Topics MODULE 3 FUNCTION OF A RANDOM VARIABLE AND ITS DISTRIBUTION LECTURES 2-6 3. FUNCTION OF A RANDOM VARIABLE 3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE 3.3 EXPECTATION AND MOMENTS OF A RANDOM VARIABLE 3.4 PROPERTIES OF RANDOM VARIABLES HAVING THE SAME DISTRIBUTION 3.5 PROBABILITY AND MOMENT INEQUALITIES 3.5. Markov Inequaliy 3.5.2 Chebyshev Inequaliy 3.5.3 Jensen Inequaliy 3.5.4 AM-GM-HM inequaliy 3.6 DESCRIPTIVE MEASURES OF PROBABILITY DISTRIBUTIONS 3.6. Measures of Cenral Tendency 3.6.. Mean 3.6..2 Median 3.6..3 Mode 3.6.2 Measures of Dispersion 3.6.2. Sandard Deviaion 3.6.2.2 Mean Deviaion 3.6.2.3 Quarile Deviaion 3.6.2.4 Coefficien of Variaion Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur
3.7 MEASURES OF SKEWNESS 3.8 MEASURES OF KURTOSIS MODULE 3 FUNCTION OF A RANDOM VARIABLE AND ITS DISTRIBUTION LECTURE 2 Topics 3. FUNCTION OF A RANDOM VARIABLE 3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE 3. FUNCTION OF A RANDOM VARIABLE Le Ω, F, P be a probabiliy space and le X be random variable defined on Ω, F, P. Furher le h: R R be a given funcion and le Z: Ω Rbe a funcion of random variable X, defined by Z ω = h X ω, ω Ω. In many siuaions i may be of ineres o sudy he probabilisic properies of Z, which is a funcion of random variable X. Since he variable Z akes values in R, o sudy he probabilisic properies of Z, i is necessary ha Z B F, B B, i.e., Z is a random variable. Throughou, for a posiive ineger k, R k will denoe he k-dimensional Euclidean space and B k will denoe he Borel sigmafield in R k. Definiion. Le k and m be posiive inegers. A funcion h: R k R m is said o be a Borel funcion if h B B k, B B m. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 2
The following lemma will be useful in deriving condiions on he funcion h: R R so ha Z: Ω R, defined by Z ω = h X ω, ω Ω, is a random variable. Recall ha, for a funcion Ψ: D D 2 and A D 2, Ψ A = ω D : Ψ ω A. Lemma. Le X: Ω R and h: R R be given funcions. Define Z: Ω R by Z ω = h X ω, ω Ω. Then, for any B R, Z B = X h B. Proof. Fix B R. Noe ha h B = x R: h x B. Clearly Therefore, Theorem. h X ω B X ω h B. Z B = ω Ω: Z ω B = ω Ω: h X ω B = ω Ω: X ω h B = X h B. Le X be a random variable defined on a probabiliy space Ω, F, P and le h: R R be a Borel funcion. Then he funcion Z: Ω R, defined by Z ω = h X ω, ω Ω, is a random variable. Proof. Fix B B. Since h is a Borel funcion, we have h B B. Now using he fac ha X is a random variable i follows ha This proves he resul. Remark. Z B = X h B F. (i) Le h: R R be a coninuous funcion. According o a sandard resul in calculus inverse image of any open inerval a, b, a < b, under coninuous funcion h is a counable union of disjoin open inervals. Since B conains all open inervals and is closed under counable unions i follows ha h a, b B, whenever a < b. Now on employing he Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 3
argumens similar o he one used in proving Theorem., Module 2 (also see Theorem.2, Module 2) we conclude ha h B B, B B. I follows ha any coninuous funcion h: R R is a Borel funcion and hus, in view of Theorem., any coninuous funcion of a random variable is a random variable. In paricular if X is a random variable hen X 2, X, max X, 0, sin X and cos X are random variables. (ii) Le h: R R be a sricly monoone funcion. Then, for a < b, h a, b is a counable union of inervals and herefore h a, b B, i.e., h is a Borel funcion. I follows ha if X is a random variable and if h: R R is sricly monoone hen h(x) is a random variable. A random variable X akes values in various Borel ses according o some probabiliy law called