HUB HUB RESEARCH PAPER. A Markov Binomial Distribution. Hogeschool Universiteit Brussel. E. Omey, J. Santos and S. Van Gulck

HUB HUB RESEARCH PAPER A Markov Binomial Distribution E. Omey, J. Santos and S. Van Gulck HUB RESEARCH PAPER 2007/5. SEPTEMBER 2007 Hogeschool Universiteit Brussel Stormstraat 2, 000 Brussel, Belgium T: +32 2 20 2 F: + 32 2 27 64 64

A MARKOV - BINOMIAL DISTRIBUTION E. Omey, J. Santos and S. Van Gulck Abstract Let fx i ; i g denote a sequence of f0; g-variables and suppose that the sequence forms a Markov Chain. In the paper we study the number of successes S n = X + X 2 + ::: + X n and we study the number of experiments Y (r) up to the r th success. In the i.i.d. case S n has a binomial distribution and Y (r) has a negative binomial distribution and the asymptotic behaviour is well known. In the more general Markov chain case, we prove a central limit theorem for S n and provide conditions under which the distribution of S n can be approximated by a Poisson-type of distribution. We also completely characterize Y (r) and show that Y (r) can be interpreted as the sum of r independent r.v. related to a geometric distribution. Keywords: Markov chain, generalized binomial distribution, central limit theorem MSC 2000 Classi cation: 60J0, 60F05, 60K99

Introduction Many papers are devoted to sequences of Bernoulli trials and they form the basis of many (known) distributions. Applications are numerous. To mention only a few: - the one-sample runs test can be used to test the hypothesis that the order in a sample is random; - the number of successes can be used for testing for trends in the weather or in the stock market; - Bernoulli trials are important in matching DNA-sequences; - the number of (consecutive) failures can be used in quality control. For further use we suppose that each X i takes values in the set f0; g and for n, let S n = P n i= X i denote the number of successes in the sequence (X ; X 2 ; :::; X n ). If the X i are i.i.d. with P (X i = ) = p and P (X i = 0) = q = p, it is well known that S n has a binomial distribution S n BIN(n; p). In the classical theory one either calculates probabilities concerning S n by using the binomial distribution or by using a normal- or a Poisson-approximation. A related variable of interest is Y (r) where for r the variable Y (r) counts the number of experiments until the r-th success. In the i.i.d. case it is well known that Y (r) has a negative binomial distribution and that Y (r) can be interpreted as the sum of i.i.d. geometrically distributed r.v. In the section 2 below we rst list the Markov chain properties that we need and then study S n (section 2.) and Y (r) (section 2.2). We nish the paper with some concluding remarks. 2 Markov Chains Let the initial distribution be given by P (X = ) = p and P (X = 0) = q = p and for i; j = 0;, let p i;j = P (X 2 = j j X = i) denote the transition probabilities. To avoid trivialities we suppose that 0 < p i;j <. The one-step transition matrix of the Markov chain is given by p0;0 p P = 0;. p ;0 p ; We list some elementary properties of this Markov chain, cf. [3, Chapter XVI.3]. First note that the Markov chain has a unique stationary vector 2

