Consider $X_{n} \sim \operatorname{Exp}(\lambda_{n})$ for each $n \in \mathbb{N}$, where $(\lambda_{n})_{n \in \mathbb{N}} \subset \mathbb{R}$ with $\lambda_{n} > \varepsilon$ for all $n \in \mathbb{N}$, and $\varepsilon > 0$ is a fixed constant. Additionally, we assume that $(X_{n})_{n \in \mathbb{N}}$ is a sequence of independent random variables. Now, let $m_{n} = \min \{X_{1}, \ldots, X_{n}\}$ for all $n \in \mathbb{N}$. Determine the almost sure limit of the sequence $(m_{n})_{n \in \mathbb{N}}$.
Attempt/Idea:
To determine the almost sure limit of $m_n$, we can use the information about the distribution of $\min\{X_1, \ldots, X_n\}$.
As mentioned earlier, we have $\min\{X_1, \ldots, X_n\} \sim \operatorname{Exp}(\lambda_1 + \ldots + \lambda_n)$. Since $\lambda_n > \varepsilon$ for all $n \in \mathbb{N}$ and a fixed $\varepsilon > 0$, we have:
$\lambda_1 + \ldots + \lambda_n > n \varepsilon$
As the exponential distribution is a continuous distribution, the probability that $\min\{X_1, \ldots, X_n\}$ exceeds a certain value $x$ is given by:
$P(\min\{X_1, \ldots, X_n\} > x) = e^{-(\lambda_1 + \ldots + \lambda_n) x}$
For $x > 0$, we can estimate this probability as:
$P(\min\{X_1, \ldots, X_n\} > x) < e^{-n \varepsilon x}$
Now, let's consider the series $\sum_{n=1}^{\infty} P(\min\{X_1, \ldots, X_n\} > x)$:
$\sum_{n=1}^{\infty} P(\min\{X_1, \ldots, X_n\} > x) < \sum_{n=1}^{\infty} e^{-n \varepsilon x}$
The right-hand side is a geometric series that converges when $e^{-\varepsilon x} < 1$. This is equivalent to $\varepsilon x > 0$, which holds true for $x > 0$. Hence, the series converges for all positive $x$.
According to the Borel-Cantelli lemma, if a series of probabilities $(p_n)$ converges, then the event that infinitely many of the independent events $A_n$ occur happens with probability 0 when $\sum_{n=1}^{\infty} p_n < \infty$, and with probability 1 when $\sum_{n=1}^{\infty} p_n = \infty$.
In our case, we have shown that the series $\sum_{n=1}^{\infty} P(\min\{X_1, \ldots, X_n\} > x)$ converges for all positive $x$. Therefore, the event $\min\{X_1, \ldots, X_n\} > x$ occurs with probability 0 for infinitely many $n$. This implies that the almost sure limit of $m_n$ for all $p$ is 0.
In summary, the answer is: The almost sure limit of $m_n$ is 0.
Does this argumentation make sense? I would appreciate some feedback on this attempt.