he probabiliy disribuion of random variable X. Clearly he probabiliy disribuion of a random variable of absoluely coninuous/discree ype is described by is disribuion funcion (d.f.) and/or by is probabiliy densiy funcion/probabiliy mass funcion (p.d.f/p.m.f.). For a given Borel funcion h: R R, in he following secion, we will derive probabiliy disribuion of h(x) using he probabiliy disribuion of random variable X. 3.2 PROBABILITY DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE In our fuure discussions when we refer o a random variable, unless oherwise saed, i will be eiher of discree ype or of absoluely coninuous ype. The probabiliy disribuion of a discree ype random variable will be referred o as a discree (probabiliy) disribuion and he probabiliy disribuion of a random variable of absoluely coninuous ype will be referred o as an absoluely coninuous (probabiliy) disribuion. The following heorem deals wih discree probabiliy disribuions. Theorem 2. Le X be a random variable of discree ype wih suppor S X and p.m.f. f X ( ). Le h: R R be a Borel funcion and le Z: Ω R be defined by Z ω = h X ω, ω Ω. Then Z is a random variable of discree ype wih suppor S Z = h x : x S X and p.m.f. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 4
f Z z = f X x, if z S Z x A z where A z = x S X : h x = z. = P X A z, if z S Z, Proof. Since h is a Borel funcion, using Theorem., i follows ha Z is a random variable. Also X is of discree implies ha S X is counable which furher implies ha S Z is counable. Fix z 0 S Z, so ha z 0 = h x 0 for some x 0 S X. Then X = x 0 = ω Ω: X ω = x 0 ω Ω: h X ω = h x 0 = h X = h x 0 = Z = z 0, and X S X = ω Ω: X ω S X ω Ω: h X ω S Z = h X S Z = Z S Z. Therefore, P Z = z 0 P X x 0 > 0, (since x 0 S X ), and P Z S Z P X S X =. I follows ha S Z is counable, P Z = z > 0, z S Z and P Z S Z =, i. e., Z is a discree ype random variable wih suppor S Z. Moreover, for z S Z, Hence he resul follows. P Z = z = P ω Ω: h X ω = z = P X = x x A z = f X x x A z = P X A z. The following corollary is an immediae consequence of he above heorem. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 5
Corollary 2. Under he noaion and assumpions of Theorem 2., suppose ha h: R R is one-one wih inverse funcion h : D R, where D = h x : x R. Then Z is a discree ype random variable wih suppor S z = h x : x S X and p.m.f. f Z z = f X h z, if z S Z Example 2. = P X = h z, if z S Z. Le X be a random variable wih p.m.f., if x 2,, 0, 7 f X x = 3, if x 2, 3 4. Show ha Z = X 2 is a random variable. Find is p.m.f. and disribuion funcion. Soluion. Since h x = x 2, x R, is a coninuous funcion and X is a random variable, using Remark. (i) i follows ha Z = h X = X 2 is a random variable. Clearly S X = 2,, 0,, 2, 3 and S Z = 0,, 4, 9. Moreover, P Z = 0 = P X 2 = 0 = P X = 0 = 7, P Z = = P X 2 = = P X, = 7 + 7 = 2 7, P Z = 4 = P X 2 = 4 = P X 2, 2 = 7 + 3 4 = 5 4, and P Z = 9 = P X 2 = 9 = P X 3, 3 = 0 + 3 4 = 3 4 Therefore he p.m.f. of Z is Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 6
f Z z =, if z = 0 7 2, if z = 7 5,, if z = 4 4 3, if z = 9 4 and he disribuion funcion of Z is F Z z = 0, if z < 0, if 0 z < 7 3, if z < 4 7, if 4 z < 9 4, if z 9 Example 2.2 Le X be a random variable wih p.m.f. f X x = x 2550, if x ±, ±2,, ±50. Show ha Z = X is a random variable. Find is p.m.f., and disribuion funcion. Soluion. As h x = x, x R, is a coninuous funcion and X is a random variable, using Remark. (i), Z = X is a random variable. We have S X = ±, ±2,, ±50 and S Z =, 2,, 50. Moreover, for z S Z, P Z = z = P X = z = P X z, z = z 2550 + z 2550 = z 275 Therefore he p.m.f. of Z is f Z z = and he disribuion funcion of Z is z, if z, 2,, 50 275, Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 7
