Why is $P(t < T \leq t + dt) = f(t)dt$? Why is $P(t < T \leq t + dt) = f(t)dt$?
https://en.wikipedia.org/wiki/Probability_density_function#Further_details
Is this perhaps some idea around:
https://mathworld.wolfram.com/DistributionFunction.html
 A: In the area of survival analysis we can define $F(t)$ - the distribution function - as the probability that item fails in time interval $(0,t]$. Then the probability density function $f(t)$ is defined in the following way:
$$ f(t) = \frac{d}{dt} F(t) =  \lim_{dt\to\ 0} \frac{F(t+dt) - F(t)}{dt} = \lim_{dt\to\ 0} \frac{P(t < T \leq t+dt)}{dt} $$
This implies that for small $dt$:
$$ P(t < T \leq t+dt) \approx f(t) dt $$
Note that there is not equal sign, it is just approximation for small $dt$.
A: This statement is not strictly true, just approximately true assuming that $dt$ is small, and f is continuous and differentiable. It becomes strictly true in the limit $dt \rightarrow 0$.
The probability density function $f(t)$ is defined,
$$
P(t < T \le t + \Delta t) = \int_t^{t+\Delta t} f(t') dt'.
$$
The RHS can be rewritten by transforming the integration variable, and with a Taylor expansion of the integrand we get,
$$
\begin{split}
P(t < T \le t + \Delta t) &= \int_0^{\Delta t} f(t + \tau) d\tau \\
&= \int_0^{\Delta t} \left[ f(t) + f'(t) \tau + \frac{1}{2} f''(t) \tau^2 + \dots \right] d\tau \\
&\approx \int_0^{\Delta t} f(t) d \tau \\
&= f(t) \Delta t.
\end{split}
$$
In the second step I just Taylor expanded $f(t + \tau)$ about $t$. If $\Delta t$ is sufficiently small, then all factors of $\tau$ at order $1$ or greater can be ignored compared to $f(t).$ You can see that in the limit as $\Delta t \rightarrow 0$, the approximate equality becomes a true equality.
