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Problem:

Two birds have landed on a power line that spans the 100' distance between utility poles.
a) What is the average distance between the birds?
b) The line runs north and south. Another bird lands on the line. What is the expected position of the north-most bird from the south-most pole?

Partial Solution:

For the first part of the problem, I defined the first bird's position as the random variable $X$ and the second birds position as the random variable $Y$. Since both birds could lie in the range [0, 100], the plot of the birds' possible positions yielded a square with area $100^2$ feet$^2$. I then defined a new random variable, $Z$, to represent the distance between the two birds, defined by $|X-Y|$.Thus, the PDF of $Z$ can be written as $f_Z(z) = \begin{cases} 1/10000, & z\leq100, \\ 0, & \text{otherwise.} \end{cases} $ From here, to find the average distance between the birds, I solved for the expected value of $Z$, using the equation $\int_{-\infty}^\infty \int_{-\infty}^\infty z f_Z(z) \,dy\,dx$, which yielded that $E[Z] = E[|X-Y|] = 100/3$ feet.

So first and foremost, did I do this correctly?

Secondly, I am completely stuck on the second part of the problem.

Community
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Bill Jenkins
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  • $Z = |X-Y|$ does not follow a uniform distribution, you have made a mistake there. – Lord Soth Aug 04 '14 at 00:42
  • For the second one, note that the north-most bird's position is $\max{X_1,X_2,X_3}$, where $X_1,X_2,X_3$ are iid random variables uniformly distributed on $[0,100]$. – Lord Soth Aug 04 '14 at 00:45
  • This may help although its not a formal proof. I generated a simple Monti Carlo. For each of 10000 samples, I generated two random numbers in the range [0-100] and then the absolute value of the difference between the pairs. The average of the pairs was about 33; which is consistent with your result. I then generated another Monti with triplets and caclulated the average of the maximum of each triplet. The result of this Monti was about 75. – Gary's Student Aug 04 '14 at 01:41

1 Answers1

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The operating principle is that : $\mathbb{E}[g(X,Y)] = \iint_{\bf X\times Y} (g(x,y)\cdot f_{X,Y}(x,y))\operatorname{d}y\operatorname{d}x$

Since we have independent uniform distributions over $(0,100)$ then: $f_{X,Y}(x,y)= \frac{1}{100^2}$.

For the first part: $|x-y|=\begin{cases} x- y & : x\geq y\\y-x & : x<y\end{cases}$

So we need to integrate the function over two ranges: $y\in [0,x), y\in[x,100]$

$\begin{align} \mathbb{E}[|X-Y|] & =\frac{1}{100^2} \int_0^{100}\int_0^{100}|x-y|\operatorname{d}y\operatorname{d}x \\ & = \frac{1}{100^2} \int_0^{100} \int_{0}^x (x-y)\operatorname{d}y +\int_x^{100}(y-x)\operatorname{d}y\operatorname{d}x \\ & = \frac{1}{100^2}\int_0^{100} \left[xy-\frac 12 y^2\right]_{y=0}^{y=x} +\left[\frac 1 2 y^2-xy\right]_{y=x}^{y=100}\operatorname{d}x \\ & =\frac{1}{100^2} \int_{0}^{100} x^2 + \frac{100^2}{2}-100x\operatorname{d}x \\ & = \frac {1}{100^2}\left[\frac 1 3 x^3 + \frac{100^2 x}{2}- \frac{100 x^2}{2}\right]_{x=0}^{x=100} \\ & = \frac{1}{100^2}\left(\frac{100^3}{3}+\frac{100^3}{2}-\frac{100^3}{2}\right) \\ & = \frac{100}{3} \end{align}$

For the second part, you want $\mathbb{E}(\max(W,X,Y))$ where $W,X,Y$ are the mutually independent positions of the three birds; in feet north of the southern pole. Thus by the same principle:

$$\begin{align}\mathbb{E}(\max(W,X,Y)) & = \frac{1}{1000000}\int_0^{100}\int_0^{100}\int_0^{100} \max(w,x,y)\operatorname{d}y \operatorname{d}x \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \int_0^{100} \left(\int_0^{\max(w,x)} \max(w,x)\operatorname{d}y+\int_{\max(w,x)}^{100}y\operatorname{d}y\right) \operatorname{d}x \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \int_0^{100} \max(w,x)^2+\frac{100^2-\max(w,x)^2}{2} \operatorname{d}x \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \int_0^{100} \frac{100^2+\max(w,x)^2}{2} \operatorname{d}x \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \int_0^{100} \frac{100^2}{2}\operatorname{d}x + \int_0^w \frac{w^2}{2}\operatorname{d}x + \int_w^{100} \frac{x^2}{2} \operatorname{d}x \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \frac{100^3}{2} + \frac{w^3}{2} + \frac{100^3-w^3}{6} \operatorname{d}w \\ & = \frac{1}{1000000}\int_0^{100} \frac{2\times 100^3}{3} + \frac{w^3}{3} \operatorname{d}w \\ & = \frac{1}{100^3}\left(\frac{2\times 100^4}{3} + \frac{100^4}{12}\right) \\ & = 75 \end{align}$$

Graham Kemp
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