How to use Bayes' Theorem to detect an event in a noisy signal I'm trying to use Bayes' Theorem to solve a question that's come up in work, but I don't know if I've done it correctly, because the result seems a bit strange.
The problem involves a stochastic variable $v$ part of which can be modelled as a Gaussian with mean $\mu$ and standard deviation $\sigma$.
There is an event $E$ which may occur with probability $p$ and which adds a value $\Delta$ to $v$. Therefore, the model for $v$ is:
$$
v \sim N(\mu, \sigma) + Bernoulli(p).\Delta
$$
What I want to know is given $v = x$, what is the probability that the event occurred?
Using Bayes' Theorem:
$$
P(E \mid v=x) = \frac{P(E)P(v=x \mid E)}{P(v=x)}
$$
and filling in, we get:
$$
P(E \mid v=x) = \frac{pN(\mu, \sigma)(x-\Delta)}{p(N(\mu, \sigma)(x-\Delta)+\Delta) + (1-p)N(\mu,\sigma)(x)}.
$$
However, I'm not sure this is correct, since for large $x$, the probability seems to go to zero, where I would expect it to approach one. (For very large measured $x$, it should be much more likely than not that $E$ occurred, shouldn't it?)
I would greatly appreciate any help or pointers, or even a confirmation that this is the correct way to apply Bayes' Theorem.

 A: Your model is
$$V=W+E\Delta \, ,\quad W\sim\mathrm{N}(\mu,\sigma) 
\, ,\quad E\sim\mathrm{Bern}(p)$$
To simplify the formulas, we can introduce standardized variables
$$X= \frac{V-\mu}{\sigma} \, ,\quad Z=\frac{W-\mu}{\sigma} \, ,\quad d = \frac{\Delta}{\sigma}$$
so that the model becomes
$$X=Z+Ed \, , \quad Z\sim\mathrm{N}(0,1)$$
where $E$ is the same as before.
Then Bayes rule gives
\begin{align}
P(E=1 \mid X=x) &= \frac{P(X=x \mid E=1)P(E=1)}{P(X=x \mid E=1)P(E=1)+P(X=x \mid E=0)P(E=0)}\\
 &= \frac{P(Z=x-d)\,p}{P(Z=x-d)\,p+P(Z=x)(1-p)} \\
 &= \frac{1}{1+\frac{P(Z=x)}{P(Z=x-d)}\frac{1-p}{p}} \\
\end{align}
Since $Z$ has a standard normal distribution, we have
$$P(Z=z)\propto \exp\left[-\tfrac{1}{2}z^2\right] \implies 
P(Z=x-d)=\exp\left[d\left(x-\frac{d}{2}\right)\right]P(Z=x)
$$
Substituting into Bayes rule, we then have
$$
P(E=1 \mid X=x) = \frac{1}{1+\alpha e^{-dx}}
$$
where the parameter
$$
\alpha = \tfrac{1-p}{p}\exp\left[\tfrac{1}{2}d^2\right]
$$
is constant (independent of $x$).
So for large $x$, the probability of an event goes to 1, as expected.

Note: This is essentially a two component Gaussian mixture model. Here we are essentially computing the E step of the standard E-M algorithm.
