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Chapter 6: Problem 14

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, its distance 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, compute, assuming independence, the distribution of its distance from the accident.

Short Answer

Expert verified
The distribution of the distance, \(Z\), between the ambulance and the accident is given by the PDF: \[f_Z(z) = \begin{cases} \frac{2}{L}, & \text{0 < }z\text{ < }\frac{L}{2} \\ \frac{2}{L}(1 - \frac{z}{L}), & \text{\(\frac{L}{2} <\) z < L} \\ 0, & \text{otherwise} \end{cases}\]

Step by step solution

01

Define the Variables

Let \(X \sim U(0, L)\) be the position of the ambulance and \(Y \sim U(0, L)\) be the position of the accident. Both variables are uniformly distributed over \((0, L)\). We want to find the distribution of \(Z = |X - Y|\).
02

Determine the Cumulative Distribution Function (CDF) for Z

For 0 < \(z\) < L/2, the CDF is given by: \[F_Z(z) = P(Z \le z) = P(|X - Y| \le z)\] \[= \int_0^L \int_{x-z}^{x+z} f(x, y) dy dx\] Where \(f(x, y)\) is the joint PDF of X and Y which is equal to \(\frac{1}{L^2}\) in this case. For L/2 < \(z\) < L, the CDF can be found similarly but we need to adjust the integration region.
03

Calculate the PDF of Z by differentiating the CDF

To get the PDF of Z, we'll differentiate the CDF with respect to z: \[f_Z(z) = \frac{d}{dz} F_Z(z)\]
04

Calculate the PDF of Z for 0 < \(z\) < L/2

First, consider the case when 0 < \(z\) < L/2. We evaluate the integral: \[f_Z(z) = \frac{d}{dz} \int_0^L \int_{x-z}^{x+z} \frac{1}{L^2} dy dx\] \[= \frac{d}{dz} \frac{1}{L^2} \int_0^L (2z) dx\] \[= \frac{d}{dz} \frac{2zL}{L^2}\] \[= \frac{2}{L}\] So, the PDF in this case is \(f_Z(z) = \frac{2}{L}\) for 0 < \(z\) < L/2.
05

Calculate the PDF of Z for L/2 < \(z\) < L

Next, consider the case when L/2 < \(z\) < L: We need to adjust the range of integration for this case. After evaluating the integral and differentiating with respect to z, we get: \[f_Z(z) = \frac{2}{L} (1 - \frac{z}{L})\] Hence, the PDF for the given range is \(f_Z(z) = \frac{2}{L} (1 - \frac{z}{L})\) for L/2 < \(z\) < L.
06

Write the PDF for all possible values of Z

Finally, we write the PDF for all possible values of Z: \[f_Z(z) = \begin{cases} \frac{2}{L}, & \text{0 < }z\text{ < }\frac{L}{2} \\ \frac{2}{L}(1 - \frac{z}{L}), & \text{\(\frac{L}{2} <\) z < L} \\ 0, & \text{otherwise} \end{cases}\] Thus, we have found the distribution of the distance between the ambulance and the accident.

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