DOC

On the hypothesis testing concerning the type of the survival function

By Alexander Ruiz,2014-11-26 12:19
14 views 0
On the hypothesis testing concerning the type of the survival function

    ON THE HYPOTHESIS TESTING CONCERNING THE TYPE

    OF THE SURVIVAL FUNCTION

     12Evgeniy Chepurin, Innesa Chepurina

    Moscow State University, Moscow, Russia 12echepurin@mail.ru, uchsov@cs.msu.ru

    The aim of this paper is to discuss the problems of the testing hypothesis concerning a

    survival distribution function form.

    Let the random variable be a lifetime of a person which exists at the moment . Tt0In other words, the random variable is the future lifetime of this entity measured from . Tt0The probability that this person’s lifetime is greater than , i.e. t

     StTt!,,;;??

    is called survival distribution function. It is supposed that

     a) T0;

     b) S(0)1;

    c) St() is a non-increasing function;

    d) S()0.)!

    Suppose also that

    e) St() is a differentiable function.

    In this paper it is assumed that any other information about survival distribution function except the characteristics a) e) is unknown.

    Let us denote by

    Sx(1)pTxTx!,?,!1| (1) ??xSx()

    and

    SxSx()(1)??qp!?!1. (2) xxSx()

     Consider the sequence of the embedded events

     (3) TkTkTxTx,?,??,??,11,????????

    xwhere and are integers. From the properties of conditional distributions, the embedding (3) k

    and the assumption that , follows that xk

     (4) SxSxTxTxpSx()1|11,!?,,?!?;;??;;x1

    x1

     (5) SxSkp()(),?jjk

    and

    x1

     (6) SxSkq()()1.!?;;?jjk

    St()Let be a veritable survival distribution function which generates our random life 0

    data concerning the life durations of some people group, for example, the participants of some

    St()pension fund. In reality an actual functional form of is unknown. 0

     1

    (::()()StStAt the same time often it is necessary to test hypothesizes for 101

    (?:()()StStttt((ttt((, under alternatives for on the basis of special censored 2011212

    St()ttsample of size . Here will be well known survival distribution function, and will N121

    St()be known constants. It is supposed that is either parametric function with known 1

    ktkt:((parameters or is defined by the life table, for integers . To obtain survival data for 12

    iN1,the members of pension plan we use the following sample scheme: participant , , is i

    [,]HHHHobserved only in the time interval , where and are calendar dates for the 1212

    beginning and the termination of the observation period in which life status information of

    statistical data members can be obtained.

    YAAA(,,,)Let be a sample data, where 12N

    AWJJBDyz(,,,,,,), iiiiiiii

    W is calendar dates of withdrawal from observation (by reason other than by death), i

    J is calendar dates of joining to pension plan, i

    J is observable calendar dates of joining to group under observation, i

    B is calendar dates of birth, i

    DDW is calendar dates of death (if ), iii

    ()yJB!?y is calendar dates of age under joining to observation group , iiii

    thz is scheduled age at termination for sample member. ii

    In our case

    ?JifJH,?ii1J ?iHifJH,?11i?

    and

    HBifWH?,,?22ii z?iWBifWH?(.iii2?

    Then let be

    yymin,(1)i

    zzmax,ni

    xintegerxy!,min, ??(1)txt((12

    xintegerxz!(max,??()n((txt12

    AxAforxintegeryxzxxyx(){:(,1)(1)!(,??,,??iiii

    ?,,??(,,?(1)(1)},xzxxyzxiii

    AAx?()n is the number of . ix

    AAx?()For let us define i

    zxwhenxzx?,,?1,?ii ??!!()x?ii11,whenzx,?i?

    and

     2

    yxwhenxyx?,,?1,?ii rrx!!()?ii0,whenyx(i?

     is the number died in interval ?(,1].xxAAx