given by (x; y) = (p ;0, p 0; )=(p 0; + p ;0 ). The eigenvalues of P are = and 2 = p 0; p ;0 = p 0;0 + p ;. Note that j 2 j <. By induction it is easy to show that the n-step transition matrix is given by P n = A + n 2B where x A = x y y, B = y y x x It follows that p (n) 0;0 = x + n 2y and p (n) ;0 = x n 2x. Using these relations for n we have. P (X n = ) = y + n 2 (px qy) = y n 2 (y p), P (X n = 0) = x + n 2 (y p). Information about moments is given in the following result. Lemma For n we have (i) E(X n ) = P (X n = ) = y n 2 (y p). (ii) V ar(x n ) = (y n 2 (y p))(x + n 2 (y p)). (iii) For n m we have Cov(X n ; X m ) = n m 2 V ar(x m ). Proof. The rst and the second part are easy to prove. To prove (iii), (n m) note that E(X n X m ) = P (X n = ; X m = ) = p ; P (X m = ). It follows (n m) that Cov(X n ; X m ) = (p ; P (X n = ))P (X m = ). Using the expressions obtained before we obtain Cov(X n ; X m ) = (y + n m 2 x y + n 2 (y p))p (X m = ) = n m 2 (x + m 2 (y p))p (X m = ), and the result follows. As a special case we consider the type of correlated Bernoulli trials studied in [2] and [9]. In this model we assume that P (X n = ) = p, P (X n = 0) = q = p and = (X n ; X n+ ) 6= 0 for all n. From this it follows that Cov(X n ; X n+ ) = pq and that P (X n = ; X n+ = ) = p(p + q). Since P (X n = ) = p we also nd that P (X n = 0; X n+ = ) = P (X n = ; X n+ = 0) = pq( It turns out that P (X n+ = j j X n = i) = p i;j (i; j = 0; ), where the p i;j are given by q + p p( ) P (p; ) =. q( ) p + q In this case we have (x; y) = (q; p) and 2 =. For n m it follows from Lemma that (X n ; X m ) = n m. 3 ).

2. The number of successes S n 2.. Moments In this section we study the number of successes S n = P n i= X i. In our rst result we study moments of S n and extend the known i.i.d. results. Proposition 2 (i) We have E(S n ) = ny (y p)( n 2)=( 2 ). (ii) We have Xn V ar(s n ) = nxy( + 2 )=( 2 ) + (A k 2 + B 2k 2 + Ck k 2), k=0 where A; B; C will be determined in the proof of the result. (iii) If P = P (p; ) we have E(S n ) = np and V ar(s n ) = pq(n( + ) 2( n )=( ))=( ). Proof. Part (i) follows from Lemma (i). To prove (ii) we start from V ar(s k+ ) = V ar(s k + X k+ ). Using Lemma, we see that V ar(s k+ ) V ar(s k ) = V ar(x k+ ) + 2 Again from Lemma we see that kx i= V ar(x i ) = xy + a i 2 b 2i 2 2 k+ i 2 V ar(x i ). where a = (y p)(y x) and b = (y p) 2. Straightforward calculations show that V ar(s k+ ) V ar(s k ) = xy( + 2 )=( 2 ) + A k 2 + B 2k 2 + Ck k 2 where C = 2a and A = (a( 2 ) 2xy 2 2b)=( 2 ), B = b( + 2 )=( 2 ). If we de ne S 0 = 0 this result holds for all k 0. Using V ar(s n ) = P n k=0 (V ar(s k+) V ar(s k )), result (ii) follows. Result (iii) is easier to prove. Using Lemma we have V ar(s k+ ) V ar(s k ) = xy( + 2 kx k+ i ) = xy( + 2 k+ )=( ). i= 4

The result follows by taking sums as before. The expression for the variance can be simpli ed asymptotically. We use the notation u(n) s cv(n) to indicate that u(n)=v(n)! c. Corollary 3 As n! we have (i) E(S n ) ny and V ar(s n ) nxy( + 2 )=( 2 ). (ii) E(S n ) ny! (p y)=( 2 ) and V ar(s n ) nxy(+ 2 )=( 2 )! c, where c = A + B 2 2 + C 2 2 ( 2 ). 2 2..2 Distribution of S n In this section we determine p n (k) = P (S n = k). It is convenient to condition on X n and to this end we de ne p i n(k) = P (S n = k; X n = i), i = 0;. Obviously p n (k) = p 0 n(k) + p n(k). Note that p () = p, p 0 (0) = q and that p (0) = p 0 () = 0. In the next result we show how to calculate p i n(k) recursively. Lemma 4 For n we have (i) p 0 n+(k) = p 0;0 p 0 n(k) + p ;0 p n(k); (ii) p n+(k) = p 0; p 0 n(k ) + p ; p n(k ). Proof. We have p 0 n+(k) = I + II where I = P (S n+ = k; X n+ = 0; X n = 0), II = P (S n+ = k; X n+ = 0; X n = ). Clearly I = P (S n = k; X n+ = 0; X n = 0). Now note that I = P (X n+ = 0 j S n = k; X n = 0)p 0 n(k) = P (X n+ = 0 j S n = k; X n = 0)p 0 n(k) = p 0;0 p 0 n(k). In a similar way we nd that II = p ;0 p n(k) and the rst result follows. The second result can be proved in a similar way. For small values of n we can use Lemma 4 to obtain the p.d. of S n. It does not seem to be easy to obtain an explicit expression for p n (k). For an alternative approach we refer to the end of section 2.2 below. 5

