|chessguy||this might be helpful http://en.wikipedia.org/wiki/Joint_probability_density_function#Probability_function_associated_to_multiple_variables|
|MrFreak||yeah i'm reading that right now|
I guess it would be like
Pr(x >= 0, 0 <= y <= e^-x) Integral 0 to infinity Integral 0 to 1 e^-x dx dy?
does that look right?
|kmh||MrFreak : well do you have info for outside the region ?|
|MrFreak||info for outside the region?|
i assume outside the region is always 0
|kmh||the probabilty outside the region is 0 ?|
you need an interval for x though
as the area under the curve needs to be limited
or bounded rather
|MrFreak||interval for x would be 0 to infinity i think|
|kmh||hmmm yes if the improper integral exists that's good enough|
|MrFreak||e^-x approaches 0 asymptotically|
|kmh||but you need the area under the curve first so that you can norm the measure to 1|
oh it is 1 anyway :)
|MrFreak||so how do i set up the integral for the area under the curve?|
|jhardin||quick question, can anyone think of any hermitian unitary matrices that aren't just a string of +-i's along the diagonal, or antidiagonal|
|kmh||MrFreak : i*m a bit rusty too with n-dim distribution, but i think it looks like that|
and 1_[0,e^(-x)](u) is the indicator function
|Copter||X_n+1= 3X_n/ (X_n^2+X_n+1), Xn=2007. I wanted to prove that its bounded above by 1.|
So I did X_n+1 <=1 , 3X_n/ (X_n^2+X_n+1) <=1 and reached (X_n-1)^2 >=0 which is true.
Can I conclude Xn is bounded by 1 from above?
|MrFreak||so how simplyy do i find the area under the curve y = e^-x|
from x = 0 to infinity?
the joint density will just be 1/area, since it's uniformly distributed
integral from 0 to infinity of e^-x?
|chessguy||that's just a simple improper integral|
|MrFreak||heh, i'm a CS major, and the last time i did an integral was calc 3 two years ago|
|chessguy||take the limit as M goes to infinity of the integral from 0 to M|
|MrFreak||i sort of hazily remember the terms integration by parts and u substitution|
|chessguy||i was a CS major too :)|