Question

(a) For the random variable X, show that E[(x – a)?] is minimized when a = E(X). (b) For random variables X and Y, show that Var(X + Y) < Var(x) + Var(Y), that is, the standard deviation of the sum is less than or equal to the sum of standard deviations. (c) For random variables X and Y, prove the Cauchy-Schwartz Inequality: [E(XY)]? 5 E(X2) E(Y2)

Answer 1

solution: - Let, 'X' Rardom Verichte & E (x-a)" - ECX-ECX) + F(X)-a) 25(X-EC%)*+$ECW-a3* + 2[{ ecx)-ECx}(600) a)] = 'ECX-E(x)} + {ECO-a??. 7 6(x-ECx) as (EC) a), 70 clearly "ECx-a) is minimum, When a - E(X). 5 For two Rivis xey to We know that, S V(x+y) – VCX) +vCy) + 2 COUC Xsy) To show : Jux+y) < Jux) + Jucy). v(x+y)s V(x) + VCV)+ a jucovcy) as variance can't Note that, or be negative. PCX = 2 ES ) cov (x,y) s JVCEDUCy) = V(x+y) = V(x): + VCy) +2.CoVCX,Y) VVO) VOUS ! P

su(x) + VCY) +a Juvcx) VcY JVCx+y) s JVX) + Suci) Hence proved. E (Qx+ By)? 70 ,H LEIR, 1. PER | = 4 x) + ECY) +24 8 ECxy) 20 Also, Elax- py)? 7,0 ,Hd,BER da ECx2) + B EC 12) -20% Ecxy) 70 Now, &? - EC42) { ß2 EC42) :: 2xß ECx4) 77-202282 2&B ECxy) s 2d782 Dividend by ddf, - ECA. JECV) s ECxy) < JECE) JECA) ☆ JE(XV)) JECX) SECV) [EC«»]= EC®) EC1) Hence proved.