5 Continuous random variables

Transcription

1 5 Cotiuous rdom vribles We devite from the order i the boo for this chpter, so the subsectios i this chpter do ot correspod to those i the text. 5.1 Desities of cotiuous rdom vrible Recll tht i geerl rdom vrible X is fuctio from the smple spce to the rel umbers. If the rge of X is fiite or coutble ifiite, we sy X is discrete rdom vrible. We ow cosider rdom vribles whose rge is ot coutbly ifiite or fiite. For exmple, the rge of X could be itervl, or the etire rel lie. For discrete rdom vribles the probbility mss fuctio is f X (x) = P(X = x). If we wt to compute the probbility tht X lies i some set, e.g., itervl [, b], we sum the pmf: P( X b) = f X (x) x: x b A specil cse of this is P(X b) = f X (x) x:x b For cotiuous rdom vribles, we will hve itegrls isted of sums. Defiitio 1. A rdom vrible X is cotiuous if there is o-egtive fuctio f X (x), clled the probbility desity fuctio (pdf) or just desity, such tht P(X t) = t Propositio 1. If X is cotiuous rdom vrible with desity f(x), the 1. P(X = x) = for y x R. 2. P( X b) = f(x) dx 3. For y subset C of R, P(X C) = C f(x) dx 1

2 4. f(x) dx = 1 Proof. First we observe tht subtrctig the two equtios gives P(X b) =, P(X ) = P(X b) P(X ) = d we hve P(X b) P(X ) = P( < X b), so Now for y P( < X b) = (1) P(X = x) P(x 1/ < X x) = x x 1/ f X (t) dt As, the itegrl goes to zero, so P(X = x) =. Property 2 ow follows from eq. (1) sice P( X b) = P( < X b) + P(X = ) = P( < X b) Note tht sice the probbility X equls y sigle rel umber is zero, P( X b), P( < X b), P( X < b), d P( < X < b) re ll the sme. Property 3 is esy if C is disjoit uio of itervls. For more geerl sets, it is ot cler wht eve mes. This is beyod the scope of this C course. Property 4 is just the fct tht P( < X < ) = 1. Cutio Ofte the rge of X is ot the etire rel lie. Outside of the rge of X the desity f X (x) is zero. So the defiitio of f x (x) will typiclly ivolves cses: i oe regio it is give by some formul, elsewhere it is simply. So itegrls over ll of R which coti f X (x) will reduce to itervls over subset of R. If you mistely itegrte the formul over the etire rel lie you will of course get osese. 2

3 5.2 Ctlog As with discrete RV s, two cotiuous RV s defied o completely differet probbility spces c hve the sme desity. Ad there re certi desities tht come up lot. So we strt ctlog of them. Uiform: (two prmeters, b R with < b) The uiform desity o [, b] is { 1 f(x) =, if x b b, otherwise We hve see the uiform before. Previously we sid tht to compute the probbility X is i some subitervl [c, d] of [, b] you te the legth of tht subitervl divided by the legth of [, b]. This is of course wht you get whe you compute d c = d c 1 b dx = d c b Expoetil: (oe rel prmeter λ > ) { λe f(x) = λx, if x, if x < Chec tht its totl itegrl is 1. Note tht the rge is [, ). Norml: (two rel prmeters µ, σ > ) ( f(x) = 1 σ 2π exp 1 ( ) ) 2 x µ 2 σ The rge of orml RV is the etire rel lie. It is ythig but obvious tht the itegrl of this fuctio is 1. Try to show it. Cuchy: f(x) = 1 π(1 + x 2 ) 3

4 Exmple: Suppose X is rdom vrible with expoetil distributio with prmeter λ = 2. Fid P(X 2) d P(X 1 X 2). Exmple: Suppose X hs the Cuchy distributio. Fid the umber c with the property tht P(X c) = 1/4. Exmple: Suppose X hs the desity { f(x) = c x(2 x) if x 2 otherwise where c is costt. Fid the costt c d the compute P(1/2 X). 5.3 Expected vlue A rigorous tretmet of the expected vlue of cotiuous rdom vrible requires the theory of bstrct Lebesgue itegrtio, so our discussio will ot be rigorous. For discrete RV X, the expected vlue is E[X] = x xf X (x) We will use this defiitio to derive the expected vlue for cotiuous RV. The ide is to write our cotiuous RV s the limit of sequece of discrete RV s. Let X be cotiuous RV. We will ssume tht it is bouded. So there is costt M such tht the rge of X lies i [ M, M], i.e., M X M. Fix positive iteger d divide the rge ito subitervls of width 1/. I ech of these subitervls we roud the vlue of X to the left edpoit of the itervl d cll the resultig RV X. So X is defied by X (ω) =, where is the iteger with X(ω) < + 1 Note tht for ll outcomes ω, X(ω) X (ω) 1/. So X coverges to X poitwise o the smple spce Ω. I fct it coverges uiformly o Ω. The expected vlue of X should be the limit of E[X ] s. The rdom vrible X is discrete. Its vlues re / with ruig from M to M 1 (or possibly smller set). So E[X ] = M 1 = M 4 f X ( )

