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    In this paper, we study the number of isolated vertices Y of K , that is, the number of vertices having noincident edges, given by

    Y =vV

    1(d(v) = 0) . (2)

    In [5], the mean and variance 2

    of Y are computed as = n(1 p)n 1 and 2 = n(1 p)n 1(1 + np(1 p)n 2 (1 p)n 2). (3)

    In the same paper, Kolmogorov distance bounds to the normal were obtained and asymptotic normalityshown when

    n2 p and np log(n) .OConnell [6] showed that an asymptotic large deviation principle holds for Y . Raic [7] obtained nonuni-

    form large deviation bounds in some generality, for random variables W with E (W ) = 0 and Var( W ) = 1,of the form

    P (W

    t)

    1 (t) et 3 ( t ) / 6

    (1 + Q(t)(t)) for all t 0 (4)where ( t) denotes the distribution function of a standard normal variate and Q(t) is a quadratic in t .Although in general the expression for (t) is not simple, when W equals Y properly standardized andnp c as n , (4) holds for all n sufficiently large with

    (t) =C 1n exp

    C 2 tn + C 3(e

    C 4 t/ n 1)

    for some constants C 1 , C 2 , C 3 and C 4 . For t of order n1/ 2 , for instance, the function (t) will be small asn , allowing an approximation of the deviation probability P (W t) by the normal, to within somefactors. Theorem 1.1 below, by contrast, provides a non-asymptotic, explicit bound, that is, it does notrequire any relation between n and p and is satised for all n. Moreover as (8) and (6) show, this bound isof order eat

    2

    over some range of t , and of worst case order ebt , for the right tail, and ect2

    for the left tail,where a, b and c are explicit, with the bounds holding for all t

    R .

    Theorem 1.1. For n {1, 2, . . .}and p(0, 1) let K denote the random graph on n vertices where each edge is present with probability p, independently of all other edges, and let Y denote the number of isolated vertices in K . Then for all t > 0,

    P Y

    t inf 0 exp(t + H ()) where H () =

    22

    0s s ds, (5)

    with the mean and variance 2 of Y given in (3), and

    s = 2 e2s 1 +pes

    1 pn

    + + 1 where = (1 p)n .For the left tail, for all t > 0,

    P Y

    t exp t2

    22

    ( + 1). (6)

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    Recall that for a nonnegative random variable Y with nite, nonzero mean , the size bias distributionof Y is given by the law of a variable Y s satisfying

    E [Y f (Y )] = Ef (Y s ) (7)

    for all f for which the expectations above exist. The main tool used in proving Theorem 1.1 is size bias

    coupling, that is, constructing Y and Y s, having the Y -size biased distribution, on the same space. In [3],size bias couplings were used to prove concentration of measure inequalities when |Y s Y | can be almostsurely bounded by a constant. Here, where Y is the number of isolated vertices of K , we consider a coupling

    of Y to Y s with the Y -size bias distribution where the boundedness condition is violated. Unlike the maintheorem in [3] which can be applied to a wide variety of situations where the bounded coupling assumptionis satised, it seems that cases where the coupling is unbounded, such as the one we consider here, needapplication specic treatment, and cannot be handled by one single general result.

    Remark 1.1. Useful bounds for the minimization in (5) may be obtained by restricting to [0, 0] for some 0 . In this case, as s is an increasing function of s, we have

    H ()

    42 0 2 for [0, 0].

    The quadratic t + 0 2 / (42) in is minimized at = 2 t 2 / ( 0 ). When this value falls in [0, 0] weobtain the rst bound in (8), while otherwise setting = 0 yields the second.

    P Y

    t exp(t

    2 2

    0 ) for t[0, 0 0 / (22)]

    exp(0 t + 0

    20

    4 2 ) for t(0 0 / (22), ).

    (8)

    Though Theorem 1.1 is not an asymptotic result, when np c as n , we have,2

    1 + cec ec , + 1 ec + 1 and s 2e2s + ce

    s

    + ec + 1 as n .

    Since limn s and limn /2 exist, the right tail decays at the rate at most e0 t . Also, the left tail

    bound (6) in this asymptotic behaves at xed t as

    limn

    exp t2

    22

    ( + 1)= exp

    t2

    21 + cec ec

    ec + 1.

    The paper is organized as follows. In Section 2 we review results leading to the construction of sizebiased couplings for sums of possibly dependent variables, and then in Section 3 apply this construction tothe number Y of isolated vertices, a sum of indicator variables; this construction rst appeared in [4]. Theproof of Theorem 1.1 is also given in Section 3.

    2 Construction of size bias couplings

    In this section we will review the discussion in [4] which gives a procedure for a construction of size biascouplings when Y is a sum; the method has its roots in the work of Baldi et al. [1]. The construction dependson being able to size bias a collection of nonnegative random variables in a given coordinate, as describedDenition 2.1. Letting F be the distribution of Y , rst note that the characterization (7) of the size biasdistribution F s is equivalent to the specication of F s by its Radon Nikodym derivative

    dF s (x) =x

    dF (x). (9)

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    Denition 2.1. Let Abe an arbitrary index set and let {X : A}be a collection of nonnegative random variables with nite, nonzero expectations EX = and joint distribution dF (x ). For A, we say that X = {X : A}has the X size bias distribution in coordinate if X has joint distribution dF (x ) = x dF (x )/ .