F Z z = 0, if z <, if z < 2 275. i i +, if i z < i +, i = 2, 3,,49 2550, if z 50 Example 2.3 Le X be a random variable wih p.m.f. f X x = n x px p n x, if x 0,,, n, where n is a posiive ineger and p 0,. Show ha Y = n X is a random variable. Find is p.m.f. and disribuion funcion. Soluion. Noe ha S X = S Y = 0,,, n and h x = n x, x R, is a coninuous funcion. Therefore Y = n X is a random variable. For y S Y P Y = y = P X = n y = n n y pn y p y = n y p y p n y. Thus he p.m.f. of Y is f Y y = n y p y p n y, if y 0,,, n, and he disribuion funcion of Y is F Y y = 0, if y < 0 p n, if 0 y < i j =0 n j p j p n j., if i y < i +, i =,2,, n, if y n The following heorem deals wih probabiliy disribuion of absoluely coninuous ype random variables. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 8
Theorem 2.2 Le X be a random variable of absoluely coninuous ype wih p.d.f. f X ( ) and suppor k S X. Le S, S 2,, S k, be open inervals in R such ha S i S j = φ, if i j and i= S i = S X. Le h: R R be a Borel funcion such ha, on each S i i =,, k, h: S i R is sricly monoone and coninuously differeniable wih inverse funcion h i ( ). Le h S j = h x : x S j so ha h S j j =,, k is an open inerval in R. Then he random variable T = h X is of absoluely coninuous ype wih p.d.f. f T = k f X j = h j d d h j I h Sj (). Proof. We will provide an ouline of he proof which may no be rigorous. Le F T ( ) be he disribuion funcion of T. For R and > 0, F T + F T = P < h X + = k j = P < h X +, X S j Fix j,, k. Firs suppose ha h j ( ) is sricly decreasing on S j. Noe ha X S j = h X h S j and h S j is an open inerval. Thus, for belonging o he exerior of h S j and sufficienly small > 0, we have P < h X +, X S j = 0. Also, for h S j and sufficienly small > 0, P < h X +, X S j = P h j + X < h j. Thus, for all R, we have P < h X +, X S j = P h j + X < h j I h Sj () = h j () h j (+) f X z dz I h Sj () 0 fx h j d d h j I h Sj. (2.) Similarly if h j is sricly increasing on S j hen, for all R, we have Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 9
P < h X +, X S j = P h j < X h j + I h Sj () h j (+) = f X z dz I h Sj () h j () 0 fx h j d d h j I h Sj. (2.2) Noe ha if h is sricly decreasing (increasing) on S j hen d d h j < > 0 on S j. Now on combining (2.) and (2.2) we ge, for all R, P < h X +, X S j 0 fx h j d d h j I h Sj, F T + F T 0 k j = Similarly one can show ha, for all R, f X h j d d h j I h Sj. F T + F T lim 0 k = f X h j j = d d h j I h Sj. (2.3) I follows ha he disribuion funcion of T is differeniable everywhere on R excep possibly a a finie number of poins (on boundaries of inervals h S,, h S k of S T ). Now he resul follows from Remark 4.2 (vii) of Module 2 and using (2.3). The following corollary o he above heorem is immediae. Corollary 2.2 Le X be a random variable of absoluely coninuous ype wih p.d.f. f X ( ) and suppor S X. Suppose ha S X is a finie union of disjoin open inervals in R and le h: R R be a Borel funcion such ha h is differeniable and sricly monoone on S X (i.e., eiher h x < 0, x S X or h x > 0, x S X ). Le S T = h x : x S X. Then T = h X is a random variable of absoluely coninuous ype wih p.d.f. f T = f X h d d h, if S T Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 0
