Q.6.14 - A First Course in Probability

Question

An ambulance travels back and forth at a constant speed along a road of length L. At a certain moment of time, an accident occurs at a point uniformly distributed on the road. [That is, the distance of the point from one of the fixed ends of the road is uniformly distributed over (0, L).] Assuming that the ambulance’s location at the moment of the accident is also uniformly distributed, and assuming independence of the variables, compute the distribution of the distance of the ambulance from the accident. 

Step-by-Step Solution

Answer

The distribution of the distance of the ambulance from the accident is $f_Z\left(a\right)=\frac{2(L-a)}{L^2}$

Step 1: Content Introduction

The length of the line connecting two places is the distance between them. If the two locations are on the same horizontal or vertical line, the distance between them can be calculated by subtracting the non-overlapping coordinates.

Step 2: Content Explanation

Define random variables X and Y to be positions of the ambulance and the accident on the road. We are given that X, Y$~$Unif(0,L) and that X and Y are independent. Hence, the joint density function of X and Y is

  $$\begin{gathered} f(x,y)=F_X\left(x\right)F_Y\left(y\right) \\ =\frac{1}{L^2} \end{gathered}$$

for $(x , y)\in (0 , L^2)$, otherwise it is equal to zero. Random variable that marks the distance between these two positions is Z:= (X - Y). Lets find the cumulative distribution of Z.

$$\begin{gathered} P(Z\le a)=1-P (Z>a) \\ =1-\frac{{(L-a)}^2}{L^2} \end{gathered}$$

Step 3: Conclusion

The last equality is got when we consider the squares (0,L)² in the two dimensional plane. Event $Z>a$ is equivalent to the set of points where X and Y differ more than a. And that sets in fact consisted of two rights triangles (one upper and one lower) with the length of side L - a.

Finally the density function of Z is $f_Z\left(a\right)=\frac{d}{da}P(Z\le a)= \frac{2(L-a)}{L^2}$