Question

In this problem, we study a simple noisy communication channel. Suppose that X is a binary signal that takes value -1 and 1 with equal probability. This signal X is sent through a noisy communication channel, and the medium of transmission adds an independent noise term. More precisely, the received signal is Y=X+N, where N is standard normal, indpendendent of X.

The decoder receives the value y of Y, and decides whether X was 1 or -1, using the following decoding rule: it decides in favor of 1 if and only if

\text {}\quad \mathbf{P}(X=1|Y=y)>2\mathbf{P}(X=-1|Y=y).

It turns out that the decoding rule can be expressed in the form: decide in favor of 1 if and only if Y>t, for some threshold t. Find the threshhold t.

As an intermediate step, find p_1\triangleq \mathbf{P}(X=1|Y=y).

Now find t.

Answer 1

To find the threshold t, we need to first find p_1, which is the probability that X=1 given Y=y.

Using Bayes' theorem, we can express p_1 as:

p_1 = P(X=1|Y=y) = (P(Y=y|X=1)*P(X=1)) / P(Y=y)

Since X is equally likely to be -1 or 1, we have P(X=1) = P(X=-1) = 0.5.

To find P(Y=y), we can use the law of total probability:

P(Y=y) = P(Y=y|X=1)*P(X=1) + P(Y=y|X=-1)*P(X=-1)

Since N is standard normal and independent of X, we have P(Y=y|X=1) = P(N=y-1) and P(Y=y|X=-1) = P(N=y+1). The probability density function of a standard normal distribution can be denoted as f(x), so we have:

P(Y=y) = f(y-1)*0.5 + f(y+1)*0.5

Now, substituting these values into the expression for p_1, we have:

p_1 = (P(Y=y|X=1)*P(X=1)) / P(Y=y)
= (P(N=y-1)*0.5) / (f(y-1)*0.5 + f(y+1)*0.5)
= P(N=y-1) / (f(y-1) + f(y+1))

Now, we need to find the threshold t such that p_1 > 2p_-1, where p_-1 = P(X=-1|Y=y).

Using Bayes' theorem, we can express p_-1 as:

p_-1 = P(X=-1|Y=y) = (P(Y=y|X=-1)*P(X=-1)) / P(Y=y)

Similarly as before, we have P(Y=y|X=-1) = P(N=y+1), so we have:

p_-1 = (P(N=y+1)*0.5) / (f(y-1)*0.5 + f(y+1)*0.5)
= P(N=y+1) / (f(y-1) + f(y+1))

Now, we want to find the threshold t such that p_1 > 2p_-1:

p_1 > 2p_-1
P(N=y-1) / (f(y-1) + f(y+1)) > 2*P(N=y+1) / (f(y-1) + f(y+1))
P(N=y-1) > 2P(N=y+1)
f(y-1) > 2f(y+1)

Now, we can solve for t by finding the value of y that satisfies this inequality. Since f(x) is the probability density function of a standard normal distribution, we can use standard tables or calculators to find the value of t.