weak convergence of $Y_n = n\cdot \min_{1,\dots,n}X_i$

Question

I try to understand wether $Y_n = n\cdot\min_{1,\dots,n}X_i$ does converge weakly (also called in distribution) and if so, to which limit.

The $X_i$ are supposed to be independently uniformly distributed on $(0,1)$.

Thank you for hints or solutions!

Answer 1

$$Y_n = n \cdot \min_{1,\ldots\,n}X_i$$

The distribution of $X_{(1)}$ is given by,

$$P(X_{(1)} > x) = \prod_{i=1}^{n}P(X_i>x) = \prod_{i=1}^{n}(1-P(X_i\leq x)) = (1-x)^n$$

$$\implies P(X_{(1)}\leq x) = 1-(1-x)^n$$ $$\implies f_{X_{(1)}}(x) = n(1-x)^{n-1}$$

Now, applying the transformation $Y = n\cdot X_{(1)}$, we get,

$$f_{Y}(y) = f_{X_{(1)}}\left(\frac{y}{n}\right)\cdot\frac{1}{n} = \left(1-\frac{y}{n}\right)^{n-1}$$

As $n \rightarrow \infty$,

$$f_{Y}(y) = e^{-y}$$

$$F_{Y}(y) = \left\{\begin{matrix} 0 & y < 0 \\ 1 - e^{-y}& y\geq0 \end{matrix}\right.$$

Which is an $\exp(1)$ distribution.