# 2.4 Bivariate distributions

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1 page Bvarate dstrbutons Defntons Let X and Y be dscrete r.v.s defned on the same probablty space (S, F, P). Instead of treatng them separately, t s often necessary to thnk of them actng together as a random vector (X, Y ) takng values n R 2. The jont probablty functon of (X, Y ) s defned as and s often wrtten as p X,Y (x, y) = P({E S : X(E) = x and Y (E) = y}), (2.28) P(X = x, Y = y) (x = x 1, x 2,...x M ; y = y 1, y 2,..y ), (2.29) where M, may be fnte or nfnte. It satsfes the two condtons Varous other functons are related to P(X = x, Y = y). P(X = x, Y = y) 0 P(X = x, Y = y) = 1. (2.30) The jont cumulatve dstrbuton functon of (X, Y ) s gven by x y F (u, v) = P(X u, Y v), < u, v, = P(X = x, Y = y). (2.31) x u,y v The margnal probablty (mass) functon of X s gven by P(X = x ) = y P(X = x, Y = y), = 1,...M. (2.32) The margnal probablty (mass) functon of Y s gven by P(Y = y j ) = x P(X = x, Y = y j ), j = 1,... (2.33) The condtonal probablty (mass) functon of X gven Y = y j s gven by P(X = x Y = y j ) = P(X = x, Y = y j ), = 1,...M. (2.34) P(Y = y j ) The condtonal probablty (mass) functon of Y gven X = x s gven by P(Y = y j X = x ) = P(X = x, Y = y j ), j = 1,... (2.35) P(X = x ) Expectaton The expected value of a functon h(x, Y ) of the dscrete r.v.s (X, Y ) can be found drectly from the jont probablty functon of (X, Y ) as follows: E[h(X, Y )] = x h(x, y)p(x = x, Y = y) (2.36) y provded the double seres s absolutely convergent. Ths s the bvarate verson of the law of the unconscous statstcan dscussed earler.

2 page 29 The covarance of X and Y s defned as Cov(X, Y ) = E[{X E(X)}{Y E(Y )}] = E(XY ) E(X)E(Y ). (2.37) The correlaton coeffcent of X and Y s defned as ρ(x, Y ) = Cov(X, Y ) Var(X).Var(Y ). (2.38) Independence In Chapter 1 the ndependence of events has been defned and dscussed: now ths concept s extended to random varables. The dscrete random varables X and Y are ndependent f and only f the par of events {E S : X(E) = x } and {E S : Y (E) = y j } are ndependent for all x, y j, and we wrte ths condton as P(X = x, Y = y j ) = P(X = x ).P(Y = y j ) for all (x, y j ). (2.39) It s easly proved that an equvalent statement s: X and Y are ndependent f and only f there exst functons f( ) and g( ) such that p X,Y (x, y) = P(X = x, Y = y) = f(x)g(y) for all x, y. (2.40) Example 2.3 A based con yelds heads n a sngle toss wth probablty p. The con s tossed a random number of tmes, where Posson(λ). Let X and Y denote the number of heads and tals obtaned respectvely. Show that X and Y are ndependent Posson random varables. Condtonng on the value of X + Y, we have P(X = x, Y = y) = P(X = x, Y = y X + Y = x + y)p(x + Y = x + y) +P(X = x, Y = y X + Y x + y)p(x + Y x + y). The second condtonal probablty s clearly 0, so But Smlarly P(X = x, Y = y) = P(X = x, Y = y X + Y = x + y)p(x + Y = x + y) = ( x+y) x p x q y λ x+y. (x + y)! e λ [usng (2.201) & = X + Y ] = (λp)x (λq) y e λ. x!y! P(X = x) = P(X = x = n)p( = n) n x = ( n ) n x x p x qn x λn n! e λ = (λp)x e λ (λq) n x x! n x (n x)! = (λp)x x! e λ.e λq = (λp)x e λp. x! P(Y = y) = (λq)y e λq. y! Then P(X = x, Y = y) = P(X = x).p(y = y) for all (x, y), and t follows that X and Y are ndependent Posson random varables (wth parameters λp and λq respectvely).