I may be worh menioning here ha, in view of Remark 4.2 (vii) of Module 2, Theorem 2.2 and Corollary 2.2 can be applied even in siuaions where he funcion h is differeniable everywhere on S X excep possibly a a finie number of poins. Example 2.4 Le X be random variable wih p.d.f. and le T = X 2 f X x = e x, if x > 0, (i) (ii) (iii) Show ha T is a random variable of absoluely coninuous ype; Find he disribuion funcion of T and hence find is p.d.f.; Find he p.d.f. of T direcly (i.e., wihou finding he disribuion funcion of T). Soluion. (i) and (iii). Clearly T = X 2 is a random variable (being a coninuous funcion of random variable X). We have S X = S T = 0,. Also h x = x 2, x S X, is sricly increasing on S X wih inverse funcion h x = x, x S T. Using Corollary 2. i follows ha T = X 2 is a random variable of absoluely coninuous ype wih p.d.f. f T = f X d d, if > 0 = e, 2 if > 0. (ii) We have F T = P X 2, R. Clearly, for < 0, F T = 0. For 0, F T = P X = f X x dx = e x dx 0 = e. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur
Therefore he disribuion funcion of T is F T = 0, if < 0 e, if 0. Clearly F T is differeniable everywhere excep a = 0. Therefore, using Remark 4.2 (vii) of Module 2, we conclude ha he random variable T is of absoluely coninuous ype wih p.d.f. f T = F T, if 0. A = 0 we may assign any arbirary nonnegaive value o f T 0. Thus a p.d.f. of T is f T = e, if > 0 2. Example 2.5 Le X be a random variable wih p.d.f. f X x = x 2 x 3, if < x <, if x < 2, and le T = X 2 (i) (ii) (iii) Show ha T is a random variable of absoluely coninuous ype; Find he disribuion funcion of T and hence find is p.d.f; Find he p.d.f. of T direcly (i.e., wihou finding he disribuion funcion of T). Soluion. (i) and (iii). Clearly T = X 2 is a random variable (being a coninuous funcion of random variable X ). We have S X =, 0 0, 2 = S S 2, say. Also h x = x 2, x S X, is sricly decreasing in S =, 0 wih inverse funcion h = ; h x = x 2, x S X, is sricly increasing in S 2 = 0, 2, wih inverse funcion h 2 = ; h S = 0, and h S 2 = 0, 4. Using Theorem 2.2 i follows ha T = X 2 is a random variable of absoluely coninuous ype wih p.d.f. f T = f X d d I 0, + f X d d I 0,4 =, if 0 < < 2, if < < 4 6 Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 2
(ii) We have F T = P X 2, R. Since P X, 2 P T 0, 4 =. =, we have Therefore, for < 0, F T = P T = 0 and, for 4, F T = P T =. For [0,4), we have Therefore, he disribuion funcion of T is F T = P X = f X x = F T = dx x dx, if 0 < 2. x 2 dx + x 3 dx, 0, if < 0, if 0 < 2 + 2, if < 4 6, if 4 if < 4 Clearly F T is differeniable everywhere excep a poins 0, and 4. Using Remark 4.2 (vii) of Module 2 i follows ha he random variable T is of absoluely coninuous ype wih a p.d.f. f T =, if 0 < < 2, if < < 4 6 Noe ha a Borel funcion of a discree ype random variable is a random variable of discree ype (see Theorem.). Theorem 2.2 provides sufficien condiions under which a Borel funcion of an absoluely coninuous ype random variable is of absoluely coninuous ype. The following example illusraes ha, in general, a Borel funcion of an absoluely coninuous ype random variable may no be of absoluely coninuous ype. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 3
Example 2.6 Le X be a random variable of absoluely coninuous ype wih p.d.f. f X x = e x, if x > 0 0, soherwise, and le T = x, where, for x R, x denoes he larges ineger no exceeding x. Show ha T is a random variable of discree ype and find is p.m.f. Soluion. For a R, we have T, a =, a + B. I follows ha T is a random variable. Also S X = 0,. Since P X S X =, we have P T 0,, 2, =. Also, for i 0,, 2,. P T = i = P( i X < i + ) i+ = f X x dx i i+ = e x dx i = e e i > 0. Consequenly he random variable T is of discree ype wih suppor S T = 0,, 2, and p.m.f. f T = P T = = e e, if 0,, 2, 0, oherwise. Dep. of Mahemaics and Saisics Indian Insiue of Technology, Kanpur 4