2..3 Central limit theorem For xed P and (q; p) and large values of n we can approximate the p.d. of S n by a normal distribution. We prove the following central limit theorem. Theorem 5 As n! we have S n E(S n ) p V ar(sn ) d ) Z N(0; ). Proof. We prove the theorem using generating functions. For jzj i and i = 0; let n (z) = P n k=0 pi n(k)z k 0 and let n(z) = n (z) + n (z). 0 Furthermore, let n (z) = ( n (z); n (z)) and note that (z) = (q; pz). Using Lemma 4 we have 0 0 n+(z) = p 0;0 n(z) + p ;0 n(z), 0 n+(z) = p 0; z n(z) + p ; z n(z). Switching to matrix notation, we nd that n+ (z) = n (z)a(z), where p0;0 p A(z) = 0; z. p ; z p ;0 It follows that n+ (z) = (z)a n (z). We can nd A n (z) by using the eigenvalues of A(z) and to this end we use the characteristic equation ja(z) Ij = 2 (p 0;0 + p ; z) + (p 0;0 p ; p ;0 p 0; )z = 0. Solving this equation gives (z) = (p 0;0 + p ; z + p D)=2 p 2 (z) = (p 0;0 + p ; z D)=2 where D = (p 0;0 p ; z) 2 +4p 0; p ;0 z. Now de ne U = U(z) and W = W (z) by the equations I = A 0 (z) = U +W and A(z) = (z)u + 2 (z)w. We nd that U(z) = (A(z) 2 (z)i)= p D and W (z) = (A(z) (z)i)= p D. Using the theorem of Cayley and Hamilton we obtain for n 0 that A n (z) = n (z)u + n 2(z)W. Note that as z! we have (z)! and 2 (z)! p 0; p ;0 =, where by assumption we have jj <. Since j 2 (z)= (z)j! jj as z!, for all " su ciently small we can nd = (") such that 0 < < and 6

such that j 2 (z)= (z)j jj + " <, for all z such that z. We conclude that for any sequence z n!, we have j 2 (z n )= (z n )j n! 0. From this it follows that A n (z n )= n (z n )! U() where both rows of U() are equal to (x; y). Using n+(z) = (z)a n (z) (; ) t it follows that n+(z n )= n (z n )!. Now we discuss the asymptotic behaviour of (z) as z!. For convenience we write (z) = (z). Note that (z) satis es () = and the characteristic equation 2 (z) (z)(p 0;0 + p ; z) + (p 0;0 p ; p ;0 p 0; )z = 0. Taking derivatives with respect z we nd that and 2(z) 0 (z) 0 (z)(p 0;0 + p ; z) (z)p ; + p 0;0 p ; p ;0 p 0; = 0 2(z) 00 (z) + 2( 0 (z)) 2 2 0 (z)p ; 00 (z)(p 0;0 + p ; z) = 0. Replacing z by z = we nd that 2() 0 () 0 ()(p 0;0 + p ; ) ()p ; + p 0;0 p ; p ;0 p 0; = 0, 2() 00 () + 2( 0 ()) 2 2 0 ()p ; 00 ()(p 0;0 + p ; ) = 0. Since () =, straightforward calculations show that 0 () = p 0; =(p 0; + p ;0 ) = y, 00 () = 2xy(p ; p 0; )=(p 0; + p ;0 ). Using the rst terms of a Taylor expansion, it follows that (z) = + y(z ) + 2 00 ()(z ) 2 ( + o()). Using twice the expansion log(x) = ( x) ( x) 2 (+o())=2, we obtain It follows that log((z)) y log(z)! ( z) 2 2 (00 () + y y 2 ). log((z)z y ) ( z) 2! xy p ; p 0; p 0; + p ;0 + xy 2 = 2, 7

