Question

Alice is trying to transmit to Bob the answer to a yes-no question, using a noisy channel. She encodes "yes" as 1 and "no" as 0, and sends the appropriate value. However, the channel adds noise; specifically, Bob receives what Alice sends plus a N(0,σ2) noise term (the noise is independent of what Alice sends). If Bob receives a value greater than 1/2 he interprets it as "yes"; otherwise, he interprets it as "no". (a) Find the probability that Bob understands Alice correctly. (b) What happens to the result from (a) if σ is very small? What about if σ is very large? Explain intuitively why the results in these extreme cases make sense.

Answer 1

Answer:

(a) Find the probability that Bob understands Alice correctly.

The probability that Alice will send 1 and Bob will receive it correctly is P, and the probability that she sends a 0 and Bob receives it correctly is 1 – p. Bob will receive the 1 correctly only if the noise is not below -½, and he will receive the 0 correctly only if the is not above ½.  

Therefore, the probability of receiving both messages correctly is:

P = P(Noise ≤ ½) x (1 – p) + P(Noise ≥ -½) x p

Since the probability of the noise is normally distributed, then:

P(Noise ≤ ½)  = P(Noise ≥ -½)  

Therefore,  

P(Noise ≤ ½)  = P(N/σ ≤ ½σ) = ½σ

(b) What happens to the result from (a) if σ is very small? What about if σ is very large? Explain intuitively why the results in these extreme cases make sense.

Think about this case intuitively, if the noise level is very low and σ is low, then the probability of understanding the message correctly is very high. If σ is low, then ½σ will be high. On the other hand, if the noise level is very high and σ is high, then the probability of understanding the message correctly will be much lower (½σ will be low). Bob will probably be guessing if its a 1 or a 0.

Just imagine when you are trying to talk with someone and there is a lot of noise, you will not be able to understand all the words correctly if the noise level is too high.

Answer 2

The required probability is $P(Bob\ interpreted\ right) = \phi (0.5/ \sigma)$

Probability:

It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

Let A is the code Alice sent and B is the code Bob received.

We have $B = A + E$  where E is the error term with $N(0,\sigma^2)$

So, $B|(A=0)\sim N(0,\sigma^2)$ and $Y|(A=1)\simN(1,\sigma^2)$

$P(Bob\ interpreted\ right)= P(Alice\ sent 0) \times P(Bob\ Received\ 0) + P ( Alice\ sent\ 1) \times P( Bob\ received\ 1)$

Let the probability that Alice sent 0 be p so $P(X=0) = p$ and $P(X=1) = 1-p$

$P(Bob\ interpreted\ right)= P (B\le 0.5|A= 0)P(A=0) + P(B > 0.5|B = 1)P(A=1)$

Let's calculate $P (B\le 0.5|A = 0); as\ B |(A = 0) \sim N(0, \sigma^2 )$

So, $P (B\le 0.5|A = 0)$ will be calculated by $Z –value = \frac{0.5}{\sigma}$

So, $P (B\le 0.5|A = 0) = \phi (0.5/ \sigma)$

So, $P(Bob\ interpreted\ right) = p \times \phi (0.5/ \sigma) + (1-p) \times \phi (0.5/ \sigma) \\= \phi (0.5/ \sigma)$

Learn more about the topic Probability:

https://brainly.com/question/6649771