Hi, Roland,

I've been looking for a probabilistic proof. The following is not a full proof, but it gives a result that is very close to what you want. Indeed, I shall prove that:

[CENTER]for any , .

[LEFT]

Thus, for instanceifconverges, then it must be toward . More generally, one would require to show that does not fluctuate too much as in order to complete the proof. Otherwise, there must be more analytical proofs.

Now for the proof: Let be a family of independent Gaussian random variables with mean and variance . Then, classicaly, is Gaussian with mean and variance . Let . For any , we have (using Fubini)

,

where is the number of indices such that the sum falls inside . If we remember that , the law of large numbers shows that is finite. Moreover, it is a consequence of therenewal theoremthat:

[CENTER] .

[LEFT]This can be understood intuitively: in average, the successive sums are distant of from the next one, so that the average number of such sums that fall inside a given interval of length must be about . This only holds for large since some mixing must have happened for the starting point to be "forgotten"... I guess you can find ressource on the renewal theorem on the web.

I hope you found this interesting.

Laurent.

Edit: In fact, I've just found there is a "renewal density theorem" proved by W.L.Smith in 1954, which is stronger that the "classical" renewal theorem I quoted, and which gives exactly your result. A reference is here (there is a proof, but it is complicated) or here (no proof, but simpler statement (n.b.: the definition is lacking)). So there exists a fully probabilistic proof, but it is not quite simple. (Maybe in the present case Smith's proof could be simplified, I didn't read it)