Normalizing probability over Expected Value I'm reading through this article regarding the Waiting Time Paradox
http://jakevdp.github.io/blog/2018/09/13/waiting-time-paradox/
In the "Digging Deeper: Probabilities & Poisson Processes" section, the author writes:

Let's denote by $p(T)$ the distribution of intervals $T$ between buses as they arrive at a bus stop.
...
When a rider arrives at a bus stop at a random time, the probability
of the time interval they experience will be affected by $p(T)$, but
also by $T$ itself: the longer the interval, the larger the
probability is that a passenger will experience it.
So we can write the distribution of arrival times experienced by
passengers: $$p_{exp}(T)∝Tp(T)$$
The constant of proportionality comes from normalizing the
distribution: $$p_{exp}(T)=\frac{T p(T)}{\int_{0}^{\infty}T p(T) dT}$$

I understand the denominator is just the Expected Value over all $T \geq 0$.I'm trying to get the intuition behind the fraction. Specifically:

*

*How should I interpret the numerator and product $Tp(T)$? It seems to be just a weighted value of some time interval $T$ and just a fraction of the Expected Value.


*Can you better explain how this definition of $p_{exp}(T)$ is the probability a person will have to wait within time interval $T$ until the next bus comes?
In regards to question #2, I tried doing a small example using finite sums. Suppose
$$T_1 = 5, \ T_2 = 15, \ p(T_1) = 1/4, \ p(T_2) = 3/4$$
So that Expected Value is $12.5$. Then:
$$p_{exp}(T_1) = 0.1$$
$$p_{exp}(T_2) = 0.9$$
So I'm not sure what I should understand about these probabilities. Does this mean the probability of a person having to wait in the 5 minute interval is 10%?
 A: Rather than thinking of $Tp(T)$ as a weighted value of $T$, maybe it's more productive to think of it as a weighted value of $p(T)$. If all intervals were the same length, then when you arrive at the bus stop your probability of being in interval $T$ would simply be proportional to $T$. But since some intervals are longer than others, you are more likely to arrive during one of those intervals simply because they are longer, quite apart from how likely those intervals are to occur. 
For example: if $T_1$ were 1 second long with probability $.99$ of occurring, and $T_2$ were ten hours long with probability $0.01$ of occurring, you would still be much more likely to arrive during $T_2$ than during $T_1$; even though $T_2$ is rare, whenever it does happen, it lasts for a very long time. So many arriving passengers arrive during $T_2$.
So: your probability of experiencing $T$ depends on (a) the probability of $T$ occurring, and (b) the length of $T$. That is why the numerator is a product of $p(T)$ and $T$.
So we know our probability of experiencing $T$ is proportional to (a) and to (b). But to calculate the actual probability, we need to normalize the product $Tp(T)$, by dividing it by the sum/integral of the probabilities of all possible intervals. When we do that, we have an expression $p_{exp}(T)=\frac{Tp(T)}{\int_0^\infty Tp(T)dT}$ which is a normalized product of the probability of $T$ and the length of $T$, which gives us the probability of experiencing $T$.
