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02_Exam_Solution

Course: ORIE 3500, Fall 2008
School: Cornell
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2 Exam - Solution ORIE 3500/5500: Engineering Probability and Statistics II Question 1 (i) (a) Conditional PDF of X given Y fX|Y (x|y) is a function with fX|Y (x|y) 0 and a fX|Y (x|y)dx = 1 such that P (X a|Y = y) = fX|Y (x|y)dx. Also fX|Y (x|y) = fX,Y (x, y) , fY (y) where fX,Y is the joint PDF of X and Y and fY is the marginal PDF of Y . (b) X and Y will be independent if fX,Y (x, y) = fX (x)fY (y), where...

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2 Exam - Solution ORIE 3500/5500: Engineering Probability and Statistics II Question 1 (i) (a) Conditional PDF of X given Y fX|Y (x|y) is a function with fX|Y (x|y) 0 and a fX|Y (x|y)dx = 1 such that P (X a|Y = y) = fX|Y (x|y)dx. Also fX|Y (x|y) = fX,Y (x, y) , fY (y) where fX,Y is the joint PDF of X and Y and fY is the marginal PDF of Y . (b) X and Y will be independent if fX,Y (x, y) = fX (x)fY (y), where fX,Y is the joint PDF of X and Y and fX , fY are the marginal PDF of X and Y. (c) Moment-generating function (transform) of a random variable X is MX (s) = E esX . (d) Moment-generating function (transform) of two random variables X and Y is a function MX,Y (s1 , s2 ) such that MX,Y (s1 , s2 ) = E es1 X+s2 Y . (ii) (a) MY (s) = E esY = E es(a+bX) = E esa esbX = esa MX (bs). (b) MX (s) = E[esX ]. So d2 d2 sX d2 MX (s) = E e = E X 2 esX E(X 2 ) = 2 MX (s)|s=0 . 2 2 ds ds ds Question 2 (i) (a) If X P oisson() then pX (k) = e k /k! for k 0. So MX (s) = k=0 esk e k /k! k=0 =e (es )k /k! = exp((es 1)). (b) If X has Normal distribution with parameters R and 2 > 0, then X = + Z 2 where Z N (0, 1). fZ (z) = ez /2 / 2. So MZ (s) = esz ez 2 /2 / 2dz / 2dz = =e s2 /2 e(zs) . 2 /2+s2 /2 So using the result of 1(ii)(a) MX (s) = es MZ (s) = es+ (ii) Let FX , FY be the CDFs of X and Y respectively. Then FY (y) = P (Y y) = P + (aX b y) = P and hence fY (y) = FY (y) = (iii) P (X A, Y B) = {xA,yB} 2 s2 /2 . X yb a = FX yb a , 1 F aX yb a = 1 fX a yb a . fX.Y (x, y)dxdy fX (x)fY (y)dxdy {xA,yB} = = {xA} fX (s)dx {yB} fY (y)dy = P (X A)P (Y B) (iv) Let W = X + Y . Using independence MZ (s) = MX (s)MY (s). Using 2(i)(b) MX (s) = eX s+X s So MZ (s) = e(X +Y )s+(X +Y )s 2 2 2 /2 2 2 /2 , MY (s) = eY s+Y s 2 2 /2 . . Question 3 (i) It is true. The joint density is fX,Y (x, y) = So fX|Y (x|y) = fX,Y (x, y) = fY (y) 1 r 2 0 if x2 + y 2 r2 . otherwise 1/r2 , (1/r2 )dx which is a constant as a function of x. Hence the conditional distribution is uniform. 2 (ii) It is true. The PDF of X is fX (x) = ex /2 / 2. Since Y = eX , X = ln Y and fY (y) = fX (ln y) d(ln y) 1 2 = e(ln y) /2 . dy 2y 2 Question 4 (i) If T1 is the required time to send the rst message and L1 is its length, then l+2 l+2 E(T1 |L1 = l) = l tfT1 |L1 (t|l)dt = (t/2)dt = [(l + 2)2 l2 ]/4 = l + 1. l So E(T1 ) = E[E(T1 |L1 )] = E[L1 + 1] = 0 (l + 1)fL1 (l)dl (l + 1)2e2l dl = 0 e2l e2l dl + 1 |0 + 2 0 e2l =0+ | + 1 = 3/2. 2 0 = 2l (ii) T = T1 + +TN . So E(T |N = n) = E(T1 + Tn ) = nE(T1 ) = 3n/2. So E(T |N ) = 3N/2. Hence E(T ) = E[E(T |N )] = E[3N/2] = n=0 (3n/2)pN (n) (3n/2) n=0 = e100 100n n! e100 100n1 (n 1)! = 150. = 100(3/2) = 150 m=0 n=1 e100 100m m! 3
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