Question

8. Let X1...., X, be i.i.d. ~E(1) random variables (i.e., they are independent and identically distributed, all with the exponential distribution of parameter 1 = 1). a) Compute the cdf of Yn = min(X1,...,xn). b) How do P({Y, St}) and P({X1 <t}) compare when n is large and t is such that t<? c) Compute the odf of Zn = max(X1...., X.). d) How do P({Zn2 t}) and P({X1 2 t}) compare when n is large and t is such that t >> log(n) ? 9. Let (22, F, P) be a probability space, X be an square-integrable random variable defined on this space and let G be a sub-o-field of F. Relying only on the definition of conditional expectation, show the following properties: a) E(E(X19)) = E(X). b) If X is independent of G, then E(X19) = E(X) a.s.

Answer 1

8.9 5 X, X2, ..., xn a {(1) x's given then CDF of Xx (1=1,2, .. n) F(xi) = 1- exp(-ne) and PDF of xi xs x's given by f(wi) = exp(-1) Let us find the CDF of Yn = Yn = min(x, .- fn) P ( Yn st) =P (min (x,,.., An) < t ) - P( min (X,,- - Xn) 7 ) Now af min (x, xo) t than all Xi 7t (for i=1,2,. n) thus P( Yn st) = = 1- P(X, 7, 4, X2 > t, –, Xnot) XI, X21 ore independent, this P(Yn st) = 1 - . Since Хи کد I P(xx>t) 221 n - - ñ (1 - P(xist)) nil 7 exp (te) 251 - expc&to) - 1- exp(-nt) (b P(XI<t) 1- exp(-t) It t <th nt << Thus for large n to tso) that xs xuf n too and nt 71.

omd 1- exp(-1) This P(x, Et P( Yn st) Zn = max(x,1x2, - - ,xn) a P(Zn , t) 1- P( Zn st) - 1-P(max (x,x2,. – An) St As max (x1, x2, ... Xulst i=1,2, .n. Thus +) = All (tilst for الله - Xust) n P( Zn >, t) = 1- P(x, St, Since Xils are = 1- Pixist) [ i=1 independent = 1 (1 - exp(-AI)" la P(x>t) & - P(x, t) exp(-x) If to logh or te tog n th (hto) exp(-x) = exp(-logn- h) cz / expl-h) This for large n ( no) and ho epp (-t) 0. Thus P (41 t) - 0 and P (2034) = 1- (1-exp(-))" Notice that as nyo (1-exp(-t)) x's in the form 10 h

Let ㅕ Ls (1- exoc-ti) log L = n log (1-exp(+1) log (1 - exp(-x) 1/ we can conite n By equation exp(-x) = 1 expl-h) log L = log (1- t exp(-h)) or In using L'hospital role of limit we optain + 1/n2 exp(-h) lim logL lim exp(-27) nto n too 2ln (1- = L = explo) - 1 Thus for lorge n (1 - explot 1h 1 and it gives Pl Zn>,t) - 1 - 1 = 0