    Just as (9) is related to (7), the random vector X

    has the X size bias distribution in coordinate if andonly if

    E [X f (X )] = E [f (X )] for all functions f for which these expectations exist.

    Letting f (X ) = g(X ) for some function g one recovers (7), showing that the th coordinate of X , that is,X , has the X size bias distribution.

    The factorizationP (X d

    x ) = P (X dx |X = x)P (X dx)

    of the joint distribution of X suggests a way to construct X . First generate X , a variable with distributionP (X dx). If X = x, then generate the remaining variates {X , = }with distribution P (X dx |X = x). Now, by the factorization of dF (x ), we have

    dF (x ) = x dF (x )/ = P (X dx |X = x)x P (X dx)/ = P (X dx |X = x)P (X dx). (10)

    Hence, to generate X with distribution dF , rst generate a variable X with the X size bias distribution,then, when X = x, generate the remaining variables according to their original conditional distributiongiven that the th coordinate takes on the value x.

    Denition 2.1 and the following special case of a proposition from Section 2 of [4] will be applied in thesubsequent constructions; the reader is referred there for the simple proof.

    Proposition 2.1. Let Abe an arbitrary index set, and let X = {X , A}be a collection of nonnegativerandom variables with nite means. Let Y =

    A X and assume A = EY is nite and positive. Let X have the X -size biased distribution in coordinate as in Denition 2.1. Let I be a random index taking values in A with distribution

    P (I = ) = / A , A.

    Then if X I has the mixture distribution

    A P (I = )L(X ), the variable Y s = A X I has the Y -sized biased distribution as in (7).In our examples we use Proposition 2.1 and the random index I , and (10), to obtain Y s by rst generating

    X I I with the size bias distribution of X I , then, if I = and X = x, generating {X : A\{}}accordingto the (original) conditional distribution P (X , = |X = x).

    3 Proof of Theorem 1.1

    We now present the proof of Theorem 1.1.

    Proof. We need a basic inequality which we state rst. For real x = y, by the convexity of the exponentialfunction,

    ey exy x

    = 1

    0ety +(1 t )x dt

    1

    0(tey + (1 t)ex )dt =

    ey + ex

    2. (11)

    Next, we construct a coupling of Y s , having the Y size bias distribution, to Y . Let K be given, and let Y bethe number of isolated vertices in K . To size bias Y , rst recall the representation of Y as the sum (2). Asthe summands in (2) are exchangeable, the distribution of the random index I in Proposition 2.1 is uniform.

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    Hence, choose one of the n vertices of K uniformly. If the chosen vertex, say V , is already isolated, we donothing and set K s = K , as the remaining variables already have the conditional distribution given that V is isolated. Otherwise obtain K s by deleting all the edges connected to K . By Proposition 2.1, the variableY s counting the number of isolated vertices of K s has the Y -size biased distribution.

    To derive the needed properties of this coupling, let N (v) be the set of neighbors of v V , and T be thecollection of isolated vertices of K , that is, with d(v), the degree of v, given in (1),N (v) = {w : X vw = 1 } and T = {v : d(v) = 0 }.

    Note that Y = |T |. Since all edges incident to the chosen V are removed in order to form K s , any neighborof V which had degree one thus becomes isolated, and V also becomes isolated if it was not so earlier. Asall other vertices are otherwise unaffected, as far as their being isolated or not, we have

    Y s Y = d1(V ) + 1( d(V ) = 0) where d1(V ) =w

    N (V )

    1(d(w) = 1) . (12)

    So in particular the coupling is monotone, that is, Y s Y . Since d1(V ) d(V ), (12) yieldsY s Y d(V ) + 1 . (13)

    By (11), using that the coupling is monotone, for 0 we haveE (eY

    s

    eY ) 2

    E (Y s Y )(eY s

    + eY )

    =2

    E (exp( Y )(Y s Y ) (exp( (Y s Y )) + 1))=

    2

    E {exp( Y )E ((Y s Y )(exp( (Y s Y )) + 1) |T )}. (14)Now using that Y s = Y when V T , and (13), we have

    E ((Y s Y )(exp( (Y s Y )) + 1) |T ) E ((d(V ) + 1)(exp( (d(V ) + 1)) + 1)1( V T )|T ) e E d(V )ed (V ) + ed (V ) + e d(V ) 1(V T )|T + 1 . (15)

    Note that since V is chosen independently of K ,

    L(d(V )1(V T )|T ) = P (V T )L( Bin (n 1 Y, p)|Bin (n 1 Y, p) > 0) + P (V T )0 , (16)where 0 is point mass at zero. Using (16) and the mass function of the conditioned binomial

    P (d(V ) = k|T , V T ) =n 1Y k

    pk (1 p)n 1 Y k

    1(1 p) n 1 Y for 1 k n 1 Y

    0 otherwise,

    it can be easily veried that the conditional moment generating function of d(V ) and its rst derivative arebounded by

    E (ed (V ) 1(V T )|T ) ( pe + 1 p)n 1Y (1 p)n 1Y

    1 (1 p)n 1Y and

    E (d(V )ed (V ) 1(V T )|T ) (n 1 Y )( pe + 1 p)n 2Y pe

    1 (1 p)n 1Y .

