Question

$\left(a\right)$A fire station is to be located along a road of length$A,A<\infty$. If fires occur at points uniformly chosen on$(0,A)$, where should the station be located so as to minimize the expected distance from the fire? That is, choose a so as to

minimize $E\left[\left|X-a\right|\right]$

when$X$is uniformly distributed over $(0,A)$

$\left(b\right)$Now suppose that the road is of infinite length— stretching from point$0$ outward to$\infty$. If the distance of fire from the point $0$is exponentially distributed with rate$\lambda$, where should the fire station now be located? That is, we want to minimize$E\left[\left|X-a\right|\right]$, where $X$is now exponential with rate$\lambda$.

Short answer

Therefore, the

$\left(a\right)E\left[\left|X-a\right|\right]=\frac{A}{2}$

$\left(b\right)E\left[\left|X-a\right|\right]=\ln \left(\frac{2}{\lambda }\right)$

Step-by-step solution

Given information:

$\left(a\right)$A fire station is to be located along a road of length$A,A<\infty$. If fires occur at points uniformly chosen$(0,A)$, That is, choose a so as to minimize$E\left[\left|X-a\right|\right]$

$\left(b\right)$) Now suppose that the road is of infinite length— stretching from point$0$outward to$\infty$. If the distance of fire from a point$0$ is exponentially distributed with the rate$\lambda$.

Part (a) Step 2 Explanation:

We have that

$$\begin{gathered} E\left[\left|X-a\right|\right]={\int }_{-\infty }^{\infty }\left(x-a\right)f\left(x\right)dx \\ ={\int }_0^a\left(a-x\right)\frac{1}{A}dx+{\int }_a^A\left(x-a\right)\frac{1}{A}dx \\ ={\left[\frac{a}{A}x-\frac{1}{2A}{x}^2\right]}_0^a+{\left[\frac{1}{2A}{x}^2-\frac{a}{A}x\right]}_a^A \\ =\left(\frac{a^2}{A}-\frac{a^2}{2A}\right)+\left(\left(\frac{A^2}{2A}-\frac{Aa}{A}\right)-\left(\frac{a^2}{2A}-\frac{a^2}{A}\right)\right) \\ =\frac{2a^2-a^2}{2A}+\frac{A^2-2Aa-a^2+2a^2}{2A} \\ =\frac{2a^2-2Aa+A^2}{2A} \end{gathered}$$

Since we want to minimize this, we take the derivative and set it equal to zero:

$$\begin{gathered} \frac{d}{da}\left(\frac{2a^2-2Aa+A^2}{2A}\right)=\frac{1}{2A}\left(4a-2A\right) \\ =\frac{2a}{A}-1=\frac{A}{2} \end{gathered}$$

. Thus, we can minimize the expected value by choosing the midpoint of the interval$(0,A)$

Part (b) step 3 Explanation:

Using integration by parts we can find that

$$\begin{gathered} E\left[\left|X-a\right|\right]={\int }_0^{\infty }\left(x-a\right)f\left(x\right)dx \\ ={\int }_0^a\left(a-x\right)\lambda e^{-\lambda x}dx+{\int }_0^{\infty }\left(x-a\right)\lambda e^{-\lambda x}dx \\ =a\left(1-e^{\lambda a}\right)+a{e}^{-\lambda a}+\frac{e^{-\lambda a}}{\lambda }-\frac{1}{\lambda }+a{e}^{-\lambda a}+\frac{e^{-\lambda a}}{\lambda }-a{e}^{-\lambda a} \end{gathered}$$

After differentiating and setting equal to zero, we discover that$e^{-\lambda a}-\frac{1}{2}=0$which gives the minimum value at$a=\ln \left(\frac{2}{\lambda }\right)$