Estimating time remaining using conditional expectation I'm interested in the following scenario:
The duration of a certain (natural) process is estimated using a random variable with mean $\mu$ and standard deviation $\sigma$, which we can approximate using normal distribution (with the stated parameters) for any practical purposes.
Let $X^t$ a random variable describing the expected amount of time left for an observed process to end, given that observation is made at time $t$ ($t$ time elapsed since the process has started).
How can I compute the probability of $X^t$ ?
 A: The probability of $X^t$ is given by a truncated normal distribution (truncation of lower tail at $t$) and you can compute it using Bayes theorem. Let's say $X=X^t + t$, where $X$ is another random variable describing the total lifetime of the observed process, then you are interested in finding the posterior probability 
$$
p\left(X=x \mid t^* \right) = \frac{p\left(t^* \mid X=x \right) \, p\left(X=x \right)}{p\left(t^*\right)}
$$
where I have used the notation $t^*$ to indicate the observation that the process is still running at time $t$ (there is probably a better notation for this, but I don't know it).
The prior probability is known 
$$
p\left(X=x \right) = \frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)
$$
where $\phi$ indicates the probability density function of the standard normal distribution.
The likelihood is 
$$
p\left(t^* \mid X=x \right) = 
\begin{cases}
    1, & \text{if } x\geq t\\
    0,              & \text{otherwise}
\end{cases}
$$
The marginal likelihood of $t^*$ is
$$
\begin{align}
p\left(t^*\right) &= \int_{-\infty}^{\infty} p\left(t^* \mid X=x \right) p\left(X=x \right) dx\\ 
&= 0 \cdot \left[ \int_{-\infty}^{t} p\left(X=x \right)dx \right] + 1 \cdot \left[\int_{t}^{\infty} p\left(X=x \right) dx \right]\\
&= 1 - \Phi\left( \frac{t - \mu}{\sigma}\right)
\end{align}
$$
where $\Phi$ is the cumulative distribution function of the standard normal distribution.
Finally, putting everything together and applying Bayes theorem yields the posterior probability density
$$
p\left(X=x \mid t^* \right) =
\begin{cases}
    \frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma \left[ 1 - \Phi\left( \frac{t - \mu}{\sigma}\right)\right]}, & \text{if } x\geq t\\
    0,              & \text{otherwise}
\end{cases}
$$
The probability density for the cases where $x \geq t$ is a truncated Gaussian distribution (see one side truncation of lower tail on Wikipedia). Based on this you can, for example, compute the expectation 
$$
\mathbf{E} \left[X \mid t^* \right] = \mu + \sigma \frac{\phi\left(\frac{t - \mu}{\sigma} \right)}{1 - \Phi\left(\frac{t - \mu}{\sigma}\right)}
$$
and, clearly, $\mathbf{E} \left[X^t \mid t^* \right] = \mathbf{E} \left[X \mid t^* \right] - t$.
