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For non-negative random variables $X_1,...X_n\overset{\text{iid}}{\sim} F$, $F$ is said to be a sub-exponential distribution if $$\lim_{x\rightarrow\infty}\frac{P(X_1+...X_n>x)}{P(X_1>x)}=n$$ for some $n\ge2$.

Result. If this holds for some value of $n\ge2$, then it holds for all $n\ge2$.

I want to prove this result using induction. Assume it holds for some $n=k\ge2$. We want to show that it holds for $n=k+1$ too i.e., $$\lim_{x\rightarrow\infty}\frac{P(S_k>x)}{P(X_1>x)}=k\Rightarrow \lim_{x\rightarrow\infty}\frac{P(S_{k+1}>x)}{P(X_1>x)}=k+1$$ where, $S_k=X_1+X_2+...X_k$.

Attempt. We decompose the sum as
$$P(S_{k+1}>x)=P(S_k+X_{k+1}>x)=P(S_k>x)+P(X_{k+1}>x)-P(S_k>x,X_{k+1}>x)+P(S_k<x, X_{k+1}<x,S_{k+1}>x)$$ Dividing both sides by $P(X_1>x)$, $$\frac{P(S_{k+1}>x)}{P(X_1>x)}=\frac{P(S_k>x)}{P(X_1>x)}+\frac{P(X_{k+1}>x)}{P(X_1>x)}-\frac{P(S_k>x,X_{k+1}>x)}{P(X_1>x)}+\frac{P(S_k<x, X_{k+1}<x,S_{k+1}>x)}{P(X_1>x)}$$ Taking limit, $$\lim_{x\rightarrow\infty}\frac{P(S_{k+1}>x)}{P(X_1>x)}=\lim_{x\rightarrow\infty}\left[\frac{P(S_k>x)}{P(X_1>x)}+\frac{P(X_{k+1}>x)}{P(X_1>x)}-\frac{P(S_k>x,X_{k+1}>x)}{P(X_1>x)}+\frac{P(S_k<x, X_{k+1}<x,S_{k+1}>x)}{P(X_1>x)}\right]\\=k+1+\lim_{x\rightarrow\infty}\left[-\frac{P(S_k>x,X_{k+1}>x)}{P(X_1>x)}+\frac{P(S_k<x, X_{k+1}<x,S_{k+1}>x)}{P(X_1>x)}\right]$$

Since $X_i$'s are iid, $P(S_k>x,X_{k+1}>x)=P(S_k>x)P(X_{k+1}>x)=P(S_k>x)P(X_1>x)$ $$\lim_{x\rightarrow\infty}\frac{P(S_{k+1}>x)}{P(X_1>x)}=k+1+\lim_{x\rightarrow\infty}\left[-P(S_k>x)+\frac{P(S_k<x, X_{k+1}<x,S_{k+1}>x)}{P(X_1>x)}\right]$$

How do I proceed from here? Any hints would be appreciated.

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2 Answers 2

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First, we observe that $$P(\max \{X_1,..,X_n \}>x)= 1-P(\max \{X_1,..,X_n \}\le x) = 1-F^n(x)$$

then it suffices to base on the definition of sub-exponential distribution:

$$\begin{align} \lim_{x\rightarrow\infty}\frac{P(X_1+...X_n>x)}{P(X_1>x)}&= \lim_{x\rightarrow\infty}\left(\frac{P(X_1+...X_n>x)}{P(\max \{X_1,..,X_n \}>x)}\cdot \frac{P(\max \{X_1,..,X_n \}>x)}{P(X_1>x)}\right)\\ &=\lim_{x\rightarrow\infty}\left(\frac{P(X_1+...X_n>x)}{P(\max \{X_1,..,X_n \}>x)}\cdot \frac{1-F^n(x)}{1-F(x))}\right)\\ &=\lim_{x\rightarrow\infty}\left(\underbrace{\frac{P(X_1+...X_n>x)}{P(\max \{X_1,..,X_n \}>x)}}_{\sim 1}\cdot \underbrace{(F^{n-1}(x)+...+1)}_{\sim n}\right)\\ &=n \end{align}$$

