Find the constant that makes f(x,y) a PDF

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SUMMARY

The discussion focuses on determining the constant k that makes the function f(x,y) = e^(-kX) a valid probability density function (PDF) over the specified ranges of X and Y. The integral of the joint PDF must equal 1, leading to the equation ∫(from 0 to 1) ∫(from 0 to ∞) e^(-kX) dx dy = 1. The conclusion reached is that k must equal 1 for the function to satisfy the properties of a PDF.

PREREQUISITES
  • Understanding of probability density functions (PDFs)
  • Knowledge of double integrals in calculus
  • Familiarity with joint distributions
  • Basic concepts of limits and integration
NEXT STEPS
  • Study the properties of probability density functions in detail
  • Learn about joint probability distributions and their applications
  • Practice solving double integrals, particularly in the context of PDFs
  • Explore the implications of dependent and independent random variables
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This discussion is beneficial for students in statistics, mathematics, or related fields who are learning about probability density functions and integration techniques. It is particularly useful for those tackling joint distributions and their properties.

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Homework Statement


Find the value of k that makes this a probability density function.
The question does not specify whether X and Y are independent or dependent. That is actually another part this question.

Homework Equations


Let X and Y have a joint density function given by
f(x,y) =e^(-kX), 0<=X<inf, 0<=Y<=1,
k is a constant, and 0 elsewhere

The Attempt at a Solution


I need quite a bit of direction with this question, I have no idea where to start.
 
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should I be posting this question in the statistics section of pf?
 
alias said:

Homework Statement


Find the value of k that makes this a probability density function.
The question does not specify whether X and Y are independent or dependent. That is actually another part this question.

Homework Equations


Let X and Y have a joint density function given by
f(x,y) =e^(-kX), 0<=X<inf, 0<=Y<=1,
k is a constant, and 0 elsewhere

The Attempt at a Solution


I need quite a bit of direction with this question, I have no idea where to start.
This is the right place for this question.

How can you tell whether a function is a "probability density function"?
 
The integral of a joint PDF = 1:
f(x,y) dxdy = 1
Sorry I don't have a better response.
 
That's the direction I was looking for.

For a PDF, the integral should be 1, right?

What does k need to be so that this integral is 1?
\int_{y = 0}^1 \int_{x = 0}^{\infty} e^{-kx} dx~dy = 1
 
I end up with k=1, still not sure if I'm right...
 
got it, thanks.
 

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