Aysmptotic Approximation

[S_David]

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?

Thanks in advance

[mathman]

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

[S_David]

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

E is the expectation, and h are random variables.

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

Thanks

[mathman]

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

[S_David]

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

as M goes to infinite.

[mathman]

as M goes to infinite.

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)?

[S_David]

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)?

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

[chiro]

I have the feeling that he is dividing by M.

[S_David]

I have the feeling that he is dividing by M.

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

[mathman]

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

[S_David]

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).

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

[mathman]

\lim_{M\xrightarrow{}\infty}\frac{1}{M}\sum_{i=1}^ M\sum_{l=0}^L|h(l)|^2=1
True. The author is making use of one of the fundamental theorems of probability theory - the law of large numbers.

[S_David]

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

Yes, right. Thanks

[mathman]

Yes, right. Thanks

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