoulli trial is an experiment with two possible outcomes S and F such that PS p and PF 1 p q In practice the parameter p is often unknown Note Use of p instead of a Greek letter is a violation of the usual convention Some books use 7 but that can lead to confusion with 7 314159 If we have several Bernoulli trials in an experiment we will assume they are independent and that p is the same on each trial Bernoulli trials are the basis of three families of distributions I Bernoulli I Binomial 39 Geometric and 39 Negative binomial A Bernoulli distribution is simply based on one trial PX 0 61 PX 1 11 EX 0q 119 19 Now we look at Binomial Distributions Binomial Experiment Consists of a known number n of Bernoulli trials The random variable of interest is X of Ss in 11 trials Explicitly the rules for a binomial experiment are 1 Trials independent Two possible outcomes S and F on each trial PS p is the same on each trial Known number n of trials k11th The binomial random variable is X of Ss Rules 13 simply describe repeated Bernoulli trials Rules 4 amp 5 describe how Bernoulli trials are used in a binomial experiment The parameters of a binomial experiment are n and p In statistical practice often 11 is known and p is not Both parameters must be known in order to nd the distribution of X A Bernoulli distribution is a special case of a binomial distribution in which n 1 Example Tossing a fair coin 5 times is a binomial experiment with parameters 11 5 and p 12 The distribution of X is given by the following output from R You should do the same computation by hand using the formula for the binomial distribution You should also compare these results with those in the appropriate column of the table on page 783 of the text cdf lt cumsumf prod lt kf parmfrowc12 plotk f typequothquot lwd2 mainquotPDFquot plot16 pbinom16 n p typequotsquot xlabquotkquot ylabquotFquot mainquotCDFquot parmfrowc1l roundcbindk f cdf prod 5 sumprod gt roundcbindk f cdf prod 5 k f cdf prod 1 0 003125 003125 000000 2 1 015625 018750 015625 3 2 031250 050000 062500 4 3 031250 081250 093750 5 4 015625 096875 062500 6 5 003125 100000 015625 gt sumprod l 25 c in 00 N of c 7 9 c 01 LL 6 q c c N c in O C c I11 1391 C 11111111 012345 10123450 k R Here is are the graphs for the distribution that a1ises from tossing a biased coin Withp 1 3 PDF CDF 0 c f m c 007 c 8 0 c c LL q c c 339 N 0 o c 81 111 011111111 012345 10123456 Notice that the table in the back ofWMS does not give this distiibution Show that the mean ofthis distIibution BINOM5 13 is 16667 Examples of distributions that are not binomial Each of the rules for a binomial experiment is important Different families of distributions result when rules fail A Suppose there is no xed n We toss a fair coin until we see the first Head X the number of the trial on which we see the rst Head This is an example of the Geometric Distribution Suppose there is no xed n We toss a fair coin until we see the 3rd Head X the number of the trial on which we see the 3rd Head This is an example of the Negative Binomial Distribution An urn contains 8 balls 4 marked S and 4 marked F We withdraw 5 balls without replacement The draws are not Bernoulli trials because they are dependent On each draw the PS 12 Unconditionally on the past X of Ss drawn But the distribution of X is not BINOM5 12 In particular PX O PX 5 O This is an example of the Hypergeometric Distribution A process begins in the AM with PS 99 But during the day PS gradually shifts to 95 We sample n 5 items during the day The number of Ss observed is not binomial because the p is not the same on all trials This is a messy situation there is no quotnamedquot distribution to model it Copyright 2004 by Bruce E Trumbo All rights reserved Department of Statistics CSU Hayward WMSWaclerly et al Mathematical Statistics with Applications 6th ed Duxbury 2002