Learning Probability: Solving a Density Function Problem

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SUMMARY

The discussion focuses on calculating probabilities using a given density function, f(x,y) = x^2 + xy/3, defined for 0 Y), Prob(Y < 1/2 and X < 1/2), and Prob(X + Y < 1). The solutions involve double integrals over defined regions, specifically integrating the density function to find the desired probabilities.

PREREQUISITES
  • Understanding of probability theory and random variables
  • Familiarity with double integrals in multiple-variable calculus
  • Knowledge of density functions and their applications
  • Ability to visualize geometric regions defined by inequalities
NEXT STEPS
  • Study the concept of joint probability distributions and their properties
  • Learn how to perform double integrals over specified regions
  • Explore the use of geometric interpretations in probability problems
  • Investigate the implications of conditional probabilities in multiple dimensions
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This discussion is beneficial for students learning probability, educators teaching probability concepts, and anyone interested in applying calculus to solve probability problems involving multiple variables.

Niels
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I'm trying to learn probability on my own and I'm stuck.

My multiple-variable-calculus is not so strong so the following problem got me stuck.

I have density function

f(x,y) = x^2 + xy/3 for 0<x<1; 0<y<2 otherwise 0

And I need to calculate Prob(X > Y). X and Y are random variables.

I know how to do Prob(X <= 0.5) etc.

Also would be nice if someone could explain Prob(Y < 1/2 and X < 1/2) and
Prob(X+Y < 1)

Thanks
/Niels
 
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Niels said:
I'm trying to learn probability on my own and I'm stuck.

My multiple-variable-calculus is not so strong so the following problem got me stuck.

I have density function

f(x,y) = x^2 + xy/3 for 0<x<1; 0<y<2 otherwise 0

And I need to calculate Prob(X > Y). X and Y are random variables.

I know how to do Prob(X <= 0.5) etc.
Just to make sure, this is just integrating over all values of Y, such that X <= 0.5, which is a rectangle.

Niels said:
Also would be nice if someone could explain Prob(Y < 1/2 and X < 1/2)
Similarly, this is an integration of the density function over the region where Y < 1/2 and X < 1/2. This is a square of sidelength 1/2 with one corner at the origin. It is a double integral that can be written as:
[tex]\int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} f(x,y) dy dx[/tex]
Do you know how to find P(X<1/2 OR Y<1/2) ?
Niels said:
and
Prob(X+Y < 1)
If you cannot picture the region, rewrite it in a friendlier form; ie., Y < -X + 1. You're then integrating the density function for all points below the line y = -x + 1. This integral can be written:
[tex]\int_0^1 \int_0^{-x+1} f(x,y) dy dx[/tex]
So your original problem, P(Y<X) is just the set of points below the line y=x.
 
Last edited:
thank you!
 

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