Solutions of Homework 3-520.

651

September 29, 2006

1 Problem 1-Stark and Woods(3.11)

Let us compute the cdf of Y Let Fx (x) = P (X ≤ x) be the cdf of X

Fy (y) = P (Y ≤ y) = P (a/X ≤ y) = P (X ≥ a/y) = 1 − Fx (a/y)

So we have the pdf of Y to be

dFy (y) a α a/α

fy (y) = = −Fx0 (a/y) ∗ −a/y 2 = afx (a/y)/y 2 = 2 2 2
=
dy πy α + (a/y) π((a/α)2 + y 2 )

hence fy (y) is again a Cauchy distribution with parameter a/α

2 Problem 2-Stark and Woods-3.23

Since X1 , X2 . . . Xn are independent and
Z ∞
P (X > x) = e−y dy = e−x
x

n
Y n
Y
1 − P (Z ≤ z) = P (Z > z) = P (min(X1 , X2 . . . Xn > z) = P (Xi > z) = e−z = e−nz
k=1 k=1

which means
Fz (z) = 1 − e−nz

or the pdf fy (y) is given by

fz (z) = Fz0 (z) = ne−nz

The plot for n=3 will look as above

1
2

3 Problem 3-Stark and Woods-4.18

3.1 Part a
Clearly X(ζ) will have distribution over the sample space −1, − 21 , 0, 12 , 1 and Y (ζ) = X(ζ)2 will have
sample space 0, 21 , 1 with distribution 15 , 25 , 25 respectively In order to show that they are independent,it
is enough to show that P (Y = y|X = x) 6= P (Y = y) for some value of Y.ie we need to provide a
counter example

1
P (Y = 0) =
5
But

P (Y = 0|X = 0) = 1
because if X is 0,then we know for sure that Y is also zero as Y = X 2 .
Hence they are dependent random variables

3.2 Part b
XY = ζ 3 which has a distribution 51 , 15 , 15 , 15 , 15 over the sample space (−1, − 81 , 0, 18 , 1)

−1 −1 1 11 1
E(XY ) = E(ζ 3 ) = + +0++ + =0
5 8 5 85 5
and
−1 −1 1 11 1
E(X) =
+ +0++ + =0
5 2 5 25 5
So E(XY ) = E(X)E(Y ) = 0 which means that ρ(X, Y ) = 0 which means that X and Y are uncorel-
lated

4 Problem 4-4.36 in Stark and Woods

The charecteristic function of a poisson distribution with parameter λ is given by
∞
X e−λ λx t t
E(etX ) = etx = e−λ eλe = eλ(e −1)
x=0
x!

Since X and Y are independent rv’s,we have that

t
−1) λ2 (et −1) t
−1)
E(et(X+Y ) ) = E(etX )E(etY ) = eλ1 (e e = e(λ1 +λ2 )(e

Since characteristic functions have unique inverse,we have the distribution of z to be(and noting that
λ1 = 2 and λ = 3) So λ1 + λ2 = 5
1
PZ (k) = e−5 5k
k!

5 Problem 5
5.1 Part a
Z +∞
λ λ
E(X) = x e−λ|x| dx
2 −∞ 2
Since x λ2 e−λ|x| is an odd function and
Z +∞
λ
x eλ|x| dx
0 2
 3

exists, we have that E(X) = 0 The mean is defined whenever the probability distribution is defined.The
probability distribution is defined when λ > 0

5.2 Part b
We need to compute Z ∞
x
dx
−∞ x2 + λ2
Again consider the integral
a
a2 + λ2
Z  
x 1
lim = log = − log(n)
a→∞ −na x2 + λ2 2 n2 a2 + λ2
which means the integral goes to different values for different limits of infinity
So the mean is not well defined.

5.3 Part c
Since
n
X
f (y) = (1 + y)n = n
Ck y k
k=0

we have that
n
X
f 0 (y) = n(1 + y)n−1 = n
Ck ky k−1
k=1
λ
Setting y = 1−λ ,we have
 n−1 Xn  k
λ λ n λ
n 1+ = Ck k
1−λ 1−λ 1−λ
k=1

which means
or
n
X
n
nλ = Ck kλk (1 − λ)n−k
k=1

So we have
n
X
n
n Ck kλk (1 − λ)n−k = nλ
k=1

and we are required to compute the LHS which is nλ hence the mean is nλ

5.4 Part d
We are required to compute
Z ∞ Z ∞
1 Γ(k + 1) k!
E(Y k ) = λ y k λe−λy dy = z k e−z dz = = k
0 λk 0 λ k λ