Question

Let Mn be the maximum of n independent U(0, 1) random variables.
a. Derive the exact expression for P(|Mn − 1| > ε).
Hint: see Section 8.4.
b. Show that limn→∞ P(|Mn − 1| > ε) = 0. Can this be derived from Chebyshev’s
inequality or the law of large numbers?

Answer 1

Solution is provided in one long image

pdf of U(a,b) is 1/(b-a) where x values lie between a and b

Expression in a) can be further expanded to derive a longer expression using binomial expansion

let XiNULOD. n independent random variables X1, X2, ..., Xn have cdf: - Mn=max (X1, X2,...,xn) Caf of Me is :- 7 Plon (m) = P(X, Em and 4 cm and ... Xem) Since all are independent e - P(Mr_m) = P(x, sm) . Plzem) .... P(Xn {m) Now, PC IP-11 E) = PL-E< Ma-1 < E) 6 8 Chec Hn C HE). = P(Mn Cl+E) - PC Mn 51-4) from CHEM - CILES Lt PC 1M2-11 > E) = ? - 0 palf of Me is given by differentiating with respect to m => tom 6) = o o .com E (Mn) = 5 mm m dm = [1-0] = 1 1 no 1 1 1 1 ntl Vor (Mn) = E(Mr) - Elma) BCM2) = ( nmn-ix m² dm = mx on mtl dem их n n+2 lt >) war Min = to Chebychef's Inequality states that p( {{ n = E(MA)) >E) & her Vor (Mn) .. Substiterting & Con, and vor come into above inequality Vor Chh & 2 - 8 Clan - 4 />E) = [ (1 toking lim nosso on both sides >> UPC). Ho hel>) lito-Go nso it PC I Mall > EE o Since probability is abrage 20 - it P (11)>() = 0 broved using thebyster's inequality nso Scanned with CamScanner