Question

13

atoms, each of which decays by emission of an a-particle after an exponentially distributed lifetime with rate 1, independently of the others. Let T, be the time of the first a-particle emission, T, the time of the second. Find: a) the distribution of T: b) the conditional distribution of T2 given T: c) the distribution of T2. 13. Let X and Y be independent random variables, X with uniform distribution on (0,3), Y with Poisson (1) distribution. Find: a) a formula in terms of X for P(X<Y); b) the conditional density of X given X < Y, and sketch its graph in the cases X = 1,2,3 c) E(X|X<Y). 14. Bayesian sufficiency. Let Sn X + + X be the number of successes in a sequence of n independent Bernoulli (p) trials X1 X2...., X, with unknown success probability p. Regard p as the value of a random variable II whose prior distribution has some density f(p) on (0,1). Show that the conditional (posterior) distribution of Il given X = 2,..., X, = In, for any particular sequence of zeros antil ones I 1..... In with I] + + In = k, depends only the observed number of successes k in the n trials, and not on the order in which the k successes and n-k failures appear. Deduce that this conditional distribution is identical to the posterior distribution of II given Sn = k. (This is another expression of the fact that S. is a sufficient statistic for p. See

Answer 1

o Habrwise 03 13 Xnu (0, 3) = Pdf of is fx (2) = { t , if oca23 otherwise Yr Poisson (x) =) redpre pmf of t us P (Icy) = { end x>0, y = 0, 1,- , otherwise P(X<Y) = P(«<y]Y=y) «P(= y) = * S dy elemento ocy <3 g 11 { ocy<3 37-1)! otherwise

Let CDF of xlxCY = F then, F(x) x <Y) = P(x Lx 1x2Y) = P(xLx, XLY) P(x &x, X&Y) P(Y=y) P (XLY) P(x<%) PP(xLY) is abrody been calculated which doesn't depends on 2. F = P(x 2 min (x,y)) P (Y=y) = 5 7 X P(4=y) P(xLY) P(xLY) OCXL1,970 ICX<2,5=1 Yz xetx P(XLY) /* P(Y=y), KX<2, J 1,0 P(XLY) Y/3x8 (7-8) 12(243, y=1,2 P(XCY) */3 x P (Y=y) , 2<x<3, 3+1,2,0 P(XLY) o otw / b. fxixcy (x laxy) = s P(124) 3 P(x<y) ocx<1, 4t0 , 1<x<2, Y + 1,0 21863, y EQ, 1, 2 n 0 1otw P(7-4) 3 P(xLY) , if 0<x<y <3 O otw 1 -:E(x1x2Y) = xxP(Y=y) da P(XRY) y²xP(Y=y) 2P(X<Y) y xe x x 3 (4-1)! 2Y! Xexxx 13 osy<3