5
$\begingroup$

Problem: The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are (0, 0), to the point (x, y) is |x|+|y|. If an accident occurs at a point that is uniformly distributed in the square, find the expected travel distance of the ambulance.

My Attempt: $E[D] = E[X] + E[Y]$; Where D is the travel distance, X is the travel distance on the x axis and Y is the travel distance on the y axis.
Because it's uniformly distributed and its a square I can do this: $E[D] = 2E[X]$.
Then $$(1) \space E[X] = \int_{a}^{b} xf_X(x) \,dx$$ $$(2) \space f(x, y) = |x| + |y|, 0 \leq x, y \leq 3/2$$ $$(3) \space f_X(x) = \int_{0}^{3/2} |x| + |y| \,dy = \frac{3}{2}x + \frac{9}{8}$$ (Because we're only working with positive values, there's no negative, so it's similar to integrating x + y) $$(4) \space E[X] = \int_{0}^{3/2} x(\frac{3}{2}x + \frac{9}{8}) \,dx = \frac{189}{64}$$ $$(5) \space E[D] = 2E[X]; E[D] = 2(\frac{189}{64}) = \frac{378}{64}$$

So the answer obviously doesn't make sense, as it's larger than the max travel distance possible, which is 3. \

The answer is 3/2 and this doesn't make sense to me either and I'll explain why.
$E[X] = \mu$ which is the average. If the average is 3/2 then on average the ambulance will drive 0.75 miles on the x axis and 0.75 miles on the y axis. If we take the hospital as the center point, then we have a square of 1.5 by 1.5 mile surrounding the hospital. The area of that square is $\frac{9}{4}$ and the area of the other square which was 3 by 3, is 9.
Then we have $9 - \frac{9}{4} = \frac{27}{4}.$
Considering we took the mean and the chance is uniform, meaning that every point in the square has an equal chance of being chosen, the area within the mean and the area outside of the mean should be equal, no?

$\endgroup$
1
  • 1
    $\begingroup$ If the hospital is at the center of the grid, then shouldn't $x\in\left[-\dfrac32,\dfrac32\right]$? Same for $y$ $\endgroup$
    – user170231
    Commented Sep 22, 2023 at 18:20

2 Answers 2

4
$\begingroup$

We have $$E[X] = \int_{-3/2}^{3/2} |x| \frac{1}{3} \,\mathrm{d}x = 2 \int_{0}^{3/2} x \frac{1}{3} \,\mathrm{d}x = \left.\frac{x^2}{3}\right|_0^{3/2} = \frac{3}{4},$$ so $E[D]=2E[X]=3/2$.

$\endgroup$
4
$\begingroup$

You might as well consider just one quadrant, say quadrant $1$. Consider travelling along X-axis, the mean distance will obviously be $\frac34$ mi, and the same along the Y-axis. And using geometric symmetry of the four quadrants, the mean distance along the X-axis and Y-axis will be $\large\frac34$ miles each, giving the mean distance of travel as $\large\frac32$ miles

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .