Q.7.73 - A First Course in Probability

Question

In Example 6b, let $S$ denote the signal sent and $R$ the signal received.

(a) Compute $E\left[R\right].$

(b) Compute $Var(R )$

(c) Is $R$ normally distributed?

(d) Compute $Cov(R, S)$

Step-by-Step Solution

Answer

a) The computation of $E\left[R\right]$ is $\mu$.

b) The computation of $Var\left(R\right)$ is ${\sigma }^2+1$

c) Yes, $R$ is normally distributed

d) The computation of $Cov(R,S)$ is ${\sigma }^2.$

Step 1: Given Information (Part a)

$S=$Signal Sent

$R=$Signal received

$E\left[R\right]=?$

Step 2: Explanation (Part a)

From the information, observe that the signal is sent from point $A$According to the distribution $S~N\left(\mu ,{\sigma }^2\right)$

While it gets to the point $B$, the little error $\epsilon$ has been made to the original signal.

So, the received signal $S$ can be written as $R=S+\epsilon$

Where, $\epsilon ~N(0,1)$

Signal that is sent is independent from the error.

Calculate $E\left(R\right),$

$E\left(R\right)=E\left(S\right)+E\left(\epsilon \right)$

$=\mu +0$

$=\mu$

Step 3: Final Answer (Part a)

Hence, the value of $E\left[R\right]$is $\mu .$

Step 1: Given Information (Part b)

$S=$Signal sent

$R=$Signal received

$Var \left(R\right)=?$

Step 2: Explanation (Part b)

Calculate variance:

$V\left(R\right)=V\left(S\right)+V\left(\epsilon \right)$

$={\sigma }^2+1$

Step 3: Final Answer (Part b)

Hence, the value of $Var\left(R\right)$ is ${\sigma }^2+1$

Step 1: Given Information (Part c)

$S=$Signal Sent

$R=$Signal received

Step 2: Explanation (Part c)

Yes.

$R$ is normally distributed.

Since $R$ can 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:

$N~N\left(\mu ,{\sigma }^2+1\right)$

Step 3: Final Answer (Part c)

Therefore,$R$ is normally distributed.

Step 1: Given Information (Part d)

$S=$Signal sent

$R=$Signal received

$Cov(R,S)=?$

Step 2: Explanation (Part d)

Calculate

$\begin{array}{r}=\mathrm{Var}\left(S\right)+\mathrm{Cov}(\epsilon ,S)\end{array}$

$=\mathrm{Var}\left(S\right)$

$={\sigma }^2$

Step 3: Final Answer (Part d)

Therefore,$Cov(R,S)$ is ${\sigma }^2.$