Question

Suppose X is a random variable that has density function f(x) = (1/2)e^−|x| for −∞ < x < ∞. Find:
(a) (2 pts) P(X < 10).
(b) (4 pts) The c.d.f. of X2.
(c) (4 pts) V ar(X)

Answer 1

exo(u oo Soln 1 c 20 f(x) = exp(-1x1) 0 P(x <10) s fral du ja 4 0301-mo) din Sh £ (exp (2)] + Į (-exp(-ni] (1) + (1 1 - exp(-10)) exp(-10) exp(x) du t I exp(-2) dn 20 - TI - MIN I 2 - 7 - -N COF of x is given by Fy (1) = P(x sx) There is two condition i If x co then n P( x5n) s £ exp(-1*1) du n - j 1/2 exp(x) du n こ 1 2 1 exp (n) 2

(ii) If no, then Fx (n = P(x <n) U I exp(-121) dn 2 1 s exp (2) du + & J exp(-aidh 2 e + = + [ exp hilo = 1 (1) + 4 1 - 1 explan) Į (-exp (on)] - PAP (-31) こ = CDF of X us given by 1 Fx (11 = exp(x) nco I exp(-1) no var (8) = Eck) - (E(X))? E(X) = (a fyldu Sov 2 n & exp (-141) da 2 8 - a exp(a) dn tl a expl-adr 2 8

Let I s n exp(n) du expcns 10 [ n exp(n) 1 Iz = n exp(-") du exp (en)] [-sexpen) 1 +(*) (-1) co ㅋ E(X) 1 (2) = Simill as illarly E(+²) = 2 こ th expt allah 0 x exp(a) dn + 2 Let is merecede { x' expanda [n exp(n) - 2 (ne4P(u) + PxPcull ] Il = - 00 = +2 Iz sxexplanlan (-a' ey P(-n) + 2 (-4 exp (-) - ppp (-) (x²) 2 - 02= 2 =) こ =2 Yon (8) = 2 Ans