Developing an optimal rule based on probability distribution (with solution) 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.

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$.