where, after simplifying, = xy p ; + p 0;0 p 0; + p ;0 = xy + 2 2. To complete the proof of Theorem 5 we replace z by z n = z =pn. In this case we nd that log((z n )zn y )! n 2 (log(z))2 and then that zn yn n+(z n )! exp((log(z)) 2 =2). It follows that (S n+ ny)= p n ) d W where W N(0; ). Using Corollary 3, the proof of the theorem is nished. From Theorem 5 we obtain the following result. Corollary 6 Let Z s N(0; ). As n! we have the following results: (i) (S n ny)= p n ) d Z, where = xy( + 2 )=( 2 ); (ii) If p 0; = p ; = p, we have (S n pn)= p d npq =) Z; (iii) If P = P (p; ), we have (S n pn)= p n =) d Z, where = pq( + )=( )). 2..4 Poisson approximation In the i.i.d. case it is well known how we can approximate a binomial distribution by a suitable Poisson distribution. The same can be done in the more general Markov setting. In what follows and also in Section 2.2 we shall use the following notations. We use U(a) to denote a Bernoulli r.v. with P (U(a) = ) = a (0 < a < ), V (a) is a Poisson(a)-variable (a > 0) and G(a) is a geometric r.v. with P (G(a) = k) = ( a)a k (k, 0 < a < ). Note that E(z U(a) ) = ( a) + az, E(z V (a) ) = exp( a + az), E(z G(a) ) = ( a)z=( az). 8

A compound Poisson distribution with generator G(a) is de ned as follows. Let G 0 (a) = 0 and for i, let G i (a) denote i.i.d. copies of G(a). Independent of the G i (a) let V (b) denote a Poisson(b)-variable. Now consider the new r.v. B(V (b)) = P V (b) i=0 G i(a). Clearly we have E(z B(V (b)) ) = exp( b + be(z G(a) )) and we say that B(V (b)) has a compound Poisson distribution with generator G(a). In the next result, in the limiting distribution all r.v. involved are independent and we use the notations introduced above. In each case we take limits as n!. Theorem 7 Suppose that np 0;! a > 0 and p ;! c (0 c < ). (i) If c = 0, then S d n ) U(p) + V (a). (ii) If 0 < c <, then S d n ) U(p)G(c) + B(V (a)). Proof. Using the notations as in the proof of Theorem 5 we have n+(z) = (z)a n (z) (; ) t where (z) = (q; pz) and A n (z) = n (z)u(z)+ n 2(z)W (z). Recall that U(z) = (A(z) 2 (z)i)= p D and W (z) = (A(z) (z)i)= p D and that (z) = (p 0;0 + p ; z + p p D)=2 and 2 (z) = (p 0;0 + p ; z D)=2, where D = (p0;0 p ; z) 2 + 4p 0; p ;0 z. Some straighforward calculations show that (z) = 2p 0; ( z)=(p 0; + p ;0 + p ; ( z) + p D). Since by assumption np 0;! a we have p 0;! 0 and p 0;0!. By assumption we have p ;! c and hence also p ;0! c. It follows that D! ( cz) 2. Using cz > 0 we readily see that n( (z))! a( z)=( cz). From this it follows that n (z)! (z) := exp( a( z)=( cz)) Next we consider n 2(z). Clearly we have 2 (z) = 2 ()z= (z). Our assumptions imply that 2 () = p 0;0 + p ;! 0. It follows that ( 2 ()z) n! 0 and hence also that n 2(z)! 0. Using 2 (z)! cz, we obtain that U(z)! U = 0 ( c)=( cz) 0 9