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  • $\begingroup$ So, the definition you're using for sub-exponential distribution is $$X_1+X_2+...X_n\sim \max\{X_1,...X_n\}$$ to prove the property that $$\lim_{x\rightarrow\infty} \frac{P(X_1+...X_n>x)}{P(X_1>x)}=n$$ for any $n\ge2$. Instead, using the latter as the definition, I want to show that if it holds for some $n=k$, it will necessarily hold for $n=k+1$. (In my class, sub-exponential distributions are defined this way which then implies that max $X_i$ dominates the sum exceeding $x$). Perhaps, this is incorrect? $\endgroup$ Commented Feb 7 at 8:40
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    $\begingroup$ The problem that should be proven is Given a sub-exponential distribution defined by the cdf $F$ (the sub-exponential distribution is defined as follows: for all $n\ge 2$, and for all $X_1,...,X_n \overset{\text{iid}}{\sim} F$, we have $X_1 +...+X_n \sim \max \{ X_1,...,X_n\} $) then for all $n \ge 2$ and for all $X_1,...,X_n \overset{\text{iid}}{\sim} F$: $$\lim_{x\rightarrow\infty}\frac{P(X_1+...X_n>x)}{P(X_1>x)}=n \hspace{1cm}$$ $\endgroup$ Commented Feb 7 at 9:00
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    $\begingroup$ What you are trying to prove is Given a distribution defined by the cdf $F$ (we ignore whether this distribution is sub-exponential or not, because the definition of "sub-exponential distribution" is never used), then for all $n \ge 2$ and for all $X_1,...,X_n \overset{\text{iid}}{\sim} F$: $$\lim_{x\rightarrow\infty}\frac{P(X_1+...X_n>x)}{P(X_1>x)}=n \hspace{1cm}$$ which does not use any property of "sub-exponential distribution", therefore, cannot be true. $\endgroup$ Commented Feb 7 at 9:00
  • $\begingroup$ I understand your point however, the definition given to me is what you've proved: For some $n\ge2$ (say, $n=2$), $X_1,...X_n\sim F$, $F$ is called a sub-exp distribution if it satisfies $$\lim \frac{P(X_1+...X_n>x)}{P(X_1>x)}=n.$$ What I want to prove is that if this holds for some particular $n$, it will hold for all $n\ge2$. $\endgroup$ Commented Feb 7 at 9:16
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    $\begingroup$ @zaira I see now. However, I'm not sure whether the statement holds true. Are you able to prove it for at least the easiest case (the statement holds true for $n=2$ and you want to prove it for $n = 3$)? Perhaps it is possible to find a counter-example. $\endgroup$ Commented Feb 7 at 13:18
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The question requires the condition to be true for some fixed $k$. The proof with the modified assumption that it holds for all $n\le k$ is below. I am not sure if the original statement is true (as pointed out by @NN2 ).

Assume, as $x\rightarrow\infty$ $$P(X_1+X_2+...X_n>x)\sim nP(X_1>x)$$ holds, for all $2\le n\le k$ for some $k\ge2$ . For $k+1$, $$P(X_1+X_2+...X_{k+1}>x)\\=P(X_1+X_2+...X_{k+1}>x,X_{k+1}>x)+P(X_1+X_2+...X_{k+1}>x,X_{k+1}\le x)\\=P(X_{k+1}>x)+\int_0^xP(X_1+X_2+...X_{k}>x-y)f(y)dy\\\overset{\text{asymp.}}{\approx} P(X_{k+1}>x)+k\int_0^xP(X_1>x-y)f(y)dy$$ Note that, $P(X_1+X_2>x)=P(X_2>x)+\int_0^xP(X_1>x-y)f(y)dy$.

As $x\rightarrow\infty$, using $P(X_1+X_2>x)\sim 2P(X_1>x)$ (by induction hypothesis) $$P(X_1+X_2+...X_{k+1}>x)\overset{\text{asymp.}}{\approx}P(X_{k+1}>x)+k[P(X_1+X_2>x)-P(X_2>x)]\\\approx P(X_{k+1}>x)+k[2P(X_1>x)-P(X_2>x)]\\=P(X_{k+1}>x)+kP(X_1>x)=(k+1)P(X_{k+1}>x)$$

update. as per Subexponential Distributions/Charles M. Goldie and Claudia Kluppelberg, the claim is true.

The limit holds for all $n\ge2$ if and only if it holds for $n = 2$. Also, it holds for $n = 2$ if it holds for some $n=2$.

The first part has been shown above.

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