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I'm trying to solve the following problem:

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:35 and 12:15 and the woman independently arrives between 11:40 and 12:55, what is the probability that the first to arrive waits no longer than 5 minutes?

I'm trying to solve it using the convolution of probability distributions formula:

$$F_{x+y}(z) = \int f_Y(y)*F_X(z-y)dy$$

However I keep getting a negative solution which obviously can't be right. Here's what I've done so far:

Let X = negative minutes man arrives after 11:30, Y = minutes woman arrives after 11:30

Then we can derive the following equations: \begin{equation} f_X(x) = \left\{ \begin{array}{lr} \frac{1}{40}, & \text{if } -45<x<-5\\ 0, & \text{otherwise } \end{array} \right\} \end{equation} \begin{equation} f_Y(y) = \left\{ \begin{array}{lr} \frac{1}{75}, & \text{if } 10<y<85\\ 0, & \text{otherwise } \end{array} \right\} \end{equation} \begin{equation} F_X(x) = \left\{ \begin{array}{lr} 0, & \text{if } x<-45\\ \frac{x}{40}, & \text{if } -45<x<-5\\ 1, & \text{if } -5<x \end{array} \right\} \end{equation}

\begin{equation} F_X(5-y) = \left\{ \begin{array}{lr} 0, & \text{if } y>50\\ \frac{5-y}{40}, & \text{if } 50>y>10\\ 1, & \text{if } 10>y \end{array} \right\} \end{equation} \begin{equation} F_X(-5-y) = \left\{ \begin{array}{lr} 0, & \text{if } y>40\\ \frac{x}{40}, & \text{if } 40>y>0\\ 1, & \text{if } 0>y \end{array} \right\} \end{equation}

Now we're looking for probability $$P(|X+Y|<5)$$ $$=F_{X+Y}(5)-F_{X+Y}(-5)$$ $$=\int f_Y(y)*F_X(5-y)dy - \int f_Y(y)*F_X(-5-y)dy $$ $$=\int_{10}^{85} \frac{1}{75}*F_X(z-y)dy - \int_{10}^{85} \frac{1}{75}*F_X(z-y)dy $$ $$=\int_{10}^{50} \frac{1}{75}*\frac{5-y}{40}dy - \int_{10}^{40} \frac{1}{75}*\frac{-5-y}{40}dy $$

But this evaluates to a negative probability! I can't figure out what I'm doing wrong here. I've seen solutions using graphing and solving for overlapping area, but I am only interested in solving using this formula. Can anyone help me?

Evyenia Coufos
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  • Your $F_X$ is wrong. If $x\in(-45,-5)$, then $F_X(x)=\frac{x}{40}<0$, which clearly shouldn't be able to happen – Lorago Dec 07 '24 at 19:54
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    The joint density of the the two arrival times is a uniform distribution over a region of the plane. Sketch this region, indicate on your diagram the region in which the two variables differ by 5 minutes or less, and you are nearly done! – Dilip Sarwate Dec 07 '24 at 20:35
  • The comment of @Dilip Sarwate is sound : it is surely the shortest way. If you want to solve your issue using convolution of distributions,, instead of starting from scratch, it can be helpful to know that the convolution of two uniform distributions has the shape of a Maya pyramid : see here for example. – Jean Marie Dec 07 '24 at 23:41
  • For an example of such a diagram, see https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar/ while for a more analytical approach see https://math.stackexchange.com/questions/2084778/a-problem-with-two-uniformly-distributed-random-variables – Henry Dec 08 '24 at 01:43

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