and hence also that that A n (z)! (z)u. It follows that so that n+(z)! (q; pz)(z)u (; ) t, n+(z)! L(z) = (z) q + z(p c). cz It remains to identify the limit L(z). If c = 0, we have L(z) = (q + pz) exp( a( z)). Now the interpretation is clear. Using the notations introduced before, let U(p) and V (a) denote independent r.v.. Clearly L(z) = E(z V (a) z U(p) ). If c = 0, we conclude that S n d ) V (a) + U(p). If 0 < c <, we nd L(z) = (q + pk(z)) exp( a( K(z)), where K(z) = ( c)z=( cz). Using the notations introduced before, we see that K(z) = E(z G(c) ). Let G 0 (c) = 0 and let G i (c) denote independent copies of G(c) and, independent of the other variables, let V (a) denote a Poissonvariable with parameter a. The random sum B(V (a)) = P V (a) i=0 G i(c) is wellde ned and has generating function E(z B(V (a)) ) = exp( a( K(z)). Finally, let U(p) denote a Bernoulli variable, independent of all other variables. We conclude that L(z) = E(z U(p)G(c)+B(V (a)) ) and as a consequence we nd that S d n ) U(p)G(c) + B(V (a)). Remark. If c =, then L(z) = q exp( a) which is the generating function of a degenerate variable. As a special case we have the following corollary. Corollary 8 (Wang [9]) Suppose that P = P (p; ). (i) If np! u > 0 and! 0, then S d n ) V (u). (ii) If np! u > 0 and! v (0 < v < ), then S d n ) B(V (u( v)) 2.2 Generalized negative binomial distribution In this section we invert S n and for each r we de ne Y (r) as the number of experiments until the r th succes. Since Y (r) = minfn : S n = rg we have P (S n r) = P (Y (r + ) n + ) and P (Y (r) = n) = P (S n = r; X n = ). The last quantitiy has been studied before. Adapting the notations, for n r and i = 0; we set p i n(r) = P (S n = r; X n = i) and p n (r) = P (S n = r). 0

The corresponding generating functions will be denoted by i r (z) and r(z). Using similar methods as before, for n r we obtain that p n(r) = p ; p n (r ) + p 0; p 0 n (r ), p 0 n(r) = p 0;0 p 0 n (r) + p ;0 p n (r). Using generating functions, this leads to We nd that r(z) = zp ; 0 r(z) = zp 0;0 r (z) + zp 0; 0 r(z) + zp ;0 0 r r(z). (z), 0 r(z) = where k(z) = p ; + zp 0; p ;0 =( zp ;0 p 0;0 z r(z) = zk(z) r r(z), (z), p 0;0 z). It follows that Using r(z) = 0 r (z) + r (z) we also nd that r(z) = (zk(z)) r (z). () r(z) = ( + zp ;0 p 0;0 z ) r (z) = u(z)(zk(z)) r (z), (2) where u(z) = + zp ;0 =( p 0;0 z). It remains to determine (z) = P n= P (S n = ; X n = )z n. Clearly we have It follows that P (S n = ; X n = ) = p, if n = P (S n = ; X n = ) = qp n 2 0;0 p 0;, if n 2. p 0; z (z) = z(p + q p 0;0 z ). This result can be interpreted in the following way. We use the notations as in the beginning of Section 2..4. Assuming that U(s) and G(t) are independent r.v., we have E(z G(t) ) = z( t)=( tz) and E(z U(s)G(t) ) = ( s) + sz( t)=( tz) (3)

