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),nN} as

X(n)=Y1+Y2++Yn,for all nN.
Find μX(n) and RX(m,n), for all n,mN.

  • Solution
    • We have
      μX(n)=E[X(n)]=E[Y1+Y2++Yn]=E[Y1]+E[Y2]++E[Yn]=0.
      Let mn, 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 mn, we have
      RX(m,n)=E[X(m)X(n)]=4n.
      We conclude
      RX(m,n)=4min(m,n).


Problem

For any kZ, define the function gk(t) as

gk(t)={1k<tk+10otherwise
Now, consider the continuous-time random process {X(t),tR} 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,tR.

  • Solution
    • Note that, for any kZ, g(t)=0 outside of the interval (k,k+1]. Thus, if k<tk+1, we can write
      X(t)=Ak.
      Thus,
      μX(t)=E[X(t)]=E[Ak]=1.
      So, μX(t)=1 for all tR. Now consider two real numbers s and t. If for some kZ, we have
      k<s,tk+1,
      then
      RX(s,t)=E[X(s)X(t)]=E[Ak2]=1+1=2.
      On the other hand, if s and t are in two different subintervals of R, that is if
      k<sk+1,andl<tl+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,tk+1,
      then
      CX(s,t)=RX(s,t)E[X(s)]E[X(t)]=211=1.
      On the other hand, if
      k<sk+1,andl<tl+1,
      where k and l are two different integers, then
      CX(s,t)=RX(s,t)E[X(s)]E[X(t)]=111=0.


Problem

Let X(t) be a continuous-time WSS process with mean μX=1 and

RX(τ)={3|τ|2τ21otherwise

  1. Find the expected power in X(t).
  2. Find E[(X(1)+X(2)+X(3))2].

  • Solution
      1. The expected power in X(t) at time t is E[X(t)2], which is given by
        RX(0)=3.
      2. 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)=33+22+21+22=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)=t2tX(u)du.

  1. Find μY(t)=E[Y(t)].
  2. Find RXY(t1,t2).

  • Solution
      1. We have
        μY(t)=E[t2tX(u)du]=t2tE[X(u)]du=t2t0du=0.
      2. We have
        RXY(t1,t2)=E[X(t1)t22t2X(u)du]=E[t22t2X(t1)X(u)du]=t22t2RX(t1u)du=t22t2δ(t1u)du={1t22<t1<t20otherwise


Problem

Let X(t) be a Gaussian process with μX(t)=t, and RX(t1,t2)=1+2t1t2, for all t,t1,t2R. 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)]=21+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+2111=2.
      Var(X(2))=E[X(2)2]E[X(2)]2=RX(2,2)μX(2)2=1+2224=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+21212=3.
      Therefore,
      Var(Y)=42+5+43=25.
      We conclude YN(4,25). Thus,
      P(Y<3)=Φ(345)=Φ(0.2)0.42




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