Let $X_1, \dots,X_n$ be a collection of independent and identically distributed random variables, exponentially distributed with parameter $\lambda$. We would like to do a hypothesis test:
$$\begin{cases} H_0: \lambda = \lambda_1 \\ H_1 : \lambda = \lambda_2\end{cases}$$
With $\lambda_1 > \lambda_2$ constants. And the critical region $R=\{\min \{X_1,\dots,X_n \}<k \}$ with a a significance level of $\alpha$. Find $k$.
I know that
$$P(\min \{X_1,\dots,X_n \}<k | H_0)=P(X_1<k;X_2<k;...;X_n<k|H_0)=\alpha$$
The random variables are independent so:
$$P(X_1<k;X_2<k;...;X_n<k|H_0)=P(X_1<k|H_0)P(X_2<k|H_0)...P(X_n<k|H_0)$$
Also, the random variables are exponentially distributed with parameter $\lambda_1$ (as I'm assuming $H_0$ is true):
$$P(X_1<k|H_0)P(X_2<k|H_0)...P(X_n<k|H_0)=(1-e^{-\lambda_1 k})^n$$
So
$$(1-e^{-\lambda_1 k})^n=\alpha \Rightarrow \boxed{k=\frac{-\ln(1-\sqrt[^n]{\alpha})}{\lambda_1}}$$
I know my mistake is somewhere when working with the $\min$ argument.
The correct answer is
$$k=\frac{-\ln(1-\alpha)}{n \lambda_1}$$
$P(\min \{X_1,\dots,X_n \}<k | H_0) \ne P(X_1<k;X_2<k;...;X_n<k|H_0)$. The event $\min \{X_1,\dots,X_n \}<k$ is the event at least one of $X_i < k$. What you have right now is the event all $X_i < k$. Instead, express the event as $$P(\min \{X_1,\dots,X_n \}<k | H_0) = 1 - P( X_1 > k; X_2>k;...;X_n>k|H_0) = \alpha$$. Thus $$ \alpha = 1 - (1 - (1 - e^{-k\lambda_1}))^n = 1 - e^{-nk\lambda_1} \implies \frac{-\ln(1-\alpha)}{n\lambda_1} = k$$.