# Aysmptotic Approximation

1. Feb 2, 2012

### EngWiPy

Hello,

I am reading a paper, and the author claimed that in asymptotic sense as M goes to infinite:

$$\sum_{i=1}^M\sum_{l=0}^L|h_i(l)|^2=M$$

where:

$$\sum_{l=0}^L\mathbb{E}\left\{|h_i(l)|^2\right\}=1$$.

How is that asymptotic follows?

2. Feb 3, 2012

### mathman

There are undefined symbols. What kind of things are hi(l)? What is the meaning of E?

3. Feb 3, 2012

### EngWiPy

E is the expectation, and h are random variables.

I got it, it is just by using the law of large numbers.

Thanks

4. Feb 4, 2012

### mathman

In the equation cited, what is asymptotic? Ratio -> 1 (true) or difference -> 0 (false)?

5. Feb 5, 2012

### EngWiPy

as M goes to infinite.

6. Feb 5, 2012

### mathman

I know you mean as M -> ∞, but my question is what is supposed to happening as M -> ∞, expression divided by M -> 1 (true) or expression minus M -> 0 (false)?

7. Feb 5, 2012

### EngWiPy

I am sorry, I did not understand you quiet well. Can you say it in different way, please?

8. Feb 5, 2012

### chiro

I have the feeling that he is dividing by M.

9. Feb 6, 2012

### EngWiPy

If he is dividing by the M the result would be 1 not M.

10. Feb 6, 2012

### mathman

Let ∑(M) be the double sum you are talking about. There are two ways of expressing the limit as M -> ∞, {∑(M)}/M -> 1 (true) or ∑(M) - M -> 0 (false).

11. Feb 6, 2012

### EngWiPy

$$\lim_{M\xrightarrow{}\infty}\frac{1}{M}\sum_{i=1}^M\sum_{l=0}^L|h(l)|^2=1$$

12. Feb 7, 2012

### mathman

True. The author is making use of one of the fundamental theorems of probability theory - the law of large numbers.

13. Feb 7, 2012

### EngWiPy

Yes, right. Thanks

14. Feb 7, 2012

### mathman

There is one caveat: hi independent of hj for i ≠ j.