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Statistics 116 - Fall 2004
Theory of Probability
Alternate Final Exam, December 7th, 2004
Solutions
Instructions: Answer Q. 1-6. All questions have equal weight. Bonus worth equivalent of one half question. The exam is open book. In addition, you are allowed a maximum of 3 pages of handwritten notes.
Q. 1) Let X ∼ Geom(p) be a Geometric random variable. Show that X has the following memoryless property: For any integers n and m, with m < n
P (X > n|X > m) = P (X > n − m).
Solution:
P (X > n) = P (first n trials were failures) = (1 − p)n. Therefore, for n > m
P (X > n|X > m) =
P (X > n, X > m) P (X > m)
=
P (X > n) P (X > m)
=
(1 − p)n (1 − p)m = (1 − p)n−m = P (X > n − m).
Q. 2) The joint density function of X and Y is given by
f (x, y) =
e−(x−y) (^2) / 2 y √ 2 πy
y^2 e−y 2
, −∞ < x < ∞, y ≥ 0.
(a) Find the conditional density fX|Y (x|y) of X given Y = x. (b) Compute Var(X|Y ). (c) Compute E(X|Y ). (d) Compute Var(X). Solution:
(a) By inspection, it is not hard to see that
fY (y) =
y^2 e−y 2 , y ≥ 0
and fX|Y (x|y) = e−(x−y)
(^2) / 2 y √ 2 πy is a Normal density with mean y and variance y. Or,
X|Y = y ∼ N (y, y).
(b) As the conditional distribution is N (Y, Y )
Var(X|Y ) = Y.
(c) Similarly, E(X|Y ) = Y. (d) Var(X) = E(Var(X|Y )) + Var(E(X|Y )) = E(Y ) + Var(Y ) = E(Y ) + E(Y 2 ) − E(Y )^2
E(Y ) =
0
y^3 e−y 2
=
E(Y 2 ) =
0
y^4 e−y 2
=
Therefore, Var(X) = 3 + 12 − 32 = 6.
Q. 4) Given a sequence (X 1 ,... , Xn) of n independent, identically distributed random variables we say a record occured at time i if Xi ≥ Xj , 1 ≤ j ≤ i − 1. That is, there is a record at time i, if, at time i, Xi is the “record” largest of the first i elements of the sequence.
(a) For 1 ≤ i ≤ n, compute pi = P (Xi is a record) (b) Let Rn be the total number of records of the sequence (X 1 ,... , Xn). Compute E(Rn).
Solution:
(a)
P (Xi is a record) = P (Xi ≥ X 1 , Xi ≥ X 2 ,... , Xi ≥ Xi− 1 ).
This just says that if we order the X’s from 1 to i then Xi is the last one. As there are i! such orderings and (i − 1)! orderings with Xi the last entry, P (Xi is a record) =
i
(b) Let
Yi =
1 Xi is a record 0otherwise.
Then,
Rn =
∑^ n
i=
Yi
and
E(Rn) =
∑^ n
i=
E(Yi) =
∑^ n
i=
P (Xi is a record) =
∑^ n
i=
i
Q. 5) A model for the movement of a stock supposes that if the present price of the stock is s, then after one time period it will be either u × s with probability p or d × s with probability 1 − p. Assuming that successive movements are independent, approximate the probability that the stock’s price will be up at least 30 percent after the next 1000 time periods if u = 1. 012 , d = 0.990 and p = 0. 51. (Hint: log(ab) = log(a) + log(b)). Solution: Define the random variables
Xi =
1 if stock rises at i-th time period 0 otherwise.
Then, the Xi’s are independent with
P (Xi = 1) = p
and the price of the stock Sn at the n-th time period is
Sn = S 0 u
Pn i=1 Xi^ dn−
Pn i=1 Xi^ = S 0 dn^
( (^) u d
)Pni=1 Xi .
The event that the stock has risen 30 percent over the next 1000 time periods is { S 1000 S 0
u d
)P^1000 i=1 Xi d^1000 ≥ 1. 3
log(u/d)
i=
Xi + 1000 log d ≥ log(1.3)
i=
Xi ≥
−1000 log(d) + log(1.3) log(u/d)
√^ i=1^ (Xi^ −^ p) 1000
p(1 − p)
−1000 log(d) − 1000 p log(u/d) + log(1.3) log(u/d)
p(1 − p)
But
i=1 Xi^ ∼^ Binom(1000,^0 .51), so, by the Central Limit Theorem, or the Normal approximation to the Binomial,
P
S 1000
S 0
−1000 log(d) − 1000 p log(u/d) + log(1.3) log(u/d)
p(1 − p)
