Question

7. Let X1, X2,.. be i.i.d. random variables, and let T(t)minn: X > t, t20. (a) Determine the distribution of T(t) (b) Show that, if p= P(X1> t)0 astoo, then pT(t)Exp(1) as to

Answer 1

= n. Then observe that X1, X2, ... , Xn-1 Would be less (a) If supoose T(t) = than t, and X, would be greater than t,but after that other X^'s does not matter. Hence t, X2t,...,Xn-1 < t,Xn > t) P(T(t)n) P(X1 = P(X1 t)P(X2 < t)P(Xn-1 t)P(X> t) — (1- Р(X, > t))"-1P(X, > t) (1 -P)"-P Therefore T(t~ Geometric(p (b) By Levy's continuity theorem function and convergence in distribution characteristic function to show the given convergence in distribution. Now let X Exp(1). Then a characteristic function of X would be we know that convergence in characteristic equivalent. Hence we will use are px(s) Ele'nX 8 e dar (cos(sa)isinsx)e dx sin(sa)e da (sr)e dr i COS Jo (T s(sx)e- dx COS = [- cos(s)e- (s) sin(sa)e*) dx Integration By Parts] -1-+ 1 - sI2 sin(sa)ed sin(sa)e d I2 T - [-sin()e1. scos(sa)(-e dax S = s/ S I, = 1- s2/ I1 = 1+ s2 I2 _ 1+ $2 1 +is x(s) = = (1 - is)-1

Now for T(t) Pr()(s) = Eleia( n-1 (1 P)"Pt n-1 =Pte'(1-P)e"]"-1 n=1 Preis 1 - (1-)eis Pt ei-(1 t -is Let Y Tt) Then Pr(s)EleY] = E[esp«T(e)] _ r()(spt) -(1 t) e-2sPt Now we have to show that if p, ->0 as t> oo then pY;(s) -> px(s) as t oO. Now Pt im py(s)lim - (1 P toc too e-2 spt 1 lim t oo _ -ispt +1 Pt 1 -i spt lim +1 t-oo Pt

Now -ispt lim 1 t-oo Pt -ispt - 1 lim Pt0 ir = s lim 0 cos (r) sin()1 lim I0 = S cos(r) sin(r i lim lim = S - sin2(r/2) lim x lim -0 4 = S /4 .2 is _ lim y(s) =px (s) 1-is t oc Hence, by Levy's continuity theorem, Y, converges distributionally to X