Probability of receiving a signal You have a signal receiver that is able to receive two signals, 1 and 2. Both signals have mean 0 and arrive with equal frequency. Signal 1 is normally distributed with variance 4, and signal 2 is normally distributed with variance 9. You receive a signal with magnitude 2. What is the probability that the signal you received is signal 1?
Do I solve this problem by finding the sd of signal 1 as $\sqrt{4}=2$, that a signal with magnitude 2 is $\frac{2-0}{2} = 1$ standard deviation away from the mean, and the probability that that occurs is the answer? I'm also confused on how to incorporate the distribution of signal 2 in my answer.
 A: Let $S$ denote the signal ($S \in \{1,2\})$, and $X$ the observed value. Let $N(x;\mu,\sigma^2)$ be the probability mass of a normal distribution with mean $\mu$ and variance $\sigma^2$.
By Bayes rule, and in the second line the Rule of Total Probability:
$$
\begin{align*}
\mathbf P[S = 1 | X = 2] & =
\frac{\mathbf P[X= 2 | S = 1] \mathbf P[S = 1]}{ \mathbf P[X=2]} \\
& = \frac{\mathbf P[X= 2 | S = 1] \mathbf P[S = 1]}{ \frac12 \big(\mathbf P[X=2|S=1] + \mathbf P[X=2|S=2]\big)} \\
& = \frac{N(2; 0, 4) \, \times\, \frac12 }{\frac12 \big(N(2; 0,4) + N(2;0,9)\big)} \\
& = \frac{ N(2; 0, 4) }{N(2; 0, 4)  + N(2; 0, 9) } \\
& \approx \frac{ 0.1210}{ 0.1210  + 0.1065 }  \\
& \approx 0.5319
\end{align*}
$$
A: There are a few ways to tackle this problem. Nevertheless, one important issue is that the magnitude of the observation is 2. That is, $|x|=2$, which does NOT distribute Gaussian. In fact, for a zero-mean Gaussian random variable $x$ with variance $\sigma^2$, the distribution of its magnitude $y=|x|$ is the half-normal distribution
$$p_Y(y)=\begin{cases}\displaystyle{\frac{\sqrt{2}}{\sqrt{\pi}\sigma}\exp\bigg(-\frac{y^2}{2\sigma^2}\bigg)}&y\geq0\\0&\text{otherwise.}\end{cases}
$$
The second issue is that the random variables are continuous, and then $\mathbb{P}(X=a)=0$ for any $a$, and also $\mathbb{P}(|X|=a)=0$. What we will do is the following first-order approximation:
$$
\mathbb{P}(X=a)=\mathbb{P}(X\leq a+\delta) - \mathbb{P}(X\leq a) \approx p_X(a)\cdot\delta
$$
which comes from the fact that the pdf is the derivative of the CDF, used for expressions such as $\mathbb{P}(X\leq a)$, and where we assume $\delta$ is small.
Hence, we now compute the probability that transmitted signal $S$ was signal 1, given that the observed magnitude is $Y=2$. Knowing that signals are transmitted with equal frequency,
$$
\begin{align}
\mathbb{P}(S=1|Y=2)&=\frac{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)}{\mathbb{P}(Y=2)}\quad\text{(Bayes' rule)}\\&=\frac{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)}{\mathbb{P}(Y=2|S=1)\mathbb{P}(S=1)+\mathbb{P}(Y=2|S=2)\mathbb{P}(S=2)}\quad\text{(total prob.)}\\
&=\frac{\mathbb{P}(Y=2|S=1)}{\mathbb{P}(Y=2|S=1)+\mathbb{P}(Y=2|S=2)}\quad\text{(equal frequency of signals)}\\
&\approx \frac{p_Y(2|S=1)}{p_Y(2|S=1)+p_Y(2|S=2)}\quad\text{(approx. and $\delta$ vanishes in ratio)}\\
&=\frac{\frac{\sqrt{2}}{\sqrt{\pi}2}\exp\Big(-\frac{2^2}{2\cdot 4}\Big)}{\frac{\sqrt{2}}{\sqrt{\pi}2}\exp\Big(-\frac{2^2}{2\cdot 4}\Big)+\frac{\sqrt{2}}{\sqrt{\pi}3}\exp\Big(-\frac{2^2}{2\cdot 9}\Big)}\quad\text{(half-normal distributions)}\\
&=0.531878531904...
\end{align}
$$
