A binary communication system is used to send one of two messages: message A is a sequence of 0s with probability 2/3, message B is a sequence of 1s with probability 2/3,
The ith received bit is “correct" with probability 3/4, and is “incorrect" with probability 1/4. We assume that conditioned on any specific message sent, the received bits, denoted by Y1,Y2,… are independent. There are a few questions:
1) Find the probability that the first and second received bits are the same.
Because P(Yi=0)=7/12, P(Yi=1)=5/12. My answer is 7/12*7/12+5/12*5/12, but it is wrong, where did I make the mistake?
2)Given that Y1,…,Y5 were all equal to 0, what is the probability that Y6 is also zero?
Because all Yis are independent, does it mean P(Y1=0)=P(Y6=0)?
3)Is Y2+Y3 independent of Y1? Is Y2−Y3 independent of Y1?
I can't figure out the relationship between them. Sometime, I think I misunderstand the meaning of independence.
4) Find the mean of K, where K=min{i:Yi=1} is the index of the first bit that is 1.
E=1/p=12/5, wrong answer again, really stuck here.:(
There are too many parts to cover in a single post but here is part 1:
1) Let the probability of message A be $p_A$ and message B be $p_B=1-p_A$, and the probability of reading a bit correctly be $p_c$ and incorrectly $p_i=1-p_c$. Then the probability of the first two bits reading the same is: $$\begin{aligned} p({\text{same)}}&=p_A(p_cp_c+p_ip_i)+p_B(p_cp_c+p_ip_i)\\ &=(2/3)\left[(3/4)^2+((1/4)^2 \right]+(1/3)\left[(3/4)^2+(1/4)^2 \right]\\ &=(3/4)^2+(1/4)^2=5/8=0.625 \end{aligned}$$
Hint for part 2: Use Bayes theorem to compute the posterior probability of a message being A given that the first five bits are read all zero (and so also one minus this is the posterior probability that the message is B). Then use this to compute the probability that the sixth bit will be read as a zero.
Hint for part 3: Independence of Z=Y2+Y3 and Y1 means that the joint pmf of Z and Y1 is equal to the product of the pmf of Z and the pmf of Y1. I used to think that you did not need to work this out explicitly it should be obviously true or false, but its not.
Hint for part 4: Given that the message is A the probability that the $k$th first 1 is $p(K=k|A)=p_c^{k-1}p_i$, and similarly if the message is B we have $p(K=k|B)=p_i^{k-1}p_c$.