LÖSNINGAR TILL TENTAMEN I TAMS32 STOKASTISKA
PROCESSER, FREDAG 8 JANUARI 2016 KL 08.00-12.00.
b2 =
1. Since CX (0) = 2, CX (1) = e−1 , and CX (2) = 0, we get that X
T
T
b
a (X1 , X0 ) , where
−1 −1 −1 1
e
e
2
−e−1
2 e−1
=
=
b
a = −1
−1
−2
0
0
−e
2
e
2
4−e
−1 1
0.19
2e
=
.
≈
−0.035
4 − e−2 −e−2
2. (a)
T
 

0.285
0.34 0.18 0.48
p(2)T = pT P 2 = 0.5 0.3 0.2 0.15 0.49 0.36 = 0.257 ,
0.458
0.35 0.10 0.55
so P (X = 2) = 0.458.
(b)
P (X1 = 1|X2 = 2, X0 = 0) =
=
P (X2 = 2, X1 = 1, X0 = 0)
P (X2 = 2|X0 = 0)P (X0 = 0)
p0 P0,1 P1,2
0.2 · 0.3
1
=
= .
2
p0 P0,2
0.48
8
(c) The chain is irreducible and aperiodic. The unique stationary and
asymptotic distribution π = (π1 , π2 , π3 )T is the solution to the system
π T = π T P,
π1 + π2 + π3 = 1.
, 10 , 24 )T ≈ (0.3061, 0.2041, 0.4898)T .
The solution is: π = ( 15
49 49 49
3. (a) Taking the variance on both sides of the equation and using the
fact that Yt and Xt are independent gives:
V (Y1 ) = (−0.6)2 V (Y0 ) + V (X0 ) = 0.36V (Y0 ) + 1.
Since V (Y1 ) = V (Y0 ), we get: V (Y1 ) =
1
1−0.36
=
25
16
= 1.5625.
(b)
P (Y12
√
> 6) = 1 − P (− 6 ≤ Y1 ≤
≈ 1 − Φ(
√
√
√
6
6
Y1
6) = 1 − P (−
≤
≤
)
5/4
5/4
5/4
2.4495
2.4495
2.4495
) + Φ(−
) = 2 − 2Φ(
) ≈ 0.050.
1.25
1.25
1.25
4. E(Yt ) = E(Xt Xt−1 ) = E(Xt )E(Xt−1 ) = 0, for each t ∈ Z.
RY (t, τ ) = E(Yt Yt+τ ) = E(Xt Xt−1 Xt+τ Xt+τ −1 ) for each t ∈ Z. We see
2
2
that RY (t, 0) = E(Xt2 Xt−1
) = E(Xt2 )E(Xt−1
) = 9, that RY (t, 1) =
2
2
E(Xt−1 Xt Xt+1 ) = E(Xt−1 )E(Xt )E(Xt+1 ) = 0, and that RY (t, −1) =
2
2
E(Xt−1
Xt Xt−2 ) = E(Xt−1
)E(Xt )E(Xt−2 ) = 0. For all other τ ∈ Z,
RY (t, τ ) = E(Xt−1 )E(Xt )E(Xt+τ )E(Xt+τ −1 ) = 0. Since neither the
expectation nor the autocorrelation function depends on t, the process
is wide sense stationary.
5. Using the ”Formel-och tabellsamling”, the frequency response of the
LTI is
1
H(f ) =
∀f ∈ R,
3 + i2πf
which gives:
SY (f ) = |H(f )|2 SX (f ) =
2
9 + 4π 2 f 2
∀f ∈ R.
From this we obtain the autocorrelation function of the output signal:
1
RY (τ ) = e−3|τ |
3
∀τ ∈ R.
Since the output signal is wide sense stationary, and the autocorrelation
function is continuous at τ = 0 (in fact, continuous everywhere), the
output signal is mean square continuous.
6. (a) Since B(t) ∼ N(0,
each t ≥ 0, we get: φB(T
E(esB(T ) ) =
R ∞t) forsB(T
R ∞) (s) =sB(t)
sB(T )
)
−t
E(E(e
|T )) = 0 E(e
|T = t)e dt = 0 E(e
)e−t dt =
R ∞ ts2 /2 −t
2
e
e dt, where the integral converges if and only if s2 < 1, or
0
√
equivalently if |s| < 2. In this case, we obtain:
φB(T ) (s) =
t=∞
−1
2
2
e−t(1−s /2) t=0 =
2
1 − s /2
2 − s2
∀|s| <
√
2.
(b) Using the ”Formel-och tabellsamling” (and the uniqueness
theorem
√
for mgf’s), it can be seen that B(T ) has a Laplace( 2, 0) distribution,
with pdf
√
1
fB(T ) (x) = √ e− 2|x|
∀x ∈ R.
2