MHB Joint Probability Density Functions

AI Thread Summary
The discussion focuses on calculating the average distance between two birds on a power line and determining the expected position of the north-most bird from the south-most pole. The initial approach involved defining the positions of the birds as random variables and deriving the probability density function (PDF) for their distance. However, a participant pointed out that the defined PDF for the distance variable Z was incorrect and suggested a more appropriate method for evaluating it. The conversation highlights the need for a correct formulation of the joint probability density function to accurately solve the second part of the problem. Overall, the thread emphasizes the importance of proper statistical methods in solving joint probability problems.
lexluger
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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 $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}zf_{Z}(z)dydx$, 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.
 
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lexluger said:
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 $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}zf_{Z}(z)dydx$, 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.

Hi lexluger! Welcome to MHB! :)

I'm afraid your distribution density function of $Z$ is not correct.
Consider that:
$$f_Z(z) = \underbrace{\int_z^{100} f_X(x) f_Y(x - z)\,dx}_{x-y\ge 0} + \underbrace{\int_0^{z} f_X(x) f_Y(x + z)\,dx}_{x-y < 0}$$

Can you evaluate that?
 
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