Joint probability uniform distribution

Question

I have a question on finding probabilities of joint distributions, specifically two independent random variables that are Uniformly distributed. The question I wish to solve is this one:

We agree to try to meet between $12$ and $1$ for lunch at our favorite sandwich shop. Because of our busy schedules, neither of us is sure when we’ll arrive; we assume that for each of us our arrival time is uniformly distributed over the hour. So that neither of us has to wait too long, we agree that we will each wait exactly $15$ minutes for the other to arrive, and then leave. What is the probability we actually meet each other for lunch? "

This is my approach, I am curious as to why my reasoning is wrong:

We wish to be within 15 minutes of each other, so we are looking for:

$P( |X - Y | < 15 ) = P ( -15 < X - Y < 15) $

Which I believe leads to the following:

I get the following lines:
$y = x + 15$
$y = x - 15$.

I then set up the following integral, and then integrate:

$${1\over 3600}\int^{60}_{0} \int ^{x + 15}_{x - 15}\,dydx $$

What am I missing? I know that my answer is wrong, but I don't know where my error is. When I sketch it out on paper, I am sketching out and integrating the diagonal of the rectangle.

Answer 1

Yes, you are correct. Sorry, both you and I were wrong. Should have paid attention to the r.v.'s support. Changes made in red.

If you want to argue formally (rather than by drawing lines), you can use the law of total probability. You can do $$ \begin{align*} \Pr(-15 < X - Y < 15) &= \Pr(-15 + Y < X < 15 + Y) \\ &= \int\limits_{0}^{60} p(y) \Pr(-15 + y < X < 15 + y) \operatorname d y \\ &= \int\limits_{0}^{60} p(y) \int\limits_{\color{red}{\max(}-15 + y\color{red}{, 0)}}^{\color{red}{\min(}15 + y\color{red}{, 60)}} p(x) \operatorname d x \operatorname d y \end{align*} $$ ... which is the same as what you obtained.

Answer 2

Regarding the calculus, I think it's wrong since you have three regions to consider for $y$:

1. $0$ to $x+15$,
2. $x-15$ to $x+15$,
3. $x-15$ to $60$.

But I don't claim to be good at calculus....

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Alternatively, you can compute the complementary probability $$P(|Y-X|<15) = 1-2P(Y>X+15) = 1-2\left[\frac{\frac{1}{2}(60-15)(45)}{60^2}\right] = \frac{7}{15}.$$

Answer 3

You do not wish the $Y$ value to exceed the bounds of its support $(0;60)$.

$$\begin{align}&{1\over 3600}\int^{60}_{0} \int ^{\min(60,x + 15)}_{\max(0,x - 15)}~\mathrm dy~\mathrm dx \\[2ex] & = {1\over 3600}\left(\int_0^{15}\int_{0}^{x+15} ~\mathrm dy~\mathrm dx +\int_{15}^{45}\int_{x-15}^{x+15} ~\mathrm dy~\mathrm dx+\int_{45}^{60}\int_{x-15}^{60} ~\mathrm dy~\mathrm dx\right)\end{align}$$