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I'm having some trouble understanding my professor's notes from how he went about solving this problem.

It is known that an FM radio station transmits a signal between 6 AM and 10 PM, and is switched off at night (i.e., between 10 PM to 6 AM).

Alice chooses a time instant that she selects uniformly randomly over a 24 hour period (i.e., continuous uniform pdf over a consecutive 24 hour interval), and asks Bob to determine whether the time instant she chose is in the DAY (i.e., between 6 AM – 10 PM) or NIGHT.

To aid Bob, she hands him a wireless signal detector tuned to the FM radio station that outputs a single voltage level as follows: It is known that when the FM station is transmitting, it appears as a 1 Volt signal at the detector in the absence of noise. When it is not transmitting, the signal detector reads 0 Volts (also in the absence of noise). However, it is known that there is noise in the channel (atmosphere) between the radio transmitter and the signal detector. This noise voltage can be modeled as a Gaussian random variable with ’0’ mean and variance equal to 1, and that this noise is additive. In other words, the detector receives the sum of the FM station voltage and the noise voltage.

Using the above description, determine a decision rule that Bob can employ in order to determine whether the time is DAY or NIGHT, such that the decision rule he uses minimizes the probability of error of his decision. (You do not need to compute the probability of error - you need to provide the optimal rule).

Solution: Let S denote whether it is Day or Night and R dnote the received signal, N denote the noise. Given R=x, the optimal rule would be to find the bigger value between

$$\begin{eqnarray} P(S = Day|R = x)\ and\ P(S = Night|R = x) \\ P(S = Day|R = x) = P(S = Night|R = x) \\ P(S = Day, R = x) = P(S = Night, R = x)\\ P(S = Day, 1 + N = x) = P(S = Night, N = x)\\ P(S = Day, N = x − 1) = P(S = Night, N = x)\\ x = −0.193\end{eqnarray} $$

My issue is that I'm not sure how he went from the second to last step to saying that the optimal rule is at x = -0.193.

Thanks a lot.

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up vote 2 down vote accepted

Since the day is twice as long as the night, we are in effect looking for $x$ such that the density of Gauss at $x$ is twice the density at $1-x$. Note that $$ \frac{e^{-x^2/2}}{e^{-(1-x)^2/2}}=e^{(1-2x)/2}$$ thus $ x = \frac12 - \ln 2$.

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Can you explain real fast how the expression on the left side of your equation was derived? After you have the left side you simplify to the right side and then set it equal to 2 in order to get $$x=\frac{1}{2} - ln2$$ Am I understanding that correctly? – Alex Ionescu Dec 12 '12 at 23:30
@Alex: Yes, that's correct. The left-hand side is the ratio of the density of the Gaussian noise at $x$ (the value it would need to have in case of night) and its density at $x-1$ (the value it would need to have in case of day). – joriki Dec 12 '12 at 23:54
@Hagen: I think this would be clearer if you wrote $x-1$ instead of $1-x$. – joriki Dec 12 '12 at 23:54

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