Stochastic system exam
12/16/2019
ESE 503
Stochastic Systems
Final
1. Suppose you have just landed on JFK after making a tight connection
in Atlanta, GA, and you are waiting for your suitcase to appear on the
carousel. Once the carousel starts moving, all the luggage is delivered
within 10 minutes. Other passengers have started to get their bags,
but you keep waiting, and waiting, and waiting. Assume that the a
priori probability that the luggage made the connection is 1/2. Let
T = {1, 2, · · · , 10} be time elapsed in minutes once the luggage started being delivered. If the luggage is on the plane, it is delivered with
equal probability within any minute from 1 to 10. Five minutes passed
(T = 5), and your luggage has not shown up yet. What are the chances
that your luggage made the connection from Atlanta to JFK?
Solution: Let X = {0, 1} be the random variable denoting if the bag was not or was on the plain, respectively.
Let Y = {0, 1} be the random variable denoting if the bag was not or was on the carousel, respectively.
We want to find
P(X = 1|T = k,Y = 0) = P(Y = 0|T = k,X = 1)P(X = 1)
P(Y = 0|T = k)
We have
P(Y = 0|T = k) = P(Y = 0|T = k,X = 1)P(X = 1)
+ P(Y = 0|T = k,X = 0)P(X = 0)
We write
P(Y = 0|T = k,X = 1) = 10 −k
10
P(Y = 0|T = k,X = 0) = 1
Thus,
P(X = 1|T = k,Y = 0) = 10−k 10 × 1
2 10−k 10 × 1
2 + 1 × 1
2
= 10 −k 20 −k
F-1
When k = 5, we get
P(X = 1|T = 5,Y = 0) = 1
3
♦
2. The lifetime of a machine is modeled by a Weibull random variable X,
where
fX(x) = λx β−1e−λx
β/β, x ≥ 0
What is the pdf of the density of the lifetime of the machine if it is
still operating at time x = T, i.e., fX(x|X > T)?
Solution: We want to find
FX(x|X > T) = P(X ≤ x∩X > T)
P(X > T)
= FX(x) −FX(T)
1 −FX(T) , x > T
Since
FX(x) =
∫ x 0 fX(u)du = 1 −e−λx
β/β
we get
FX(x|X > T) = e−λT
β/β −e−λx β/β
e−λT β/β
, x > T
= 1 −eλT β/β−λxβ/β, x ≥ T
Clearly, the density of X is then
fX(x|X > T) = dFX(x|X > T)
dx
= λxβ−1e −λ
β (x β−Tβ), x ≥ T
♦
3. Let X1,X2, and X3 be three independent exponential random variables
with pdfs given by
fX1 (x) = λ1e −λ1x, x ≥ 0
fX2 (x) = λ2e −λ2x, x ≥ 0
fX3 (x) = λ3e −λ3x, x ≥ 0
We define the random variable Y = min(X1,X2,X3). Find the pdf of
Y , fY (y).
F-2
Solution:
FY (y) = P(min{X1,X2,X3}≤ Y )
= 1 −P(X1 > y,X2 > y,X3 > y)
= 1 −P(X1 > y)P(X2 > y)P(X3 > y)
= 1 −e−λ1ye−λ2ye−λ3y
= 1 −e−(λ1+λ2+λ3)y, y ≥ 0
For the pdf of Y we get
fY (y) = dFY (y)
dy
= (λ1 + λ2 + λ3)e −(λ1+λ2+λ3)y, y ≥ 0
♦
4. Let
fX,Y (x,y) =
{ 2e−xe−y, 0 ≤ y ≤ x < ∞ 0, otherwise
Find E(Y |x) and Var(Y |x).
[Note: ∫ xexdx = ex(x− 1) and
∫ x2exdx = ex(x2 − 2x + 2).]
Solution: We write
fY (y|x) = fX,Y (x,y)
fX(x)
where
fX(x) =
∫ ∞ 0
fX,Y (x,y)dy
=
∫ x 0
2e−xe−ydy
= 2e−x(1 −e−x), 0 ≤ x < ∞
and
fY (y|x) = 2e−xe−y
2e−x(1 −e−x)
= e−y
1 −e−x , 0 ≤ y ≤ x.
This conditional density is a truncated exponential.
F-3
Next we find E(Y |x) and write
E(Y |x) = ∫ x 0 y
e−y
1 −e−x dy
= 1
1 −e−x ( e−y(−1 −y)
)∣∣x 0
= 1 − xe−x
1 −e−x
where we used ∫ xe−xdx = e−x(−x− 1)
For the variance we use
Var(Y |x) = E(Y 2|x) −E ( Y |x
)2 Thus, now we find E(Y 2|x). We write
E(Y 2|x) = ∫ x 0 y2
e−y
1 −e−x dy
= 1
1 −e−x ( e−y(−y2 − 2y − 2)
)∣∣x 0
= 2 −e−x(x2 + 2x + 2)
1 −e−x
where we used ∫ x2e−xdx = e−x(−x2 − 2x− 2)
Finally,
Var(Y |x) = E(Y 2|x) −E ( Y |x
)2 =
2 −e−x(x2 + 2x + 2) 1 −e−x
− (
1 − xe−x
1 −e−x
)2 = 1 −
x2e−x
(1 −e−x)2
♦
5. Let the observation Zn be given by
Zn = Sn + Wn
where Sn is a sinusoid defined by Sn = a cos(ωn + Φ), where Φ is
a uniformly distributed random variable on [0, 2π) and a and ω are
constants. The process Wn is a noise process with autocorrelation
F-4
rw[k]. The processes Sn and Wn are independent processes. Find the
autocorrelation function of Zn. Is Zn a WSS process?
Solution: For the mean of Zn we have
E(Zn) = E(Sn) + E(Wn) = 0
= E ( a cos(ωn + Φ)
) + E(Wn)
=
∫ 2π 0
a cos(ωn + φ) 1
2π dφ + 0
= 0
and
E(Zn+kZn) = E ((Sn+k + Wn+k)(Sn + Wn))
E(Sn+kSn) + E(Sn+kWn) + E(Wn+k)(Sn) + E(Wn+k)Wn)
= E(Sn+kSn) + rw[k]
For E(Sn+kSn) we write
E(Sn+kSn) =
∫ 2π 0
a cos(ω(n + k) + φ)a cos(ωn + φ) 1
2π dφ
= a2
2π
∫ 2π 0
cos(ω(n + k) + φ) cos(ωn + φ)dφ
= a2
2π
∫ 2π 0
( cos(ω(2n + k) + 2φ)
2 +
cos(2πk)
2
) dφ
= a2
2π
( 0 +
2π cos(2πk)
2
) = a2
2 cos(2πk)
So the process Zn is WSS with autocorrelation function
rz[k] = a2
2 cos(2πk) + rw[k]
We note that Zn is a WSS process.
♦
F-5