Suppose that the time until a device fails $X$ follows an exponential distribution, $X \sim \exp(\lambda).$ Now, I understand that the hazard function $h(x)$ is the conditional probability of the failure of the device at or after time $x$, given that it did not fail before time $x$. For the exponential distribution, this is just constant $1/\lambda$.
But what about the other case, that is, given that a device failed at or before time $T$, what is the rate at which it fails after $t,t\le T$?
I think in terms of probability, this is $P(X \ge t| X\le T, t<T)=\frac{e^{- \lambda t} - e^{- \lambda T}}{1-e^{- \lambda T}}$, and then the rate would be $E[X\le t|X \le T]$ ?
The hazard (rate) function is: $$h(t)=\lim_{\Delta t \to 0} \frac{R(t)-R(t+\Delta t)}{\Delta t \cdot R(t)} = \frac{f(t)}{1-F(t)} $$ where $R(t)=1-F(t)$ is the survival function, i.e. the probability of no failure before time $t$. So $h(t)$ is the conditional density of failure at time $t$ given that it did not fail before time $t$ (rather than the "conditional probability of the failure of the device at or after time $t$, given that it did not fail before time $t$"). For the exponential distribution $f(t) = \lambda e^{-\lambda t}$ the hazard rate function is constant and equal to $\lambda$.
I do not think "the other case" is clearly defined. However, if you ask the "rate" at which it fails at time $t$ given that it fails no sooner than $t$ and no later than $T$, this would be: $$h_T(t) = \frac{f(t)}{F(T) - F(t)}.$$ For the exponential distribution it would be $\lambda / (1 - e^{-(T-t)})$.