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Random variables $X_1, X_2, \dots$ are independent and have a common exponential distribution with the parameter $3$. Prove that the sequence of random variables $$ \frac{X_1 + \dots + X_n + 3n} {X_1^2 + \dots + X_n^2} $$ converges almost surely and calculate its limit.

I knew that when the random variables have a exponential distribution with the parameter $3$, then every $X_i$ has density function: $$f(x) = \begin{cases} 1 − \exp{\frac{−x}{3}},x>0 \\ 0, x \le 0 \end{cases}$$

but I don't know what it gives me to solve the problem.

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    $\begingroup$ Do you mean uniform distribution, or do you mean exponential distribution? Also you wrote the CDF (not pdf) of a random variable possessing an $\exp(1/3)$ distribution. $\endgroup$
    – Matthew H.
    Jun 28, 2022 at 13:41
  • $\begingroup$ @MatthewH. Sorry, I rewrote the content incorrectly. Exponential distribution of course, I edited the post. $\endgroup$
    – john1235
    Jun 28, 2022 at 13:45
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    $\begingroup$ Two facts are needed: (a) Strong Law of Large Numbers and (b) If $Y_n\to Y$ and $Z_n\to Z$ a.s. then $Y_n/Z_n\to Y/Z$ a.s. $\endgroup$
    – jlammy
    Jun 28, 2022 at 13:48

1 Answer 1

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We have a sequence of i.i.d. (independent identically distributed) random variables $X_1,X_2,...$, where each $X_i\sim Exp(3)$. We know that their pdf is $f(x)=\lambda e^{-\lambda x}I_{(0,+\infty)} = 3e^{-3x}I_{(0,+\infty)}$. Consider the sequence of random variables: $$ \frac{\sum_{i=1}^n X_i + 3n}{\sum_{i=1}^n X_i^2} $$ Let's rewrite it as: $$ \frac{\sum_{i=1}^n X_i + 3n}{\sum_{i=1}^n X_i^2}\frac{n}{n} =\frac{n}{\sum_{i=1}^n X_i^2}\frac{\sum_{i=1}^n X_i + 3n}{n} = \frac{n}{\sum_{i=1}^n X_i^2}(\frac{\sum_{i=1}^n X_i}{n} + 3) $$ Let's focus on each term indipendently. By the Strong Law of Large Numbers, we know that: $$ \frac{\sum_{i=1}^n X_i^2}{n} \to E(X_i^2) = \frac{2}{\lambda^2} = \frac{2}{9} $$ $$ \frac{\sum_{i=1}^n X_i}{n} \to E(X_i) = \frac{1}{\lambda} = \frac{1}{3} $$ almost surely. Then, by using the continous mapping theorem, we can say that: $$ \frac{n}{\sum_{i=1}^n X_i^2}(\frac{\sum_{i=1}^n X_i}{n} + 3) \to \frac{9}{2}(\frac{1}{3}+3) = 15 $$ almost surely.

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