Inequality Truth: X1 < X2 & X > 0

[EngWiPy]

Hi,

Is the following inequality true for x>0:

Pr[X1<x]<Pr[X2<x] for X1<X2?

 

[disregardthat]

No, they can be equal for some distributions.

 

[EngWiPy]

No, they can be equal for some distributions.

Ok, suppose we have n independent and identically distributed random variables x1,x2,...,xn. There is a fact that for positive numbers, the sum of the largest L<n numbers is greater than or equal the arithmetic mean multiplied by L, i.e.:

$$\sum_{i=1}^Lx_{(i)}\geq\frac{L}{n}\sum_{i=1}^nx_i$$

where $$x^{(i)}$$ are the order statistics in descending order. Then is it true to say that:
$$\text{Pr}\left[\sum_{i=1}^Lx_{(i)}<x\right]\leq\text{Pr}\left[\frac{L}{n}\sum_{i=1}^nx_i<x\right]$$

 

[Office_Shredder]

Yes, that's always true. What disregardthat was pointing out is simply that the probabilities might be equal, which you have in this post but didn't have in your OP. For example consider the stupid random variable which always takes the value of 1. Then the sum of the L largest is L, and L/n*mean is L as well. so your probabilities are either both 0 or both 1 depending on what the value of x is

 

[EngWiPy]

Yes, that's always true. What disregardthat was pointing out is simply that the probabilities might be equal, which you have in this post but didn't have in your OP. For example consider the stupid random variable which always takes the value of 1. Then the sum of the L largest is L, and L/n*mean is L as well. so your probabilities are either both 0 or both 1 depending on what the value of x is

I forgot to include it. Thanks