Learning Probability: Solving a Density Function Problem

  • Thread starter Thread starter Niels
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
The discussion revolves around calculating probabilities using a given density function f(x,y) = x^2 + xy/3 for specified ranges of x and y. The user seeks to determine Prob(X > Y) by integrating the density function over the appropriate region, specifically below the line y = x. Additionally, they inquire about calculating Prob(Y < 1/2 and X < 1/2) and Prob(X + Y < 1), with guidance provided on setting up the necessary double integrals for these probabilities. The conversation emphasizes the importance of visualizing the integration regions to simplify the calculations. Understanding these concepts is crucial for mastering probability in the context of random variables.
Niels
Messages
10
Reaction score
0
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
 
Physics news on Phys.org
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:
\int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} f(x,y) dy dx
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:
\int_0^1 \int_0^{-x+1} f(x,y) dy dx
So your original problem, P(Y<X) is just the set of points below the line y=x.
 
Last edited:
thank you!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top