Download Exercise 1. Let N1 and N2 be two independent Poisson distributed

Exercise 1.
Let N1 and N2 be two independent Poisson distributed random variables. Where N1 ∼
P ois(λ1 ) and N1 ∼ P ois(λ1 ).
a) Compute the expectation of N1
b) Compute V ar(N1 ) = E(N12 ) − E(N1 )2
c) Show that N1 + N2 ∼ P ois(λ1 + λ2 )
Solution.
∞
∞
X
X
λk−1
λk1 −λ1
1
= λ1
a) E(N1 ) =
e−λ1
k e
k!
(k
−
1)!
k=1
k=0
= λ1
∞
X
λk
1 −λ1
k=0
b) E(N12 ) =
∞
X
=
=
λ21
e
= λ1 .
∞
k2
k=0
λ21
k!
λk
λk1 −λ1 X
e
(k(k − 1) + k) 1 e−λ1
=
k!
k!
k=1
∞
∞
X
X
λk−1
λk−2
1
1
−λ1
e
e−λ1
+ λ1
(k
−
2)!
(k
−
1)!
k=1
k=2
∞
X
λk
1 −λ1
k=0
k!
e
+ λ1
∞
X
λk
1 −λ1
k=0
k!
e
= λ21 + λ1 .
Then the variance V ar(N1 ) = λ21 + λ1 − λ21 = λ1
c) P (N1 + N2 = n) =
∞
X
P (N1 = k, N2 = n − k) =
k=−∞
n
X
λk1 −λ1
∞
X
P (N1 = k)P (N2 = n − k)
k=0
λ2n−k −λ2
=
e
e
k!
(n − k)!
k=0
n e−(λ1 +λ2 ) X n k n−k
λ1 λ2
=
n!
k
k=0
=
(λ1 + λ2 )n −(λ1 +λ2 )
e
, for n ≥ 0
n!
For n < 0, P (N1 + N2 = n) = 0. Hence N1 + N2 ∼ P ois(λ1 + λ2 )
1
Exercise 2.
X1 , . . . , Xn are independent expentially distributed random variables, with Xi ∼ exp(λ).
P
a) Compute the expectation of X1 b) Deduce that nk=1 Xk ∼ Gamma(n, λ),
given that the density of Gamma(n, λ) distributed random variable is
λn
xn−1 e−λx 1(0,∞) (x)
f (x) =
(n − 1)!
(Hint: use induction.)
Solution.
a) The expectation
∞
Z
E(X) =
xλe
−λx
dx = −xe
∞ Z
+
−λx 0
0
∞
e−λx dx
0
e−λx ∞ 1
= 0+
= .
λ 0
λ
and
Z
2
∞
E(X ) =
2
−λx
x λe
∞ 2 Z ∞
xλe−λx dx
+
λ 0
0
2 −λx dx = −x e
0
2
.
λ2
=
Hence for the variance we have V ar(X1 ) =
b) We use induction for this proof,
2
λ2
−
1
λ2
=
1
.
λ2
n = 1 : it holds that Gamma(1, λ) = Exp(λ).
n → n + 1 : After the induction step
n
X
X1 ∼ Gamma(n, λ)
k=1
then
Pn+1
Pn
Xk + Xn+1 and the density
Z ∞
fX1 +...+Xn+1 (x) =
fX1 +...+Xn (t)fXn+1 (x − t)
−∞
Z x
λn
=
tn−1 e−λt λe−λ(x−t)
(n
−
1)!
0
Z x n−1
t
λn+1 n
−λx n+1
= e λ
dt = e−λx
x .
n!
0 (n − 1)!
k=1
Xk =
k=1
Exercise 3.
A calculator is operated by two different batteries, it works well as long as both batteries
function properly. The life span X of battery A is normally distributed with µ = 30 hours
and σ = 3 hours. The life span Y of battery B is normally distributed with µ = 35 hours
and σ = 4 hours. X and Y are independent random variables.
