Limit of a convolution and sum of distribution functions

I need to prove an induction step. $X_i$ are independently distributed with the distribution function $1-F_i=x^{-\alpha}L_{i}(x)$ where $\alpha \geq 0$ and $L_{i}(x)$ is regularly varying (If the limit $g(a)=\lim\limits_{x\rightarrow\infty}\frac{L(ax)}{L(x)}$ is finite and nonzero for $a >0$, then L is regularly varying).

$\lim\limits_{x\rightarrow \infty} \frac{P(X_1+...+X_n>x) }{P(X_1 > x)+...+P(X_n>x)} = 1$ is true.

Now we have to show:

$\lim\limits_{x\rightarrow \infty} \frac{P(X_1+...+X_{n+1}>x) }{P(X_1 > x)+...+P(X_{n+1}>x)} = 1.$

How do we show this?

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Are the $X_i$ independent? What other information is there? What have you tried? –  Macro May 24 '12 at 3:16
Yes, they are. I don't know what to try. –  Chris May 24 '12 at 3:24
Are the $X_i$ also identically distributed? –  Zen May 25 '12 at 2:35
No, they are not. I added some information, but I don't think that would be necessary just for the induction step. –  Chris May 25 '12 at 2:56
I think the problem statement is wrong. There is a "$1 -$" missing. It should be $1-F_i(x)=x^{-\alpha} L_i(x)$. This makes more sense to me. –  Zen May 26 '12 at 2:43

I have found some info on this problem. This question is about the proof of a theorem due to Feller, to be found on volume 2 of his "Introduction to Probability Theory and its Applications" (p. 278-279). Here is a restatement.

$\mathbf{Theorem.}$ Let $X_1,\dots,X_n$ be independent random variables with distribution functions satisfying $1-F_i(x)\sim x^{-\alpha}L_i(x)$, where $L_i$ is slowly varying at infinity. Then, the convolution $G_n:=F_1\star\dots\star F_n$ has a regularly varying tail such that $$1-G_n(x)\sim x^{-\alpha}(L_1(x)+\dots+L_n(x)) \, .$$

Feller proves the case with two random variables and just states that the general result follows by induction. By the way, his proof of the $n=2$ case is a gem.

So we already know from Feller that the theorem holds for two random variables. To prove the induction step, suppose that the theorem holds for $n-1$ random variables, which means that $$1-G_{n-1}(x)\sim x^{-\alpha}(L_1(x)+\dots+L_{n-1}(x)) \, .$$ Since the sum of slowly varying functions is a slowly varying function itself, we have that $X_1+\dots+X_{n-1}$ is a random variable, independent of $X_n$, whose distribution function $G_{n-1}$ satisfies the tail hypothesis of the theorem, that is, $1-G_{n-1}(x)\sim x^{-\alpha}M(x)$, where the slowly varying $M=L_1+\dots+L_{n-1}$. By the associativity of the convolution, we know that $$G_n = F_1\star\dots\star F_{n-1}\star F_n = (F_1\star\dots\star F_{n-1})\star F_n = G_{n-1}\star F_n\, ,$$ and we are back to the (already proved by Feller) case of two random variables satisfying the hypotheses of the theorem. Therefore, $$1 - G_n(x) \sim x^{-\alpha}(M(x)+L_n(x)) = x^{-\alpha}(L_1(x)+\dots+L_{n-1}(x)+L_n(x)) \, .$$

Hence, the tail of $G_n$ satisfies the necessary property, the theorem holds for $n$ random variables, and we are done with the induction step.

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That's a good job Zen. –  Michael Chernick May 26 '12 at 23:30
Tks, Michael! Check out Feller's proof of the $n=2$ case. He does it like a boss. –  Zen May 26 '12 at 23:43
(+1) Unfortunately, I don't have Feller vol. 2 handy and am curious to see his proof. Slow variation is our friend; the simplest way I can think to make it work is to fix $\epsilon > 0$ and choose judiciously $\newcommand{\one}{\delta_1}\newcommand{\two}{\delta_2} \one = \one(\epsilon) > 0$ and $\two = \two(\epsilon) > 0$. Then, we "split" on them as follows, $$\{X_1 > (1+\one)x, X_2 > -\one x\} \cup \{-\one x < X_1 < (1+\one)x, X_2 > (1+\one)x \} \subset \{X_1 + X_2 > x\} \subset \{X_1> \two x, X_2 \leq \two x\} \cup \{X_1 \leq \two x, X_2 > \two x\} \cup \{X_1 > \two x, X_2 > \two x\} \>.$$ –  cardinal May 27 '12 at 2:33
All the unions are disjoint (provided $\two < 1$) and by independence, several terms converge to either one or zero as $x \to \infty$. The remaining terms seem to work out by using $L(\delta x)/L(x) \to 1$. –  cardinal May 27 '12 at 2:35
Zen and everybody... you guys are really really great. Are you aware of that? Thank you so much, I am beyond words! –  Chris May 28 '12 at 1:34
$\frac{P(X_1+...+X_n>x-X_{n+1}) }{P(X_1 > x)+...+P(X_{n+1}>x)}$. The problem is the last term $P(X_{n+1}>x)$ in the denominator. How do I rewrite this so that I can use somehow the induction hypothesis? –  Chris May 25 '12 at 3:28