# Is This Lower Bound Right?

1. Apr 13, 2012

### EngWiPy

Hello,

I have this quantity:

$$\frac{1}{\sum_{m=1}^NX_m^{-1}}\geq\frac{1}{N\underset{m}{\text{max }}X_m^{-1}}=\frac{\underset{m}{\text{min}}X_m}{N}$$

Is that true?

2. Apr 13, 2012

### chiro

The inequality:

$$\frac{1}{\sum_{m=1}^NX_m^{-1}}\geq\frac{1}{N\underset{m}{\text{max }}X_m^{-1}}$$

Is definitely true because a simple supremum argument (NxMax(X_m) >= Sum of all X_m's) which means the reciprocal will change it from >= to <=.

In terms of the equality, I don't think this is right. As a simple counterexample let the realization of four random variables be X = {1,2,3,8}. Min(X) = 1, Max(X) = 8. Min(X)/4 = 1/4 and 1/(4xMax(X)) = 1/32 which is clearly in violation of your expression.

However if the realization is {1,1,1,1} then you get Min(X) = 1, Max(X) = 4 which means Min(X)/N = 1/4 and 1/(4x1) = 1/4, but is a very unique case.

Did you mean to have some kind inequality for the rightmost term of your expression?

3. Apr 13, 2012

### EngWiPy

In you calculation, you did not consider the inverse. Using your example: Xm={1,2,3,4}, then Xm^-1={1,1/2,1/3,1/4}. Than: min(X)/4=1/4, and 1/(4max(X^-1))=1/4. Right?

4. Apr 13, 2012

### chiro

Well if you want to do it that way, then define a new variable Y = reciprocal of X realizations and apply the reasoning that I did above.

Also you made a mistake with your example since the minimum value in your set is 1/4 so dividing that by 4 gives you 1/16. Also the maximum value is 1 which means 1/(4 x max) = 1/4.

You need to be clear about what you are describing if this is not the case.

5. Apr 13, 2012

### EngWiPy

From the reciprocal set, I choose the max not the min, so it is 1.

6. Apr 13, 2012

### chiro

So does this mean your inequality is wrong above? If so can you please change it to what you are trying to describe (for example if the max should be min, then please change it).

7. Apr 13, 2012

### EngWiPy

No, it remains the same. Look at it again, please. I have max for the inverse and min for the original set, not both are max!!

8. Apr 13, 2012

### chiro

It's nearly midnight here, I'll take a look tomorrow.

9. Apr 13, 2012

### chiro

OK I see what you are saying now (very subtle!).

Yeah I think you are right with the equality involving the min and max. The easiest way to prove this is to just use standard inequalities that if x >= y then 1/x <= 1/y and then from this show how this relates to the supremum and the infinimum (or whatever the minimum is) through the changes in the inequality. Also because you will be dealing with reciprocal the N needs to be shifted to make it a strict inequality.