Prove that: F(x1,x2, xn)<=min Fi(xi)

  • Thread starter Thread starter aashish.v
  • Start date Start date
aashish.v
Messages
13
Reaction score
0
1. Prove that: F(x1,x2,...xn)<=min Fi(xi)

Where F is a DF on (x1...xn)

I know that this is very intuitive. But I am not able to find proper mathematical argument for that.
 
Last edited:
Physics news on Phys.org
Assuming "DF" means (cumulative) distribution function, try writing out the left and right sides in terms of probabilities.
 
aashish.v said:
1. Prove that: F(x1,x2,...xn)<=min Fi(xi)

Where F is a DF on (x1...xn)

I know that this is very intuitive. But I am not able to find proper mathematical argument for that.

Check your PMs. You *must* show your efforts before we can offer tutorial help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top