10.1.6 Solved Problems
Problem
Let Y1, Y2, Y3, ⋯ be a sequence of i.i.d. random variables with mean EYi=0 and Var(Yi)=4. Define the discrete-time random process {X(n),n∈N} as
X(n)=Y1+Y2+⋯+Yn,for all n∈N.
Find
μX(n) and
RX(m,n), for all
n,m∈N.
- Solution
-
We have
μX(n)=E[X(n)]=E[Y1+Y2+⋯+Yn]=E[Y1]+E[Y2]+⋯+E[Yn]=0.
Let m≤n, then
RX(m,n)=E[X(m)X(n)]=E[X(m)(X(m)+Ym+1+Ym+2+⋯+Yn)]=E[X(m)2]+E[X(m)]E[Ym+1+Ym+2+⋯+Yn]=E[X(m)2]+0=Var(X(m))=Var(Y1)+Var(Y2)+⋯+Var(Ym)=4m.
Similarly, for m≥n, we have
RX(m,n)=E[X(m)X(n)]=4n.
We conclude
RX(m,n)=4min(m,n).
Problem
For any k∈Z, define the function gk(t) as
gk(t)=⎧⎩⎨⎪⎪10k<t≤k+1otherwise
Now, consider the continuous-time random process
{X(t),t∈R} defined as
X(t)=∑k=−∞+∞Akgk(t),
where
A1,
A2,
⋯ are i.i.d. random variables with
EAk=1 and
Var(Ak)=1. Find
μX(t),
RX(s,t), and
CX(s,t) for all
s,t∈R.
- Solution
-
Note that, for any k∈Z, g(t)=0 outside of the interval (k,k+1]. Thus, if k<t≤k+1, we can write
X(t)=Ak.
Thus,
μX(t)=E[X(t)]=E[Ak]=1.
So, μX(t)=1 for all t∈R.
Now consider two real numbers s and t. If for some k∈Z, we have
k<s,t≤k+1,
then
RX(s,t)=E[X(s)X(t)]=E[A2k]=1+1=2.
On the other hand, if s and t are in two different subintervals of R, that is if
k<s≤k+1,andl<t≤l+1,
where k and l are two different integers, then
RX(s,t)=E[X(s)X(t)]=E[AkAl]=E[Ak]E[Al]=1.
To find CX(s,t), note that if \
k<s,t≤k+1,
then
CX(s,t)=RX(s,t)−E[X(s)]E[X(t)]=2−1⋅1=1.
On the other hand, if
k<s≤k+1,andl<t≤l+1,
where k and l are two different integers, then
CX(s,t)=RX(s,t)−E[X(s)]E[X(t)]=1−1⋅1=0.
Problem
Let X(t) be a continuous-time WSS process with mean μX=1 and
RX(τ)=⎧⎩⎨⎪⎪3−|τ|1−2≤τ≤2otherwise
- Find the expected power in X(t).
- Find E[(X(1)+X(2)+X(3))2].
- Solution
-
- The expected power in X(t) at time t is E[X(t)2], which is given by
RX(0)=3.
- We have
E[(X(1)+X(2)+X(3))2]=E[X(1)2+X(2)2+X(3)2+2X(1)X(2)+2X(1)X(3)+2X(2)X(3)]=3RX(0)+2RX(−1)+2RX(−2)+2RX(−1)=3⋅3+2⋅2+2⋅1+2⋅2=19.
Problem
Let X(t) be a continuous-time WSS process with mean μX=0 and
RX(τ)=δ(τ),
where
δ(τ) is the Dirac delta function.
We define the random process
Y(t) as
Y(t)=∫tt−2X(u)du.
- Find μY(t)=E[Y(t)].
- Find RXY(t1,t2).
- Solution
-
- We have
μY(t)=E[∫tt−2X(u)du]=∫tt−2E[X(u)]du=∫tt−20du=0.
- We have
RXY(t1,t2)=E[X(t1)∫t2t2−2X(u)du]=E[∫t2t2−2X(t1)X(u)du]=∫t2t2−2RX(t1−u)du=∫t2t2−2δ(t1−u)du=⎧⎩⎨⎪⎪10t2−2<t1<t2otherwise
Problem
Let X(t) be a Gaussian process with μX(t)=t, and RX(t1,t2)=1+2t1t2, for all t,t1,t2∈R. Find P(2X(1)+X(2)<3).
- Solution
-
Let Y=2X(1)+X(2). Then, Y is a normal random variable. We have
EY=2E[X(1)]+E[X(2)]=2⋅1+2=4.
Var(Y)=4Var(X(1))+Var(X(2))+4Cov(X(1),X(2)).
Note that
Var(X(1))=E[X(1)2]−E[X(1)]2=RX(1,1)−μX(1)2=1+2⋅1⋅1−1=2.
Var(X(2))=E[X(2)2]−E[X(2)]2=RX(2,2)−μX(2)2=1+2⋅2⋅2−4=5.
Cov(X(1),X(2))=E[X(1)X(2)]−E[X(1)]E[X(2)]=RX(1,2)−μX(1)μX(2)=1+2⋅1⋅2−1⋅2=3.
Therefore,
Var(Y)=4⋅2+5+4⋅3=25.
We conclude Y∼N(4,25). Thus,
P(Y<3)=Φ(3−45)=Φ(−0.2)≈0.42
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Practical uncertainty: Useful Ideas in Decision-Making, Risk, Randomness, & AI
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