3 page 30 It can also be shown readly that f X and Y are ndependent, so too are the random varables g(x) amd h(y ), for any functons g and h: ths result s used frequently n problem solvng. If X and Y are ndependent, E(XY ) = x,y = x,y xyp(x = x, Y = y) [(2.36)] xyp(x = x)p(y = y) [ndependence, (2.39)] = x xp(x = x) y yp(y = y).e., by (2.10), E(XY ) = E(X)E(Y ) f X, Y are ndependent. (2.41) The converse of ths result s false.e. E(XY ) = E(X)(E(Y ) does not mply that X and Y are ndependent. It follows mmedately that Once agan, the converse s false. Cov(X, Y ) = ρ(x, Y ) = 0 f X, Y are ndependent. (2.42) A generalsaton of the result for E(XY ) s: f X and Y are ndependent, then, for any functons g and h, E{g(X)h(Y )} = E{g(X)}E{h(Y )}. (2.43) Condtonal expectaton Referrng back to the defnton of the condtonal probablty mass functon of X, t s natural to defne the condtonal expectaton (or condtonal expected value) of X gven Y = y j as E(X Y = y j ) = = x P(X = x Y = y j ) x P(X = x, Y = y j )/P(Y = y j ) (2.44) provded the seres s absolutely convergent. Ths defnton holds for all values of y j (j = 1, 2,...), and s one value taken by the r.v. E(X Y ). Snce E(X Y ) s a functon of Y, we can wrte down ts mean usng (2.14): thus E[E(X Y )] = j = j = j = j = = E(X Y = y j ).P(Y = y j ) x P(X = x Y = y j )P(Y = y j ) [(2.44)] P(X = x, Y = y j ) x P(Y = y j ) [(2.34)] P(Y = y j ) x P(X = x, Y = y j ) x P(X = x, Y = y j ) j x P(X = x ) [(2.30b)].e.

4 page 31 E[E(X Y ) = E(X). (2.45) Ths result s very useful n practce: t often enables us to compute expectatons easly by frst condtonng on some random varable Y and usng E(X) = j E(X Y = y j ).P(Y = y j ). (2.46) (There are smlar defntons and results for E(Y X = x ) and the r.v. E(Y X).) Example 2.4 (Ross) A mner s trapped n a mne contanng 3 doors. The frst door leads to a tunnel whch takes hm to safety after 2 hours of travel. The second door leads to a tunnel whch returns hm to the mne after 3 hours of travel. The thrd door leads to a tunnel whch returns hm to the mne after 5 hours. Assumng he s at all tmes equally lkely to choose any of the doors, what s the expected length of tme untl the mner reaches safety? Let X: tme to reach safety (hours) Y : door ntally chosen (1,2 or 3) Then ow E(X) = E(X Y = 1)P(Y = 1) + E(X Y = 2)P(Y = 2) + E(X Y = 3)P(Y = 3) = 1 3 {E(X Y = 1) + E(X Y = 2) + E(X Y = 3)}. E(X Y = 1) = 2 E(X Y = 2) = 3 + E(X) E(X Y = 3) = 5 + E(X) (why?) E(X) = 1 { E(X) E(X)} or E(X) = It follows from the defntons that, f X and Y are ndependent r.v.s, then E(X Y ) = E(X) and E(Y X) = E(Y ) (both constants). (2.47) 2.5 Transformatons and Relatons In many stuatons we are nterested n the probablty dstrbuton of some functon of X and Y. The usual procedure s to attempt to express the relevant probabltes n terms of the jont probablty functon of (X, Y ). Two examples llustrate ths. Example 2.5 (Dscrete Convoluton) Suppose X and Y are ndependent count random varables. Fnd the probablty dstrbuton of the r.v. Z = X + Y. Hence show that the sum of two ndependent Posson r.v.s s also Posson dstrbuted. The event Z = z can be decomposed nto the unon of mutually exclusve events: (Z = z) = (X = 0, Y = z) (X = 1, Y = z 1) (X = z, Y = 0) for z = 0, 1, 2,... z Then we have P(Z = z) = P(X = x, Y = z x) or, nvokng ndependence [(2.39)], x=0

5 page 32 z P(Z = z) = P(X = x).p(y = z x). (2.48) x=0 Ths summaton s known as the (dscrete) convoluton of the dstrbutons p X (x) and p Y (y). ow let the ndependent r.v.s X and Y be such that X Posson(λ 1 ); Y Posson(λ 2 ).e. P(X = x) = λ 1 x e λ 1, x > 0; P(Y = y) = λ 2 x e λ 2, y > 0. x! y! Then, for Z = X + Y,.e. Example 2.6 P(Z = z) = = z x=0 z x=0 λ 1 x e λ 1 x!. λ 2 z x e λ 2 (z x)! z! x!(z x)! λ 1 x λ z x 2. e (λ1+λ2) z! = (λ 1 + λ 2 ) z e (λ 1+λ 2 ), z = 0, 1, 2,... z! Z Posson(λ 1 + λ 2 ). Gven count r.v.s (X, Y ), obtan an expresson for P(X < Y ). events: Agan, we decompose the event of nterest nto the unon of mutually exclusve (X < Y ) = (X = 0, Y = 1) (X = 0, Y = 2) (X = 1, Y = 2) (X = 1, Y = 3) (X = 2, Y = 3) (X = 2, Y = 4) = (X = x, Y = y) x=0,..., ;y=x+1,..., P(X < Y ) = x=0 y=x+1 P(X = x, Y = y).

