M
MoleMan
Member
Hi,
I'm struggling to understand (mathematically) why the integral of [1 - CDF(x)] dx is equal to the integral of [xPDF(x)] dx.
I can visualise why this is true; as the probability drops, you multiply (integrate) by x for every increment of [1 - CDF(x)] and every increment of [1 - CDF(x)] is the same as the derivative i.e. PDF(x) hence leading from [1 - CDF(x)] dx to [xPDF(x)] dx but this is the 'logical' explanation.
When attempting to do the proof i.e. integrate [xPDF(x)] dx = xCDF(x) - [CDF(x)] dx ... well you can see the right hand side is not [1 - CDF(x)] dx because of the xCDF(x) term (this should be just x).
Can someone please explain the missing step?
Best,
MoleMan
I'm struggling to understand (mathematically) why the integral of [1 - CDF(x)] dx is equal to the integral of [xPDF(x)] dx.
I can visualise why this is true; as the probability drops, you multiply (integrate) by x for every increment of [1 - CDF(x)] and every increment of [1 - CDF(x)] is the same as the derivative i.e. PDF(x) hence leading from [1 - CDF(x)] dx to [xPDF(x)] dx but this is the 'logical' explanation.
When attempting to do the proof i.e. integrate [xPDF(x)] dx = xCDF(x) - [CDF(x)] dx ... well you can see the right hand side is not [1 - CDF(x)] dx because of the xCDF(x) term (this should be just x).
Can someone please explain the missing step?
Best,
MoleMan