2
a) What is the probability the battery A’s life span lies between 15 and 20
hours?
b) With probability 0.853141, What is the maximum life span of battery B?
c) With what probability does the calculator still work after 24 hours?
d) What is the probability that the life span of battery A is shorter than that
of battery B
Solution.
Z ∼ N (0, 1) with distribution function Φ
a)
27 − 30 i
3
3
= Φ(−1) − Φ(−5) = Φ(5) − Φ(1)
= 1 − 0.841345 = 0.158655
P [15 ≤ X ≤ 27] = P
h 15 − 30
≤Z≤
b)
h
a − 35 i
P [Y ≤ a] = 0.853141 → P Z ≤
= 0.853141
4
a − 35
= →
= 1.05
4
= a = 1.05 · 4 + 35 = 39.2
c)
P [{X > 24} ∩ {Y > 24}] = P [X > 24] · P [Y > 24]
h
24 − 35 i
24 − 30 i h
·P Y >
= P X>
3
4
= 0.974338
d) P [X < Y ] = P [X − Y < 0] =?
X − Y ∼ N (−5, 25), P [X < Y ] = P [X − Y < 0] = 0.841345.
Exercise 4.
A random variable X with density
f (x) =
a
x3
0
for x ≥ 1
else
for some a > 0.
a) Determine a such that f is a density function.
b) Determine the density of the random variable 3X + 2.
c) Determine the density of X 2 .
d) Compute the expectation of 3X + 2, (i) with the density of X, (ii) with
3
the results in b).
e) Compute the expectation of X 2 , (i) with the density of X, (ii) with the
results in c).
Solution.
a) For f a density function, it holds that
Z
Z ∞
f (x)dx =
1=
−∞
∞
1
a
a ∞ a
dx
=
−
=
x3
2x2 1
2
Hence a = 2
b) From the script, the density is given by
2
1 x − 2
= 1[5,∞] (x)
f3X+2 (x) = fX
3
3
3
3 · x−2
3
Therefore f3X+2 (x) =
18
1
(x)
(x−2)3 [5,∞]
c) For the density of X 2
1
1 2
fX 2 (x) = 1[1,∞] √ 3/2 = 2 1[1,∞] (x)
x
2 xx
d) With the density of X,
Z
∞
E(3X + 2) =
1
2
(3X + 2) 3 dx = 2 +
x
∞
Z
1
6
dx = 8.
x2
With the density of 3X + 2
Z
E(3X + 2) =
5
∞
18
y
dy = 2 +
(y − 2)3
e) With the density of X,
∞
Z
2
x2
E(X ) =
1
With the density of X
Z
∞
5
18
dy = 8
(y − 2)2
2
dx = ∞.
x3
2
2
Z
E(X ) =
1
∞
y
dy = ∞
y2
Exercise 5.
You are addicted to the windows card game spider. You must work on a stochastic exercise
sheet, but first you want to play the game a little longer. As compromise you plan to stop
after the first victory. You win a game with probability p.
a) Compute the expected value of the number of games.
b) What is the variance of the number of games.
4
Solution.
X = The number of games is a geometrically distributed random variable. Therefore
∞
X
f (x) =
xk =
k=0
1
for |x| < 1 and
1−x
∞
X
1
f (x) =
=
kxk−1
(1 − x)2
k=1
0
∞
X
2
=
k(k − 1)xk−2
f (x) =
3
(1 − x)
k=2
00
a) The expectation is
E(X) =
∞
X
kp(1 − p)k−1 =
k=1
p
1
=
(1 − (1 − p))2
p
b) We have that
2
E(X ) =
∞
X
k 2 p(1 − p)k−1
k=1
= p(1 − p)
∞
X
k−2
k(k − 1)(1 − p)
+p
∞
X
k(1 − p)k−1
k=1
k=2
2p(1 − p)
p
+
3
(1 − (1 − p))
(1 − (1 − p))2
2(1 − p) + p
=
p2
=
hence the variance
V ar(X) =
2−p
1
1−p
− 2 =
.
2
p
p
p2
5