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Let $X_1,X_2,... $ be independent exponentially distributed random variables such that $X_n$ has parameter $\lambda_n$. Let $S_n:=\sum_{i=1}^n X_i$. Show that if $\sum_{n=1}^\infty \frac{1}{\lambda_n}<\infty$, then $S_n\rightarrow S$ almost surely, where $S$ is some random variable which is almost surely finite.

It seems this can be proved by using Kolmogorov three series theorem but I'm not sure if this is the easiest way to prove it. Could somebody give any hints, thank you.

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    $\begingroup$ Yes, Kolmogorov's three-series theorem is (almost always) the easiest way to show that a sum converges almost surely. Which of the three series are you having trouble proving convergence of? $\endgroup$ – Marcus M Dec 6 '15 at 21:36
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    $\begingroup$ $E(S)=\lim_{n\to\infty}E(S_n)=\sum_{n=1}^{\infty}\lambda_n^{-1}<\infty$ - hence $S<\infty$ a.s. $\endgroup$ – A.S. Dec 6 '15 at 21:42
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HINTS: Let $A = 1$.

i) To show that $\sum \mathbb{P}(|X_n| \geq 1)$ converges, note that $e^{-x} < \frac{1}{x}$ for all $x > 0$.

ii) To see that $\sum \mathbb{E}\left(X_n \mathbf{1}_{X_n \leq 1} \right)$ converges, note that $\mathbb{E}\left(X_n \mathbf{1}_{X_n \leq 1} \right) \leq \mathbb{E} X_n$.

iii) To see that $\sum \text{Var}\left(X_n\mathbf{1}_{X_n \leq 1}\right)$ converges, note that $\text{Var}\left(X_n\mathbf{1}_{X_n \leq 1}\right) \leq \text{Var}\left(X_n\right)$ and that if $\{a_n\}$ is positive, then $\sum a_n < \infty$ implies $\sum a_n^2 < \infty$.

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  • $\begingroup$ Thank you very much, now I understand the convergence. :) $\endgroup$ – Jiangnan Yu Dec 6 '15 at 22:32
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$$\infty > \sum_{n} \frac{1}{\lambda_n} = \sum_{n} E[X_n] = E[\sum_{n} X_n]$$

$$\to \infty > \sum_{n} X_n = \lim_m \sum_{n=1}^{m} X_n = \lim_m S_m = \lim_n S_n$$

Choose $S_n = \sum_{n} X_n$

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