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=...
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.
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.
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...
I am doing a some practice questions for stats and i tried to integrate this to get 1 but i can't so what are the appropriate limits and how would i go about finding the marginal distribution of x and y? Thanks