Using these notations we can identify k(z) and (z). Using (3) we obtain that k(z) = E(z U(p ;0)G(p 0;0 ) ), (4) (z) = ze(z U(q)G(p 0;0) ) (5) Using (), (4) and (5) we obtain the following result. Theorem 9 (i) We have Y () d = + U(q)G(p 0;0 ). (ii) For r 2, we have Xr Y (r) = d ( + U i (p ;0 )G i (p 0;0 )) + + U(q)G(p 0;0 ), i= where all r.v. U(q), U i (p ;0 ), G(p 0;0 ) and G i (p 0;0 ) involved are independent. Using r (z) = zk(z) r (z) it also follows for r 2 that Y (r) d = Y (r ) + + U(p ;0 )G(p 0;0 ) and that Y (r) Y (r ) d = + U(p ;0 )G(p 0;0 ) where Y (r ), U(p ;0 ), and G(p 0;0 ) are independent r.v.. Together with Y () d = + U(q)G(p 0;0 ) we obtain the following probabilistic interpretation of the formulas. At the start, the rst value is either a succes or a failure. If we start with a failure (which happens with probability q) we have to wait geometrically long until we have a rst succes. Given another succes position in the sequence, either the next result is a succes or the next result is a failure (which happens with probability p ;0 ) and then we have to wait geometrically long until we have a new success. Although we start from a sequence of (Markov-)dependent variables, it turns out that the times between consecutive successes are independent variables! Now we take a closer look at p n (r) = P (S n = r) and use (2). Note that r() = u() = =y and observe that yu(z) = y + yzp ;0 p 0;0 z = y + xzp 0; p 0;0 z = E(zU(x)G(p 0;0) ) 2

It follows that y r(z) = yu(z)(zk(z)) r (z) and hence that yp (S n = r) = P (U(x)G(p 0;0 ) + Y (r) = n) where U(x), G(p 0;0 ) and Y (r) are independent. In the next result we formulate the central limit theorem for Y (r) and a Poisson-type of approximation. The results easily follow from the representation obtained in Theorem 9. Corollary 0 (i) As r!, we have Y (r) r=y d q ) Z N(0; ). rp ;0 p ; =p 2 0; (ii) If rp ;0! a and p 0;0! c, 0 c <, then Y (r) r d ) B(V (a)). Remark. For r 0, let M(r) = maxfn : S n rg and K(r) = maxfn : Y (n) rg. Clearly M(r) = n if and only if Y (r +) = n+ so that Y (r +) = M(r)+. Using fm(r) ng = fs n rg = fy (r + ) n + g and fk(r) ng = fy (n) rg, it follows that fk(n) rg = fy (r + ) n + g. As a consequence, we nd that K(n) = d S n. Now note that K(r) corresponds to the renewal counting process associated with the sequence A; A ; A 2 ; ::: where A = d +U(q)G(p 0;0 ) and where the A i are i.i.d. random variables with = d + U(p ;0 )G(p 0;0 ). Standard (delayed) renewal theory could now be A i used to prove the central limit theorem for S n. 3 Concluding remarks. Earlier we have observed that for n, P (X n = ) = y n 2 (y p) and P (X n = 0) = x + n 2 (y p). From this it is easy to see that for n; m we have P (X n+m = ) = y( n 2) + n 2P (X m = ) P (X n+m = 0) = x( n 2) + n 2P (X m = 0) Recall that (x; y) was the stationary vector. In what follows we assume that 2 > 0. Now let B n U( n 2) and B s U(y) denote Bernoulli r.v.. The formulas obtained above show that X n+m d = ( B n )B + B n X n where B n 3

