Download Solutions

Statistics 116 - Fall 2004
Theory of Probability
Assignment # 9
Solutions
Q. 1)
(Ross # 7.51) The joint density of X and Y is given by
(
f(x; y) =
e y
y
0 < x < y; 0 < y < 1
otherwise.
0
Compute E[X 3 jY = y].
Solution:
Since fX jY (xjy) =
Ry
31
3
0 x y dx = y =4.
R ye
0 e
y =y
y =ydx
= y1 for 0 < x < y, then E[X 3 jY = y] =
Q. 2) (a) If X Poisson(), compute MX (t).
(b) Show that if Xn Binomial(n; =n) then MXn (t) converges to MX (t)
for every t, where X Poisson().
Solution:
(a)
MX (t) = E[etX ] =
(b)
MXn (t) = E[etXn ] =
1
X
k=0
etk
1
k
etk e = exp( + et )
k!
k=0
X
n k
lambda t
p (1 p)n k = (pet +1 p)n = (1+
(e 1))n
k
n
, which in the limit, when n goes to 1, gives us exp((et
Q. 3)
(Ross # 8.9)
1)).
If X is a Gamma random variable with parameters (n; 1).
Approximately how large need n be so that
X
P 1 > 0:01 < 0:01?
n
Answer:
1
(a) using Chebyshev's inequality;
(b) the Central Limit Theorem.
Solution:
(a) Since E[X=n] = 1, by Chebyshev inequality we have :
P fj
X
n
1j > 0:01g <
Var(X=n)
= 104 =n
(0:01)2
. Then 104 =n < 0:01 if and only if n > 106 .
P
(b) Let Xi Gamma(1; 1) for i = 1; :::n. Then X = ni=1 Xi Gamma(n; 1)
P fj
X
X + X2 + ::: + Xn
pn
1j > 0:01g = P fj 1
n
n
j > 0:01png = P fjzj > 0:01png
.
By central limit theorem z has standard
normal distribution. So,
p
look at the table to nd that 0:01 n > 2:58, therefore n > (258)2 '
66564.
Q. 4)
(Ross # 8.11) Many people believe that the daily change of price of a
company's stock on the stock market is a random variable with mean 0
and variance 2 . That is, if Yn represents the price of the stock on the
n-th day, then
Yn = Yn 1 + Xn ;
n1
where X1 ; X2 ; : : : ; are independent and identically distributed random
variables with mean 0 and variance 2 . Suppose that the stock's price
today is 100. If 2 = 1, what can you say about the probability that
the stock's price will exceed 105 after 10 days, without using the Central
Limit Theorem?
Solution:
Note, that Y10 Y0 = X1 + X2 + ::: + X10 , and E[Y10 Y0 ] = 0 as well as
var(Y10 Y0 ) = 10. Moreover, Y10 Y0 has normal distribution with the
mean and variance above.
You can solve this problem by three dierent ways now.
1. Using the fact of standard normal distribution :
P fY10
Y0 > 5g = P
Y10
Y0
5
>p
10
10
p
2. Using ONE sided Chebyshev inequality :
P fY10
Y0 > 5g 2
2
2
= ;
2
2
+a
7
' 0:057:
where 2 is the variance of Y10
sided Chebyshev inequality :
P fY10
Y0 and equals to 10. 3. Using TWO
Y0 > 5g P fjY10
=
3
Var
10
:
25
Y0 j > 5g
P10
i=1 Xi
25