Find the distribution for $X_{(1)}$ where $X_1,...,X_n \sim Unif(-\theta,0)$

Question

I am trying to find the distribution for $X_{(1)}$ where $X_1,...,X_n \sim Unif(-\theta,0)$.

My understanding is that

$$f_X(x)=\frac{1}{\theta} \Bbb I \{ -\theta \le x \le 0 \}$$

so

$$F_X(x) = \frac{x+\theta}{\theta}$$

So, the order statistics pdf for the minimum should be

$$f_{X_{(1)}}(x) = n (\frac{1}{\theta})(1-\frac{x+\theta}{\theta})^{n-1}$$ or

$$\therefore \space = \frac{n}{\theta ^n}(-x)^{n-1}$$

am I correct?

Answer 1

Your answer is correct. If this is homework, it might be beneficial to the grader to display all of your logic:

For any real-valued i.i.d. random variables $X_{1},\ldots,X_{n}$, letting $X_{(1)}\equiv\min\{X_{1},\ldots,X_{n}\}$, we get \begin{multline*} 1-F_{X_{(1)}}(x)=\mathbb{P}(X_{(1)}>x)=\mathbb{P}(X_{1}>x,\ldots,X_{n}>x)\\ =\mathbb{P}(X_{1}>x)\cdots\mathbb{P}(X_{n}>x)=\mathbb{P}(X_{1}>x)^{n}=\left(1-F_{X_{1}}(x)\right)^{n} \end{multline*} and hence $$ F_{X_{(1)}}(x)=1-\left(1-F_{X_{1}}(x)\right)^{n}. $$ In our case, $X_{1}\sim U(-\theta,0)$ so that $$ F_{X_{1}}(x)=\int_{-\theta}^{x}\frac{1}{\theta}dt=\frac{x+\theta}{\theta}\qquad\text{for }-\theta<x<0. $$ Therefore, $$ F_{X_{(1)}}(x)=1-\left(-\frac{x}{\theta}\right)^{n}\qquad\text{for }-\theta<x<0. $$ Taking a derivative, we can conclude that $$ f_{X_{(1)}}(x)=-\frac{n}{x}\left(-\frac{x}{\theta}\right)^{n}\qquad\text{for }-\theta<x<0. $$