Question

A random variable Y has a uniform distribution over the interval (0,, e,). Derive the variance of Y 1 Find E(Y2 in terms of (0, 02). 1' E(Y)2 Find E(Y2) in terms of (e, 02) ECY2) = Find V(Y) in terms of (01, 02). V(Y) =

Answer 1

of O, 2 O2 a random, voriable y is said to have a continuous uniform probability distribution on the interval (0,02) Yf and only if the density function of Y is f (g) = 0x0 C o elsewhere Given f (y) is a continuous . random variable y. To find the variance of first let us calculate the Ely), and it () : The expected value of a continuous random variable y is Ely) = 9 yf cy) dy; provided the integral exists E (Y) : 9 gf (y) dy : < (9) = $ y Cydy + 3yf ()dy i ut cydy EM) • you to get a betonj dy s I vody

E (9)= 0 + (sten] gydy to EM) [es-e ] ecv) = () (034) E1%) = (10.) (-6)(0 +)] E CY) = ECY) = (10,40] 1 2 Now to calculate. Elys, E03 . 5 y2 + (y) dy, provided the integral exists E (12) ġ yaf cy) dy ' E (72) = $y* f(9)dyt jaf95dy+ 4°464)dy E ) • I wody i oda)dyt [ stody E () = 0+ com} $ g?dy to E (92) = (0,4 0, E (Y2) dy to : :

of Y is given by = (y) = 60-6] (430) = 4*) = (aa) (1950) Cele carejo d)]. - E 689 = {*+ (%) (a) + )..... The variance VCY) = E (Y-M)2 ViY) - € (32) - [E (Y)}} vw = (2+ (0) (0) + 8)]. Contoya vey) = (0** (9) C6)+8)]- [6 10:32 ] vM) = {edt (@ytotox)]-(63+224 20:01] 3 3 4 NLY) = [4 (02 + (0) (01)+03)-3(03 +0% +20,0)) 12 V (Y) = [(4 02 + 4 (01)(0) + 403) -30% +30° +60,00) 12

v (Y) = -(403 - 303 +4 (0)(01)-6020 +407-303)7 12 VIY) - 92) (0) VIV) = (163 +260) (0+ 8? = 10:0"] 12