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In Example 6b, let Sdenote the signal sent and Rthe signal received.

(a) ComputeE[R].

(b) ComputeVar(R)

(c) Is Rnormally distributed?

(d) ComputeCov(R,S)

Short Answer

Expert verified

a) The computation of E[R]is μ.

b) The computation of Var(R)is σ2+1

c) Yes, Ris normally distributed

d) The computation ofCov(R,S)isσ2.

Step by step solution

01

Given Information (Part a)

S=Signal Sent

R=Signal received

E[R]=?

02

Explanation (Part a) 

From the information, observe that the signal is sent from point AAccording to the distribution S~Nμ,σ2

While it gets to the point B, the little error εhas been made to the original signal.

So, the received signal Scan be written as R=S+ε

Where, ε~N(0,1)

Signal that is sent is independent from the error.

Calculate E(R),

E(R)=E(S)+E(ε)

=μ+0

=μ

03

Final Answer (Part a)

Hence, the value ofE[R]isμ.

04

Given Information (Part b)

S=Signal sent

R=Signal received

Var(R)=?

05

Explanation (Part b)

Calculate variance:

V(R)=V(S)+V(ε)

role="math" localid="1647434029229" =σ2+1

06

Final Answer (Part b)

Hence, the value ofVar(R)isσ2+1

07

Given Information (Part c)

S=Signal Sent

R=Signal received

08

Explanation (Part c) 

Yes.

Ris normally distributed.

Since Rcan be written was the sum of two independent normally distributed random variables.

Hence, the sum is also normally distributed random variables with parameters are as follows:

N~Nμ,σ2+1

09

Final Answer (Part c)

Therefore,Ris normally distributed.

10

Given Information (Part d)

S=Signal sent

R=Signal received

Cov(R,S)=?

11

Explanation (Part d)

Calculate

=Var(S)+Cov(ε,S)

role="math" localid="1647434711595" =Var(S)

=σ2

12

Final Answer (Part d)

Therefore,Cov(R,S)is σ2.

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Most popular questions from this chapter

Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point, either you can accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope? Let Aand B, A<B, denote the (unknown) amounts of the checks and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of (A+B)/2. Consider the following strategy: Let F(·)be any strictly increasing (that is, continuous) distribution function. Choose an envelope randomly and open it. If the discovered check has the value x, then accept it with probability F(x)and exchange it with probability 1F(x).

(a) Show that if you employ the latter strategy, then your expected return is greater than (A+B)/2. Hint: Condition on whether the first envelope has the value Aor B. Now consider the strategy that fixes a value x and then accepts the first check if its value is greater than x and exchanges it otherwise. (b) Show that for any x, the expected return under thex-strategy is always at least (A+B)/2and that it is strictly larger than (A+B)/2if xlies between Aand B.

(c) Let X be a continuous random variable on the whole line, and consider the following strategy: Generate the value ofX, and if X=x, then employ the x-strategy of part (b). Show that the expected return under this strategy is greater than (A+B)/2.

7.4. If X and Y have joint density function fX,Y(x,y)={1/y,if0<y<1,0<x<y0,otherwisefind

(a) E[X Y]

(b) E[X]

(c) E[Y]

Let X1,X2,,Xn be independent and identically distributed positive random variables. For kn findEi=1kXii=1nXi.

Let X1,...be independent random variables with the common distribution functionF, and suppose they are independent of N, a geometric random variable with a parameter p. Let M=max(X1,...,XN).

(a) FindP{Mx}by conditioning onN.

(b) FindP{Mx|N=1}.

(c) FindP{Mx|N>1}

(d) Use (b) and (c) to rederive the probability you found in (a)

If 101items are distributed among 10 boxes, then at least one of the boxes must contain more than 10 items. Use the probabilistic method to prove this result.

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