How Do You Derive the Marginal Distributions of a Uniformly Distributed Disc?

[lordslytherin]

1. Homework Statement
Here is the link to the old thread, https://www.physicsforums.com/showthread.php?t=349730
i tried posting but it doesn't seem active. I don't understand how they get the second pdf as i tried it and got the first pdf. I also don't know how to do the double integral as what would the limits be?

2. Homework Equations
2. Homework Equations
Limits for checking pdf
Pdf itself

3. The Attempt at a Solution
i got the same as rosh300 in the first attempt.
Any help would be appreciated

f_(r,theta)={1.5(r^2)/pi, for r E [0,1] and theta E [0,2pi]??
0 otherwise
by doing the double integral formula this gives 1 as required. However i don't know how to find the marginal distributions of X and Y

 

[gabbagabbahey]

. I also don't know how to do the double integral as what would the limits be?

Open up any decent multivariable calculus textbook, or see here.

f_(r,theta)={1.5(r^2)/pi, for r E [0,1] and theta E [0,2pi]??
0 otherwise
by doing the double integral formula this gives 1 as required.

What is the definition of uniform (in this context)? Does $\frac{3}{2}\pi r^2$ seem uniform to you?

However i don't know how to find the marginal distributions of X and Y

Well, what is the definition of a marginal distribution? Apply that definition.

 

[lordslytherin]

Open up any decent multivariable calculus textbook, or see here.



What is the definition of uniform (in this context)? Does $\frac{3}{2}\pi r^2$ seem uniform to you?



Well, what is the definition of a marginal distribution? Apply that definition.

So it should be integral between 0 and 1 of the integral on (1/pi) between +and - (1-(x^2))^.5 dydx but when i worked this out the answer was 0.5 rather than 1.

No 3/2(pi)r^2 doesn't seem uniform but when i changed it to polar coordinates the double integral gave a value of 1.

 

[gabbagabbahey]

So it should be integral between 0 and 1 of the integral on (1/pi) between +and - (1-(x^2))^.5 dydx but when i worked this out the answer was 0.5 rather than 1.

Show the steps of your calculation.

No 3/2(pi)r^2 doesn't seem uniform but when i changed it to polar coordinates the double integral gave a value of 1.

Doesn't the problem ask for a uniform distribution?

 

[lordslytherin]

Show the steps of your calculation.



Doesn't the problem ask for a uniform distribution?

firsly i did:
integral of (1/pi) dy evaluated between (1-(x^2))^0.5 and -(1-(x^2))^0.5 giving (2/pi)(1-(x^2)^0.5
then integral of (2/pi)(1-(x^2)^0.5 dx between 0 and 1 is (2/pi) x (pi/4) = 0.5


I see what you mean about the uniform part, i was just trying lots of methods to get it to work.

 

[gabbagabbahey]

then integral of (2/pi)(1-(x^2)^0.5 dx between 0 and 1[/color] is (2/pi) x (pi/4) = 0.5

Why are you only integrating from 0 to 1, y ranges from -1 to 1 for an entire unit circle (centered at the origin).

 

[lordslytherin]

Why are you only integrating from 0 to 1, y ranges from -1 to 1 for an entire unit circle (centered at the origin).

that's why you're a homework helper, damn that's annoying thanks very much. so the marginal distribution for y is f(y)= integral of (1/pi)dx between 1 and -1 which gives 2/pi

and the marginal distribtuion of x is f(x) =integral of (1/pi)dy between (1-(x^2)^0.5 and -(1-(x^2)^0.5= (2/pi)(1-(x^2))^0.5

so they are independent?