Interpretation of the slope in a probability distribution.

Question

When considering a certain probability-distribution $P(t)$, then what information does the derivative $\frac{\mathrm d P(t)}{\mathrm{dt}}$ give, in general?

Context:

I raised this question on account of a physics exercise, which I could not rigorously complete. In short the question goes like,

assume that the probability of an electron, in an ensemble of electrons, suffering a collision in an interval $\mathrm{dt}$ is $\mathrm{dt} / \tau$. Show that an electron picked at random at a given moment had no collisions during the the next $t$ seconds with probability $e^{-t/\tau}$.

Noting that $P(t)$ gives the probability that certain electrons have not yet suffered a collision, the difference $\frac{\mathrm d P(t)}{\mathrm{dt}} \cdot \mathrm{dt}$ gives the probabilty that those electrons have suffered a collision in an interval $\mathrm{dt}$. We now devide by $P(t)$, to treat the case for one particular electron. So we get, $$ \frac{\frac{\mathrm d P(t)}{\mathrm{dt}}\cdot \mathrm{dt}}{P(t)} = -\frac{\mathrm{dt}}{\tau} \implies P(t) = e^{-t/\tau}. $$

PS: Please tell me if this question is too off topic and belongs in the physics section.

Answer 1

Your calculation is a variant of a standard approach to the derivation of the Poisson process model. Let us pick an electron at time $t = 0$ and ask for the probability that it suffers no collisions in the interval $(0,T]$. The assumptions are that

(i) in a very short interval of time $\Delta t$, the electron c an suffer at most one collision, and the probability of a collision is $\frac{\Delta t}{\tau}$

and

(ii) the numbers of collisions suffered by the electron in any two disjoint intervals of time $(t_1,t_2]$ and $(t_3, t_4]$ (where $t_1 < t_2 \leq t_3 < t_4$) are independent random variables.

Thus, with $P(T)$ denoting the probability that the electron suffers no collisions in the interval $(0,T]$, we have that $$\begin{align} P(T+\Delta T) &= P\{\text{no collisions in interval}~ (0,T+\Delta T]\}\\ &= P\{\text{no collisions in interval}~ (0,T]~\text{and}~\text{no collisions in interval}~ (T,T+\Delta T]\}\\ &= P\{\text{no collisions in interval}~ (0,T]\}\cdot P\{\text{no collisions in interval}~ (T,T+\Delta T]\}\\ &= P(T)\left(1-\frac{\Delta T}{\tau}\right) \end{align}$$ leading to $\displaystyle\frac{P(T+\Delta T)-P(T)}{\Delta T} = -\frac{1}{\tau}P(T) ~\Rightarrow~ \frac{\mathrm dP(T)}{\mathrm dT} = -\frac{1}{\tau}P(T) ~\Rightarrow~ P(T)= e^{-T/\tau}$. If the assumptions make sense in physics (as opposed to mathematics), then you can make the following interpretation for the derivative of $P(T)$.

The tangent to the $P(T)$ curve (whose slope is, of course, $\frac{\mathrm dP(T)}{\mathrm dT}$, and is a negative number), crosses the horizontal axis at $T + \tau$ showing that at the current rate of change, the probability of no collisions will reduce to $0$ in a further $\tau$ seconds. Now, $\tau$ is the expected time till the first collision occurs after $t=0$. In fact, even though there have been no collisions thus far in our model, $\tau$ is still the expected time till the next collision. This is the memoryless property of the exponential distribution.