Finding the probability of the Exp($\lambda$) distributed minimum of X (given a pdf)

Question

I am trying to find the following probability of the $\min(x)$ of a exponential distribution:

$$P\left(\frac1{\lambda} \ge \frac{n\min(x)}{\ln(5)}\right) = ?$$

I have the following pdf of $\min(x)$ as: $$f_{\min}(x) = n\cdot\lambda\cdot(e^{-\lambda x})^n$$

Trying all kinds of methods like inserting $n\cdot\lambda / \ln(5)$ for $x$ in $f_{\min}(x)$ yielded unsuccessful.

I know that $1/\lambda$ is the 'true' estimator and $n\cdot \min(x)$ is the estimator.

But first of all I am not sure where to start, and secondly I am not sure where the $\ln(5)$ plays its role.

Secondly,

Knowing that $α=4/5$, how doI get the corresponding, non-asymptotic, one-sided confidence interval at level $1−α$ for $1/λ$?

Answer 1

1) If $T$ is $min(X_1,...,X_n)$, the CDF is $F_T(t)=1-e^{-\lambda n t}$

2) Transform your probability into $\mathbb{P}[T<k]$ and immediately you get that the requested probability is $F_T(\frac{ln5}{\lambda n})=1-\frac{1}{5}=\frac{4}{5}$

@Colinois, please for the future use MathJax as explained well in the link

Answer 2

Yes, simply solve in $T$

$\mathbb{P}[\frac{1}{\lambda} > \frac{nT}{ln5}]=\mathbb{P}[T< \frac{ln5}{\lambda n}]$

This is the definition of $F_T(\frac{ln5}{\lambda n})$

That's all