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I need to find $\lim_{n\to\infty} (X_{1} +...+ X_{n})$, where each $X_{i}$ is an $exp(\lambda)$ RV.

What I've tried to do is use the law of large numbers which gives us that:

$\lim_{n\to\infty} (\frac{X_{1} +...+ X_{n}}{n})= \frac{1}{\lambda}$ and so,

$\lim_{n\to\infty} (X_{1} +...+ X_{n})= \lim_{n\to\infty} (\frac{n(X_{1} +...+ X_{n})}{n}) = (\lim_{n\to\infty} n)\times(\lim_{n\to\infty}\frac{(X_{1} +...+ X_{n})}{n}) = \infty$

Is this correct?

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    central limit theorem? – 2017-02-11
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    "Is this correct?" Yes. Adding some "almost surely" here and there might help to make this even clearer. – 2017-02-11
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    @Guangliang No thanks. – 2017-02-11

1 Answers 1

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I hope this hint is helpful:

Assuming you're looking for the limit in distribution and that $X_1, X_2, ...$ are independent, you have for fixed $N \in \mathbb{N}$ that the cdf of $\sum_{i=1}^NX_i$ equals (see Erlang distribution)

\begin{align*} F_{\sum_{i=1}^NX_i}(x) = \bigg[1 - e^{-\lambda x}\sum_{i=0}^{N-1}\frac{(\lambda x)^i}{i!}\bigg]\,\boldsymbol{1}_{[0,\infty)}(x). \end{align*}

Then for all $x \geq 0$ you get $F_{\sum_{i=1}^NX_i}(x) \rightarrow_N 1- e^{-\lambda x}e^{\lambda x} = 0$ and for all $x < 0$ you have by definition of $F_{\sum_{i=1}^NX_i}\colon \,F_{\sum_{i=1}^NX_i}(x) = 0.$

Since the properties of an cdf aren't fulfilled, I would say that $\sum_{i=1}^NX_i$ doesn't converge in distribution.