91影视

Here, it is given that a fire station is to be located along a road of length $A,A<鈭�$.

Fires occur at points uniformly chosen on $(0,A)$.

Let $A$be the fire station and $X$be the place where the fire has occurred.

$X$is uniformly distributed over $\left(O,A\right)$.

$鈬�f_X\left(x\right)=\left\{\begin{array}{l}\frac{1}{A}0鈮�x鈮�A\\ 0\text{Otherwise}\end{array}\right.$

$$\begin{gathered} \left|X-a\right|=X-a,ifa鈮�X鈮�A \\ \left|X-a\right|=a-X,if0鈮�X鈮�a \end{gathered}$$

Now,

localid="1646883589256" $$\begin{gathered} E\left[\left|X-a\right|\right]={鈭�}_0^a\left(a-x\right)f_X\left(x\right)dx+{鈭�}_a^A\left(X-a\right)f_X\left(x\right)dx \\ =\frac{1}{A}\left[{\left(ax-\frac{x^2}{2}\right)}_0^a+{\left(\frac{x^2}{2}-ax\right)}_a^A\right] \\ =\frac{1}{A}\left[a^2+\frac{A^2}{2}-aA\right] \end{gathered}$$

Differentiating $E\left[\left|X-A\right|\right]$w.r.t $a$and equating it with zero, we get

role="math" localid="1646884916713" $$\begin{gathered} \frac{d}{da}\left(E\left|X-a\right|\right)=0 \\ \frac{2a}{A}+0-1=0 \\ \frac{2a}{A}=1 \\ a=\frac{A}{2} \end{gathered}$$

Therefore, the fire station should be located at the mid point of the length of the road to minimize the expected distance.

Here, it is given that the road is of infinite length stretching from point $0$outward to $鈭�$.

The distance of a fire from point $0$is exponentially distributed with rate $位$.

The road is of infinite length. $X$is exponentially distributed with $位$as parameter.

$$\begin{gathered} f_X\left(x\right)=\left\{\begin{array}{l}位e^{-位x}x......0\\ 0\text{Otherwise}\end{array}\right. \\ \left|X-a\right|=\left\{\begin{array}{l}X-aa鈮�X鈮�鈭�\\ a-X0鈮�X鈮�a\end{array}\right. \end{gathered}$$

$$\begin{gathered} E\left[\left|X-a\right|\right]={鈭�}_0^a\left(a-x\right)\left(位e^{-位x}\right)dx+{鈭�}_a^{鈭�}\left(x-a\right)\left(位e^{-位x}\right)dx \\ =位\left[a{鈭�}_0^a\left(e^{-位x}\right)dx-{鈭�}_0^{a}x\left(e^{-位x}\right)dx+{鈭�}_a^{鈭�}x\left(e^{-位x}\right)-a{鈭�}_a^{鈭�}\left(位e^{-位x}\right)dx\right] \\ =位\left[a{\left({\frac{e}{-位}}^{-位x}\right)}_0^a-{\left({\frac{xe}{-位}}^{-位x}-{\frac{e}{{位}^2}}^{-位x}\right)}_0^a+{\left({\frac{xe}{-位}}^{-位x}-{\frac{e}{{位}^2}}^{-位x}\right)}_a^{鈭�}-a{\left({\frac{e}{-位}}^{-位x}\right)}_a^{鈭�}\right] \\ =位\left[\frac{a}{位}\left(1-e^{-位a}\right)+\left(\frac{a{e}^{-位a}}{位}+\frac{e^{-位a}}{{位}^2}-\frac{1}{{位}^2}\right)+\left(\frac{a{e}^{-位a}}{位}+\frac{e^{-位a}}{{位}^2}\right)-\frac{a{e}^{-位a}}{位}\right] \\ =位\left[\frac{a}{位}\left(1-2e^{-位a}\right)+\left(\frac{2a{e}^{-位a}}{{位}^2}+\frac{2e^{-位a}}{{位}^2}-\frac{1}{{位}^2}\right)\right] \\ =位\left[\frac{a}{位}+\frac{2e^{-位a}}{{位}^2}-\frac{1}{{位}^2}\right] \\ =\left[a+\frac{2e^{-位a}}{位}-\frac{1}{位}\right] \end{gathered}$$

Differentiating $E\left|X-a\right|$w.r.t $a$and equating it with $0$, we get

$$\begin{gathered} \frac{d}{da}\left(E\left|X-a\right|\right)=0 \\ 鈬�1-2e^{-位a}=0 \\ 鈬�1=2e^{-位a} \\ 鈬�e^{-位a}=\frac{1}{2} \\ 鈬�e^{位a}=2 \\ 鈭�a=\frac{\log 2}{位}. \end{gathered}$$

Therefore, the fire station should be located at $a=\frac{\log 2}{位}$to minimize the expected distance.