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, independent of X.

Question

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
P(X=1|Y=y)>2P(X=0|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 p1≜P(X=1|Y=y).

p1= unanswered

Now find t.

t= unanswered

Answer 1

p1=(0.5)*(1/sqrt(2*pi))*exp(-x^2/2)

Answer 2

p1 = 0.5 / (1 + e^-2x)

Answer 3

p1 = 1/(1 + e^(-2*y)) <-- answer should be in terms of y not x

Answer 4

t?
I would try solving P(X=1|Y=t)>2P(X=0|Y=t)
t>(ln2)/2

Answer 5

1. (1/2)*e^((2*y-1)/2) I don't understand the other answer given here, but I've got this
2. ln(2)/2