6 page Multvarate dstrbutons Defntons The basc defntons for the multvarate stuaton where we consder a p-vector of r.v.s (X 1, X 2,..., X p ) are obvous generalsatons of those for the bvarate case. Thus the jont probablty functon s P(X 1 = x 1, X 2 = x 2,..., X p = x p ) = P({E S : X 1 (E) = x 1 and X 2 (E) = x 2 and... and X p (E) = x p }) (2.49) and has the propertes and x 1 P(X 1 = x 1,..., X p = x p ) 0 for all (x 1,..., x p ) P(X 1 = x 1,..., X p = x p ) = 1. (2.50) x p The margnal probablty functon of X s gven by P(X = x ) = x 1 x 1 x +1 x p P(X 1 = x 1,..., X p = x p ) for all x. (2.51) The probablty functon of any subset of (X 1,..., X p ) s found n a smlar way. Condtonal probablty functons can be defned by analogy wth the bvarate case, and expected values of functons of (X 1,..., X p ) are found as for bvarate functons Multnomal dstrbuton Ths s the most mportant dscrete multvarate dstrbuton, and s deduced by arguments famlar from the case of the bnomal dstrbuton. Consder n repeated ndependent trals, where each tral results n one of the outcomes E 1,..., E k wth P(E occurs n a tral) = p, k p = 1. Let X = number of tmes the outcome E occurs n the n trals. Then the jont probablty functon of (X 1,..., X k ) s gven by P(X 1 = x 1,..., X k = x k ) = where the x 1,..., x k are counts between 0 and n such that n! x 1!x 2!...x k! p 1 x 1...p k x k, (2.52a) For consder the event k x = n. E 1...E 1 E 2...E 2... E k...e k x 1 x 2 x k tmes (2.52b) It has probablty p 1 x 1 p 2 x 2...p k x k, x = n, p = 1.

7 page 34 Any event wth x 1 outcomes E 1, x 2 outcomes E 2,... and x k outcomes E k n a gven order also n! has ths probablty. There are x 1!...x k! dfferent arrangements of such a set of outcomes and these are mutually exclusve: the event (X 1 = x 1,..., X k = x k ) s the unon of these mutually exclusve arrangements. Hence the above result. The margnal probablty dstrbuton of X s Bnomal wth parameters n and p, and hence E(X ) = np Var(X ) = np (1 p ). (2.53) Also, we shall prove later that Cov(X, X j ) = np p j, j. (2.54) Independence For convenence, wrte I = {1,..., p} so that we are consderng the r.v.s {X : I}. These r.v.s are called ndependent f the events {X = x }, I are ndependent for all possble choces of the set {x : I} of values of the r.v.s. In other words, the r.v.s are ndependent f and only f P(X = x for all J) = J for all subsets J of I and all sets {x : I}. P(X = x ) (2.55) ote that a set of r.v.s whch are parwse ndependent are not necessarly ndependent Lnear combnatons Lnear combnatons of random varables occur frequently n probablty analyss. The prncpal results are as follows: [ n ] n E a X = a E(X ) (2.56) (whether or nor the r.v.s are ndependent); [ n ] n Var a X = a 2 Var(X ) + 2 a a j Cov(X, X j ) <j (2.57) n = a 2 Var(X ) f the r.v.s are ndependent; n m n m Cov a X, b j X j = a b j Cov(X, X j ) (2.58) j=1 j=1 where Cov(X, X ) = Var(X ) by defnton.

8 page Indcator random varables me probablty problems can be solved more easly by usng ndcator random varables, along wth the above results concernng lnear combnatons. An ndcator random varable X for an event A takes the value 1 f A occurs and the value 0 f A does not occur. Thus we have: P(X = 1) = P(A) P(X = 0) = P(A) = 1 P(A) E(X) = 1.P(X = 1) + 0.P(X = 0) = P(A) E(X 2 ) = 1 2.P(X = 1) P(X = 0) = P(A) Var(X) = P(A) [P(A)] 2. (2.59) Clearly 1 X s the ndcator r.v. for the event A. Let Y be the ndcator r.v. for the event B. Then the varous combnatons nvolvng A and B have ndcator r.v.s as follows: Event Indcator r.v. A B XY A B (1 X)(1 Y ) A B 1 (1 X)(1 Y ) A B (A, B mutually exclusve) X + Y EXAMPLES Example 2.6 Derve the generalsed addton law (1.16) for events A 1, A 2,...A n usng ndcator r.v.s. Let X be the ndcator r.v. for A. Then we deduce the followng ndcator r.v.s: 1 X for A ; (1 X 1 )...(1 X n ) for A 1... A n = A 1... A n 1 (1 X 1 )...(1 X n ) for A 1... A n. Hence P(A 1...A n ) = E[1 {1 (1 X 1 )...(1 X n )}] = E[ X X X j + X X j X k... + ( 1) n+1 X 1...X n ] <j <j<k = E(X ) E(X X j ) ( 1) n+1 E(X 1...X n ) <j = P(A ) P(A A j ) ( 1) n+1 P(A 1... A n ). <j The last lne follows because, for example, E(X X j ) = 1.1.P(X = 1, X j = 1) + 0 = P(A A j )

