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