課程名稱︰機率導論
課程性質︰必修
課程教師︰陳 宏
開課學院：理學院
開課系所︰數學系
考試日期（年月日）︰2013/06/13
考試時限（分鐘）：1:20-2:10pm
試題 :
Introductory Probability
Quiz 4
Thursday 1:20-2:10pm, June 13th, 2013
1. (25 points) Suppose X_1,…,X_n are independent and have an exponential
distribution with parameter λ. Let M_n = max{X_1,…,X_n}. Show that if
a < λ < b then P(M_n ≦ aln(n)) → 0 and P(M_n ≦ bln(n)) → 1.
Solution: Note that
P(M_n≦alnn) = (1 - x/n^a)^n = exp(nln(1-n^-a)),
both n^-a and n^-b are between 0 and 1/2 as n large enough, and
-x - x^2 < ln(1 - x) < -x when 0 < x < 1/2.
We conclude the proof.
2. (25 points) Suppose X and Y are independent and uniform on (0,1). Show that
Z = XY has density function f(z) = ln(1/z) for 0 < z < 1 and 0 otherwise.
Solution: Let U = X and Z = XY. The Jacobian of this transform is
│ 1 Y │
│ │
│ 0 X │
We have
f_U,Z(u,z) = 1/x ˙ f_X(x)f_Y(z/x)
which leads to
1
f_Z(z) = ∫ 1/x dx = -lnz
z
3. (25 points) Let X_1,X_2,… denote independent and identically distributed
random variables with a distribution function F(x). Define Y_i = 1,
when X_i ≦ x_0. Otherwise, Y_i = 0. Note that F(x) is continuous at x_0.
n
(a) (8 points) Describe the distribution function of Σ Y_i.
i=1
n
Solution: P(Y_i=1)=F(x_0). Σ Y_i is a binomial distribution Bin(n,F(x_0)).
i=1
n
(b) (7 points) Show that Σ Y_i/n converges to F(x_0) in probability.
i=1
Solution: It follows from the law of large numbers.
(c) (10 points) When n goes to infinity, show that
n
(√n)˙(Σ Y_i/n - F(x_0)) converges to a normal distribution N(0,σ^2).
i=1
Solution: It follows from the central limit theorem and
σ^2 = F(x_0)(1 - F(x_0)).
4. (25 points) Suppose X and Y have joint density
f(x,y) = (y^2/4)exp(-x), if x > 0 and -x < y < x.
(a) (13 points) Are X and Y independent?
(b) (12 points) Are X and Y uncorrelated?
Solution: First, plot the region determined by x>0 and -x<y<x which is a region
falling in the first and fourth quadrants with boundary formed by y=x and y=-x.
We then have
x
f_X(x) = ∫(y^2/4)exp(-x)dy = (x^3/6)exp(-x), x > 0.
-x
∞
f_Y(y) = ∫(y^2/4)exp(-x)dx = (y^2/4)exp(-|y|), -∞ < y < ∞
|y|
Since f_X,Y(x,y) ≠ f_X(x)f_Y(y), they are not independent.
Obsevre that E[XY] = 0, E[X] = 4, and E[Y] = 0. X and Y are uncorrelated.
5. (25 points) Note that the moment generating function for a Poisson random
variable with λ is M_X(t) = exp(λ(e^t - 1)).
(a) (10 points) Prove that the sum of two independent Poisson random
variables with λ = 1 is a Poisson random variable with λ = 2 using
moment generating function.
Solution: Let X_1 and X_2 denote those two independent random variables.
Hence, M_X1(t) = exp(e^t - 1) and M_X2(t) = exp(e^t - 1). Since X_1 and X_2
are indenpendent, we have
M_X1+X2(t) = M_X1(t)M_X2(t) = exp(2(e^t - 1)).
We conclude that X_1 + X_2 is also a Poisson random variable with λ = 2.
(b) (15 points) Suppose Y = Poisson(100). Use central limit theorem to
provide an estimate of P(85≦X≦115). 請說明理由。你可以使用(a)來說明。
Solution: By the argument in proving (a), we can write Y as the sum of
X_1,X_2,…,X_100. Note that E(X_i) = 1 and Var(X_i) = 1 and write
100
P(85≦X≦115) = P(85≦Σ X_i≦118).
i=1
By the central limit theorem, the distribution of (√100)[Y/(100-1)] is
close to a standard normal random variable Z. Hence, P(85≦Y≦115) can then
be approximated by
100
P(√100 [0.85-1]≦√100 [(Σ X_i/100)] - 1≦√100[1.15-1])
i=1
≒ P(-1.5≦Z≦1.5) = 0.93319 - 0.06681 = 0.86638.
If you use continuity correlation, we should work on P(84.5≦Y≦115.5)
instead.