9 page 36 Example 2.7 (Lft problem) Use ndcator r.v.s to solve the lft problem wth 3 people and floors (Ex. 1.8). and Let Then X = (1 Y 1 )(1 Y 2 )(1 Y 3 ) { 1, f exactly one person gets off at each floor X = 0, otherwse { 1, f no-one gets off at floor Y = 0, otherwse. and P(one person gets off at each floor) = P(X = 1) = E(X) = E[1 {Y 1 + Y 2 + Y 3 } + {Y 1 Y 2 + Y 1 Y 3 + Y 2 Y 3 } Y 1 Y 2 Y 3 ] = 1 p 1 p 2 p 3 + p 12 + p 13 + p 23 p 123 where the requred probablty s p = P(Y = 1) = p j = P(Y = 1, Y j = 1) = p 123 = P(Y 1 = 1, Y 2 = 1, Y 3 = 1) = 0 ( 2 3 ( 1 3 ) 3, = 1, 2, 3 ) 3, j ( ) 2 3 ( ) = Example 2.8 Consder the generalsaton of the tokens-n-cereal collectng problem (Ex. 1.2) to dfferent card types. (a) Fnd the expected number of dfferent types of cards that are contaned n a collecton of n cards. (b) Fnd the expected number of cards a famly needs to collect before obtanng a complete set of at least one of each type. (c)fnd the expected number of cards of a partcular type whch a famly wll have by the tme a complete set has been collected. (a) Let and let Then ow X = number of dfferent types n a collecton of n cards { 1, f at least one type card n collecton I = 0, otherwse. X = I I. = 1,..., n. E(I ) = P(I ( = 1) = ) 1 P(no type cards n collecton of n) 1 n = 1, = 1,...,. E(X) = E(I ) = [ 1 ( 1 ) n ].

10 page 37 (b) Let and X = number of cards collected before a complete set s obtaned, Y = number of addtonal cards that need to be obtaned after dstnct cards have been collected, n order to obtan another dstnct type ( = 0,..., 1). When dstnct cards have already been collected, a new card obtaned wll be of a dstnct ( ) type wth probablty ( )/. Y s a geometrc r.v. wth parameter,.e. Hence from (2.21b) ow (c) Let Then E(X) = ( ) ( ) k 1 P(Y = k) =, k 1, = 1 =0 E(Y ) =. X = Y 0 + Y Y 1. E(Y ) = ( ). X = number of cards of type acqured. [ ] E(X) = E X = E(X ). By symmetry, E(X ) wll be the same for all, so E(X ) = E(X) ( = ) from part (b).

11 page 38 Example 2.9 Suppose that (X 1,..., X p ) has the multnomal dstrbuton P(X 1 = x 1,..., X k = x k ) = n! x 1!x 2!...x k! p 1 x 1...p k x k, where k x = n, k p = 1. Show that Cov(X, X j ) = np p j, j. Consder the rth tral: let Then ow { 1, f rth tral has outcome E I r = 0, otherwse. Cov(I r, I rj ) = E(I r.i rj ) E(I r ).E(I rj ). E(I r.i rj ) = 0.0P(I r = 0, I rj = 0) +0.1P(I r = 0, I rj = 1) +1.0P(I r = 1, I rj = 0) +1.1P(I r = 1, I rj = 1) = 0, j (snce P(I r = 1, I rj = 1) = 0 when j). Also, from the ndependence of the trals, Cov(I r, I rj ) = E(I r ).E(I rj ) = p p j, j. Cov(I r, I sj ) = 0 when r s. ow the number of tmes that E occurs n the n trals s X = I 1 + I I n. Cov(X, X j ) = Cov( n = n = n r=1 n r=1s=1 r=1 = n r=1 I r, n s=1 I sj ) Cov(I r, I sj ) Cov(I r, I rj ) ( p p j ) = np p j, j. (Ths negatve correlaton s not unexpected, for we antcpate that, when X s large, X j wll tend to be small).

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