5 Now So f X ( ) = P(X = ) = P( X(ω) < + 1 ) = +1 E[X ] = = M 1 = M M 1 = M f X(x) dx Whe is lrge, the itegrls i the sum re over very smll itervl. I this itervl, x is very close to /. I fct, they differ by t most 1/. So the limit s of the bove should be M 1 = M +1 x = M M x = x The lst equlity comes from the fct tht f X (x) is zero outside [ M, M]. So we me the followig defiitio Defiitio 2. Let X be cotiuous RV with desity f X (x). The expected vlue of X is provided E[X] = x x < (If this lst itegrl is ifiite we sy the expected vlue of X is ot defied.) The vrice of X is σ 2 = E[(X µ) 2 ], µ = E[X] provided the expected vlue is defied. 5

6 Ed of September 3 lecture Just s with discrete RV s, if X is cotiuous RV d g is fuctio from R to R, the we c defie ew RV by Y = g(x). How do we compute the me of Y? Oe pproch would be to wor out the desity of Y d the use the defiitio of expected vlue. We hve ot yet see how to fid the desity of Y, but for this questio there is shortcut just s there ws for discrete RV. Theorem 1. Let X be cotiuous RV, g fuctio from R to R. Let Y = g(x). The E[Y ] = E[g(X)] = g(x) Proof. Sice we do ot ow how to fid the desity of Y, we cot prove this yet. We just give o-rigorous derivtio. Let X be the sequece of discrete RV s tht pproximted X defied bove. The g(x ) re discrete RV s. They pproximte g(x). I fct, if the rge of X is bouded d g is cotious, the g(x ) will coverge uiformly to g(x). So E[g(X )] should coverges to E[g(X)]. Now g(x )] is discrete RV, d by the lw of the ucoscious sttistici E[g(X )] = x g(x) f X (x) (2) Looig bc t our previous derivtio we see this is E[g(X )] = = M 1 = M M 1 = M g( +1 ) +1 g( ) f X(x) dx which coverges to g(x) (3) 6

7 Exmple: Fid the me d vrice of the uiform distributio o [, b]. The me is µ = xf(x) dx = For the vrice we hve to first compute E[X 2 ] = x b dx = 1 b b = + b 2 (4) x 2 f(x) dx (5) We the subtrct the squre of the me d fid σ 2 = (b ) 2 /12. Exmple: Fid the me d vrice of the orml distributio. Exmple: Fid the me of the Cuchy distributio The gmm fuctio is defied by Γ(w) = x w 1 e x dx (6) The gmm distributio hs rge [, ) d depeds o two prmeters λ >, w >. The desity is { λ w f(x) = Γ(w) xw 1 e λx if x (7) otherwise I oe of the homewor problems we compute its me d vrice. You should fid tht they re µ = w λ, σ2 = w λ 2 (8) Exmple: Let X be expoetil with prmeter λ. Let Y = X 2. Fid the me d vrice of Y. 5.4 Cumultive distributio fuctio I this sectio X is rdom vrible tht c be either discrete or cotiuous. Defiitio 3. The cumultive distributio fuctio (cdf) of the rdom vrible X is the fuctio F X (x) = P(X x) 7

8 Why itroduce this fuctio? It will be powerful tool whe we loo t fuctios of rdom vribles d compute their desity. Exmple: Let X be uiform o [ 1, 1]. Compute the cdf. GRAPH!!!!!!!!!!!!!!!!!!!! Exmple: Let X be discrete RV whose pmf is give i the tble. x f X (x) 1/8 1/8 3/8 2/8 1/8 GRAPH!!!!!!!!!!!!!!!!!!!! Exmple: Compute cdf of expoetil distributio. Theorem 2. For y rdom vrible the cdf stisfies 1. F(x) is o-decresig, F(x) lim x F(x) =, lim x F(x) = F(x) is cotiuous from the right. 4. For cotiuous rdom vrible the cdf is cotiuous. 5. For discrete rdom vrible the cdf is piecewise costt. The poits where it jumps is the rge of X. If x is poit where it hs jump, the the height of the jump is P(X = x). Proof. 1 is obvious... To prove 2, let x. Assume tht x is icresig. Let E = {X x }. The E is i icresig sequece of evets. By the cotiuity of the probbility mesure, P( =1 E ) = lim P(E ) Sice x, every outcome is i E for lrge eough. So =1 E = Ω. So lim F(x ) = lim P(E ) = 1 (9) 8