    By the mean value theorem applied to the function f (x) = xn 1Y , for some (1 p, 1) we have1 (1 p)n 1Y = f (1) f (1 p) = ( n 1 Y ) pn 2Y (n 1 Y ) p(1 p)n .

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    Hence, recalling 0,

    E (d(V )ed (V ) 1(V T )|T ) (n 1 Y )( pe + 1 p)n pe

    1 (1 p)n 1Y

    (n 1 Y )( pe + 1 p)n pe

    (n

    1

    Y ) p(1

    p)n

    = where = e 1 +pe

    1 pn

    . (17)

    Similarly applying the mean value theorem to f (x) = ( x + 1 p)n 1Y , for some (0, pe ) we have

    E (ed (V ) 1(V T )|T ) (n 1 Y )( + (1 p))n 2Y pe

    1 (1 p)n 1Y

    (n 1 Y )( pe + (1 p)) n 2Y pe

    1 (1 p)n 1Y , (18)

    as in (17).Next, to handle the second to last term in (15) consider

    E (d(V )1(V T )|T ) (n 1 Y ) p

    1 (1 p)n 1Y (n 1 Y ) p

    (n 1 Y ) p(1 p)n= where = (1 p)n . (19)

    Applying inequalities (17),(18) and (19) to (15) yields

    E ((Y s Y )(exp( (Y s Y )) + 1) |T ) where = 2 e + + 1 . (20)Hence we obtain, using (14),

    E (eY s

    eY ) 2

    E (eY ) for all 0.Letting m() = E (eY ) thus yields

    m () = E (Y eY ) = E (eY s

    ) 1 + 2

    m(). (21)

    SettingM () = E (exp( (Y )/ )) = e/ m(/ ),

    differentiating and using (21), we obtain

    M () =1

    e/ m (/ )

    e/ m(/ )

    e/ (1 +

    2 )m(/ )

    e/ m(/ )

    = e/ 22

    m(/ ) = 22

    M (). (22)

    Since M (0) = 1, (22) yields upon integration of M (s)/M (s) over [0, ],

    log(M ()) H () so that M () exp(H ()) where H () =

    22

    0s s ds.

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    Hence for t 0,P (

    Y t) P (exp(

    (Y )

    ) et ) et M () exp(t + H ()) .As the inequality holds for all 0, it holds for the achieving the minimal value, proving (5).

    For the left tail bound let < 0. Since Y s

    Y and < 0, using (11) and (13) we obtainE (eY eY

    s

    ) ||2

    E (eY + eY s

    )(Y s Y ) ||E (eY (Y s Y ))= ||E (eY E (Y s Y |T )) ||E (eY E ((d(V ) + 1)1( V T ))|T )) .

    Applying (19) we obtain

    E (eY eY s

    ) ( + 1) ||E (eY ),and therefore

    m () = E (eY s

    ) (1 + ( + 1) ) m().Hence for < 0,

    M () =1

    e/ m (/ )

    e/ m(/ )

    e/ ((1 + ( + 1) / )m(/ ))

    e/ m(/ )

    =( + 1)

    2M ().

    Dividing by M () and integrating over [ , 0] yields

    log(M ()) ( + 1) 2

    22. (23)

    The inequality in (23) implies that for all t > 0 and < 0,

    P (Y

    t) exp( t +( + 1) 2

    22).

    Taking = t 2/ (( + 1)) we obtain (6).

    References

    [1] Baldi, P. , Rinott, Y. and Stein, C. (1989). A normal approximations for the number of localmaxima of a random function on a graph, Probability, Statistics and Mathematics, Papers in Honor of Samuel Karlin , T. W. Anderson, K.B. Athreya and D. L. Iglehart eds., Academic Press, 59-81.

    [2] Barbour, A.D. , Karo nski, M. and Ruci nski,A. (1989). A central limit theorem for decomposablerandom variables with applications to random graphs, J. Combinatorial Theory B , 47 , 125-145.

    [3] Ghosh, S. and Goldstein, L. (2009). Concentration of measures via size biased couplings, Probab.Th. Rel. Fields , to appear.

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    [4] Goldstein, L. and Rinott, Y. (1996). Multivariate normal approximations by Steins method andsize bias couplings, Journal of Applied Probability , 33 ,1-17.

    [5] Kordecki, W. (1990). Normal approximation and isolated vertices in random graphs, Random Graphs87 , Karo nski, M., Jaworski, J. and Ruci nski, A. eds., John Wiley & Sons Ltd.,1990, 131-139.

    [6] OConnell, N. (1998). Some large deviation results for sparse random graphs, Probab. Th. Rel. Fields ,110 , 277-285.

    [7] Rai c, M. (2007). CLT related large deviation bounds based on Steins method, Adv. Appl. Prob. , 39 ,731-752.

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