# 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
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

You would use induction. Assume it is true for n and then show for n+1. Also write P{X1+X2+...+Xn +Xn=1>x] as P{X1+X2+...+Xn>x-Xn+1] which equals ∫P{X1+X2+...+Xn>x-y]f(y) dy where fis the density for Xn+1. Then try to express the ratio as a factor times the ratio in the form of the induction hypothesis for n. Then limit of product should be the product of the limits evaluate the two limits. One goes to 1 by the induction hypothesis and then the other factor should also converge to 1.

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 I need some more input here, I am not quite getting there. – Chris May 25 '12 at 2:57 Can't help without knowing where you are stuck. Were able to take the left hand side and factor it into two terms with one being the ratio in the asumption that the result is true for n? – Michael Chernick May 25 '12 at 3:10 $\frac{P(X_1+...+X_n>x-X_{n+1}) }{P(X_1 > x)+...+P(X_{n+1}>x)}$. 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 multiply numerator and denominator by P(X1>x)+P(X2>x)+...+ P(Xn>x). – Michael Chernick May 25 '12 at 3:32 I agree with Michael. If you can prove that $\lim_{x\to\infty}\frac{\sum_{i=1}^n P(X_i>x)}{\sum_{i=1}^{n+1} P(X_i>x)}=1$, then you're done. Use the info on $F_{X_i}$ to prove this. – Zen May 25 '12 at 20:16