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Statistics 116 - Fall 2004
Theory of Probability
Assignment # 8
Solution
show (and briefly explain) all of
your work
Q. 1)
(Ross # 6.39) Choose a number at random from the set of numbers
f1; 2; 3; 4; 5g. Now choose a number at random from the subset f1; : : : ; X g.
Call this second number Y .
(a) Find the joint p.m.f. of X and Y .
(b) Find the conditional mass function of X given that Y = i. Do it for
i = 1; 2; 3; 4; 5.
(c) Are X and Y independent? Why?
Solution:
(a) P fX = j; Y = ig = 15 1j .
(b) P fX = j jY = ig = P51=51j=5k , for i j
k=i
5.
(c) No, because the distribution of random variable Y depends on the
value of random variable X .
Q. 2)
(Ross # 6.43) The joint density of X and Y is
y )e x ;
f (x; y) = c(x
2
2
0 x < 1; x y x
Find the conditional distribution of Y given X = x, i.e. nd fY jX (y j!x).
Solution:
fY jX (yjx) = x3 (x y ), x < y < x
FY jX (yjx) = x3 (x y y =3 + 2x =3), x < y < x.
3
2
2
4
3
2
3
3
4
1
Q. 3)
(Ross # 7.3) If X and Y are independent uniform (0,1) random variables,
show that
E (jX
Solution:
E (jX
Y j ) =
Z
1
0
RR
1
0
jx
2
;
( + 1)( + 2)
Y j ) =
1
0
jx yj dydx.
yj dy =
=
Z
Zx
!
x
0
jx
u du +
0
= (x+1 + (1
Hence
E (jX
Q. 4)
Y j ) =
1
a+1
Z
1
yj dy +
Z
1
x
> 0:
Z
1
x
jx yj dy =
u du =
0
x) )=(a + 1):
+1
(x+1 + (1
x) )dx =
+1
0
2
:
(a + 1)(a + 2)
Ross # 7.25) Let X ; X ; : : : be a sequence of independent and identi1
2
cally distributed continuous random variables. Dene the random variable
N 2 as the rst point in which the sequence (X1 ; X2 ; : : : ) stops decreasing, i.e. if N = n, then
X
1
X X Xn < Xn :
2
3
1
(a) Compute the p.m.f. of N by rst computing P (N
(b) Show that E (N ) = e.
n).
(a) P fN ng = P fX1 X2 ::: Xn g = n1!
1 1
(b) E [N ] = 1
n=1 P fN ng = n=1 n! = e
P
Q. 5)
P
(Ross # 7.40) The joint density function of X and Y is given by
1
f (x; y) = e y x=y ;
y
( +
)
x > 0; y > 0:
(a) Find E (X ):
(b) Find E (Y ):
(c) Show that Cov(X; Y ) = 1:
R
(a) fY (y ) = e y y1 e x=y dx = e y .
(b) In addition, the conditional distribution of X given Y = y is exponential with mean y . Hence, E [Y ] = 1, E [X ] = E [E [X jY ]] = E [Y ] = 1.
(c) Since,
E [XY ] = E [E [XY jY ]] = E [Y E [X jY ]] = E [Y ] = 2
(the last equality follows because Y is exponential with mean 1).
Hence, Cov(X; Y ) = 2 1 = 1.
2
2
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