9 The proof tht the limit s x is is similr. GAP Now cosider cotiuous rdom vrible X with desity f. The F(x) = P(X x) = x f(t) dt So give the desity we c compute the cdf by doig the bove itegrl. Differetitig the bove we get F (x) = f(x) So give the cdf we c compute the desity by differetitig. Theorem 3. Let F(x) be fuctio from R to [, 1] such tht 1. F(x) is o-decresig. 2. lim x F(x) =, lim x F(x) = F(x) is cotiuous from the right. The F(x) is the cdf of some rdom vrible, i.e., there is probbility spce (Ω, F,P) d rdom vrible X o it such tht F(x) = P(X x). f The proof of this theorem is wy beyod the scope of this course. 5.5 Fuctio of rdom vrible Let X be cotiuous rdom vrible d g : R R. The Y = g(x) is ew rdom vrible. We wt to fid its desity. This is ot s esy s i the discrete cse. I prticulr f Y (y) is ot x:g(x)=y f X(x). Key ide: Compute the cdf of Y d the differetite it to get the pdf of Y. Exmple: Let X be uiform o [, 1]. Let Y = X 2. Fid the pdf of Y.!!!!!!! GAP Exmple: Let X be uiform o [ 1, 1]. Let Y = X 2. Fid the pdf of Y. 9

10 1 λ!!!!!!! GAP Exmple: Let X be uiform o [, 1]. Let λ >. Y = l(x). Show Y hs expoetil distributio.!!!!!!! GAP Exmple: The stdrd orml distributio is the orml distributio with µ = d σ = 1. Let X hve orml distributio with prmeters µ d σ. Show tht Z = (X µ)/σ hs the stdrd orml distributio.!!!!!!! GAP Propositio 2. (How to write geerl rdom umber geertor) Let X be cotiuous rdom vrible with vlues i [, b]. Suppose tht the cdf F(x) is strictly icresig o [, b]. Let U be uiform o [, 1]. Let Y = F 1 (U). The X d Y re ideticlly distributed. Proof. P(Y y) = P(F 1 (U) y) = P(U F(y)) = F(y) (1) Applictio:My computer hs routie to geerte rdom umbers tht re uiformly distributed o [, 1]. We wt to write routie to geerte umbers tht hve expoetil distributio with prmeter λ. How do you simulte orml RV s? Not so esy sice the cdf cot be explicitly computed. More o this lter. 5.6 More o expected vlue Recll tht for discrete rdom vrible tht oly tes o vlues i, 1, 2,, we showed i homewor problem tht E[X] = P(X > ) (11) = There is similr result for o-egtive cotiuous rdom vribles. Theorem 4. Let X be o-egtive cotiuous rdom vrible with cdf F(x). The E[X] = provided the itegrl coverges. [1 F(x)] dx (12) 1

11 Proof. We use itegrtio by prts o the itegrl. Let u(x) = 1 F(x) d dv = dx. So du = fdx d v = x. So [1 F(x)] dx = x(1 F(x)) x= + xf(x) dx = E[X] (13) Note tht the boudry term t is zero sice F(x) 1 s x. We c use the bove to prove the lw of the ucoscious sttistici for specil cse. We ssume tht X d tht the fuctio g is from [, ) ito [, ) d it strictly icresig. Note tht this implies tht g hs iverse. The E[Y ] = = = [1 F Y (x)] dx = [1 P(g(X) x)] dx = [1 P(Y x)] dx (14) [1 P(X g 1 (x))] dx (15) [1 F X (g 1 (x))] dx (16) Now we do chge of vribles. Let s = g 1 (x). So x = g(s) d dx = g (s)ds. So bove becomes Now itegrte this by prts to get [1 F X (s)] g(s) s= + [1 F X (s)] g (s) ds (17) which proves the theorem i this specil cse. g(s) f(s) ds (18) 11