Joint PDF of a spatial distribution

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Homework Help Overview

The problem involves computing the joint probability density function for the spatial distribution of residences in a uniformly populated square town measuring 1.5 miles by 1.5 miles, with a given density of 2000 residences per square mile.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between uniform distribution and the number of residences, considering how to express the joint PDF in terms of area and density. There are attempts to derive the cumulative distribution function (CDF) and relate it to the joint PDF.

Discussion Status

Participants are exploring different interpretations of how to express the joint PDF and its relation to the total number of residences. Some have suggested integrating the density function over a specified area, while others are clarifying the meaning of terms used in the context of the problem.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the uniform distribution and how to accurately represent the number of residences within a given area. The total number of residences in the town is noted to be 4500, which factors into the calculations being discussed.

Kalinka35
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Homework Statement


In a square town that is 1.5 miles by 1.5 miles, the places of residence are uniformly distributed (2000 per square mile) over the whole town.
Compute the joint probability density function for the spatial distribution of places of residence (fX, Y).


Homework Equations





The Attempt at a Solution


I've set it up so the lower left corner of the town is at the origin.
I understand how a uniform distribution works, but the addition of the 2000 residences per square mile is really throwing me off.
I am thinking of trying to find the CDF first and then differentiating but I don't have an idea of how I would begin to find that.
 
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well as its a uniform distribution, then number of houses will be directly proportional to area

so i think maybe rather than writing the pdf in terms of probability, you could use the (2000 per square mile) to write number of houses as a function of x & y

then f_{X,Y}(X=x,Y=y)dxdy would represent the number of houses contianed in x & x+dx, and y + y+dx
 
Okay, so the number of houses would be given by 2000xy, if I am not mistaken.
<br /> f_{X,Y}(X=x,Y=y)dxdy <br /> isn't clear to me. Wouldn't you need to multiply by 2000?
 
Kalinka35 said:
Okay, so the number of houses would be given by 2000xy, if I am not mistaken.

what do you mean by 2000xy?
 
Since there are 2000 houses per square mile, the total number of houses over a certain area would be 2000 time that area. The area would be given by xy. So in this case we have 2000(1.5)(1.5)=4500.
Is that what you meant?
 
ok, so that sounds like the amount of houses given X<x, and Y<y
 
think about how this realtes to the density function, and how the density function is related to area
 
Well, what I've come up with is that for small dx and dy, the number of houses in that small region, R, is 2000 dxdy.
So we essentially want to find the number of residences in R over the total residences in the town. The total residences in the town is known to be 4500 so to find the residences in R would you integrate 2000dxdy? I guess that doesn't really make sense to me.
 
sounds reasonable to me, think its coming together...

notice the pdf is 2000dxdy, which is the number of house in the area dxdy at (X=x,Y=y), in this case constant

integrating gives the cdf, which you gave earlier

F_{X,Y}(X\leq x,Y \leq y) = \int_0^y \int_0^x f_{X,Y}(X=x,Y=y)dxdy =\int_0^y \int_0^x 2000dxdy = 2000xy

so instead of working in probability, its just shifted to number of houses
 

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