# Not preferenced weight function

1. Oct 17, 2005

### Iraides Belandria

Dear people of this forum

Let us suppose we have infinite functions of the form

F= w1 X1 + w2X2+ w3 X3 + w4 X4+............
Where w1,w2,w3, w4 are variable weights and X1, X2, X3, X4 are fixed temperatures.

Now, I have to choose one of the above infinite functions with the requirement that this selected function should be fair, honest, in the sense that it is not preferenced to one side or another . ¿What weigths should I use, w1=w2=w3=w4 ?

2. Oct 17, 2005

### mathman

If you have an infinite set of x's, you cannot have equal weights. For a finite set you can.

3. Oct 17, 2005

### Iraides Belandria

Would you please, explain me why it is so?

4. Oct 18, 2005

### EnumaElish

What do you mean by "infinite functions"? Is it

$$\lim_{n\rightarrow\infty}\sum_{k=1}^nw_kX_k$$

?

5. Oct 18, 2005

### mathman

Since the weights must sum to 1, if you have n weights, equal means all 1/n. Try to do it with infinite number - you can't. You will have an infinite number of zeros.

6. Oct 24, 2005

### Iraides Belandria

In relation to my question, If I use equal weights , do this means that, statiscally, we are using a distibution which is not oriented to one side or another, or that it is not biased?

7. Oct 25, 2005

### HallsofIvy

Yes, in order that the distribution not be biased toward one outcome, all weights must be equal. As was pointed out before, that is not possible with an infinite number of possible outcomes.