is independent of B and X n. In particular it follows that fx n g satis es the stochastic di erence equation: X d n+ = ( B )B + B X n. 2. A correlated binomial distribution has been introduced and studied in [], [4], [6], [7]. Examples and applications can be found e.g. in quality control, cf. [5]. One of the models can be described as follows. Let X; X ; X 2 ; :::; X n denote i.i.d. Bernoulli variables with X s U(p) and let U() and U() denote independent Bernoulli variables, independent of the X i. Now de ne Y i = U()X i + ( U())U() and the corresponding sum T n = nx Y i = U()S n + n( i= U())U(). The following interpretation can be given. In quality control we can decide to check all produced units. The alternative is to check just one unit and then accept or reject all produced units. In the rst scenario S n counts the number of defects or successes. In the second scenario we conclude that we have either 0 or n defects or successes. Clearly S n BIN(p; n) and for T n we have P (T n = k) = P (S n = k) + ( )P (nu() = k). It follows that P (T n = k) = P (S n = k) for k n, P (T n = 0) = P (S n = 0) + ( )( ), P (T n = n) = P (S n = n) + ( ). As to Y i we have Y i d = U() where = p + ( ). For the variance V ar(y i ) = 2 we nd that 2 = ( )(p ) 2 + p( p) + ( )( ). For i 6= j we obtain that Cov(Y i ; Y j ) = ( )(p ) 2 + ( )( ). It follows that (Y i ; Y j ) = = if = p. If 6= p we nd that (Y i ; Y j ) = = =(c + ) where c = p( p)=(( )(p ) 2 + ( )( )). We see that (Y i ; Y j ) is the same for all i 6= j. Clearly this implies that V ar(t n ) = n( + (n )=2) 2. In the next result we formulate two asymptotic results. Proposition (i) As n!, (T n ( U())nU() npu())= p np( p) d ) B()Z 4

where Z N(0; ). (ii) If np! a > 0 and! 0, then T n d ) Y where Y d = U()V (a) and U() and V (a) are independent. 3. From the physical point of view it seems reasonable to study random vectors (X ; X 2 ; :::; X n ) with a joint probability distribution of the following form: for x i = 0; we have nx nx nx P (X i = x i ; i = ; 2; :::; n) = C exp( a i x i + a i;j x i x j ). The second term represents the interaction between particles. If = 0, the X i are independent and no interaction appears. 4. In our next paper [8] we study runs, singles and the waiting time until the rst run of r consecutive successes for Bernoulli-sequences that are generated by a Markov chain. 4 References [] P. Altham, Two generalizations of the binomial distribution. Applied Statistics 27(978) 62-67. [2] A.W.F. Edwards, The meaning of binomial distribution. Nature, London 86 (960), 074. [3] W. Feller, "An Introduction to Probability Theory and its Applications", Vol. I (3rd edition). John Wiley and Sons, New York, 968. [4] L.L. Kupper and J.K. Haseman, The use of the correlated binomial model for the analysis of certain toxicological experiments. Biometrics 34(978), 69-76. [5] C.D. Lai, C.D., K. Govindaraju and M. Xie, E ects of correlation on fraction non-conforming statistical process control procedures. Journal of Applied Statistics 25(4) (998), 535-543. [6] R.W. Madsen, Generalized binomial distributions. Comm. in Statistics: Theory and Methods 22()(993), 3065-3086. [7] S.A. Mingoti, A note on sample size required in sequential tests for the generalized binomial distribution. Journal of Applied Statistics 30(8) (2003), 873-879. [8] E. Omey and S. Van Gulck, Runs, singles and waiting times for Bernoulli trials governed by a Markov chain. Forthcoming (2008). 5 i= i= j=

[9] Y.H. Wang, On the limit of the Markov binomial distribution. J. Appl. Prob. 8(98) 937-942. Acknowledgement. The authors are grateful to the referees for their detailed comments and useful suggestions to improve the paper. E. Omey and S. Van Gulck EHSAL Stormstraat 2, 000 Brussels - Belgium. E-mail: {edward.omey;stefan.vangulck}@ehsal.be. J. Santos Public University of Navarre 3006 Pamplona - Spain. E-mail: santos@si.unavarra.es. 6