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My teacher gave me a demonstration of the idea behind the bus waiting paradox, but I'm having some issues understanding one of the initial assumptions.

Convention here is that X is the time between two arrivals, while Y is the waiting time.

The demonstration starts by saying that

$$ E[Y]=\int_{0}^\infty E[Y|A_t]*P(A_t)\label{a}\tag{1} $$

where $A_t$ is the event defined as

$$ A_t= \text{"arrival happening during a realization } x \text{ of } X : t \le x \le t+dt" $$

He also adds that

$$ E[Y|A_t]=\int_{u=0}^t u*f_{Y|A_t}(u)*du=\frac1t*\int_{u=0}^t u*du = \frac t2\label{b}\tag{2} $$

What I don't understand is how both $(1)$ and $(2)$ work, from a formal point of view, even if I have a rough idea.

So, what I'd like to know is if someone can provide all the steps to prove the two equalities.

EDIT - adding the full demonstration just for clarity

Thesis:

$$ E[Y]=\frac{E[X^2]}{2*E[X]}=\frac12*E[x]+\frac{VAR(X)}{2*E[X]} $$

Demonstration:

$$ P(A_t)=c*t*f_X(t)*dt; c=\frac{1}{E[X]} $$

$$ E[Y]=\int_{0}^\infty E[Y|A_t]*P(A_t) $$

$$ E[Y|A_t]=\frac t2 $$

$$ E[Y]=\int_{0}^\infty E[Y|A_t]*P(A_t)=\int_{0}^\infty \frac t2 * \frac{t*f_X(t)}{E[X]}*dt=\frac{E[X^2]}{2E[X]} $$

StepTNT
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1 Answers1

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(1) is just the Law of Total Probability in disguise, and (assuming $X$ has a density function $f_X$) can be less obtusely written as $$ E[Y]=\int_0^\infty E[Y\mid X=t]\,f_X(t)\,dt. $$ But $X$ is the sum of two iid random variables, $Y$ and $Z$ (say). By symmetry $$ E[Y\mid X=t]=E[Z\mid X=t], $$ but also $$ E[Y\mid X=t]+E[Z\mid X=t]=E[Y+Z\mid X=t]=E[X\mid X=t]=t. $$ It follows that $E[Y\mid X=t]=t/2$. This is (2).

John Dawkins
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