Showing that $Y_n =\min X_i$ converges in law to a null random variable.

Question

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent random variables with the same probability distribution, whose cumulative distribution function $F$ is the same and verifies:

$$F(x) = 0 \text{ if } x<0 $$ $$0<F(x) <1 \text{ if } 0<x<1 $$ $$F(x) > =1 \text{ if } x>1$$ We define: $$Y_n = \underset{i \in \{1,\dots n \} }{\min X_i} $$ Show that $Y_n$ converges in law to a random variable who's equal to $0$

I first tried to understand what a random variable who's equal to 0 means. It means that if $F_Y$ is its cumulative distribution function, then $F_Y(t) = 0$ for all $t<0$ and $F_Y(t) = 1$ for all $t \geq 0$.

If $F_{Y_{n}}$ is the cumulative distribution function of $Y_n$, then it is equal to $$F_{Y_n}(t) = 1 - \prod^{n}_{k=1}(1 - F_k(t)) $$

Thus of $t < 0$ I have $F_{Y_n}(t) = 0$. Now if $t>0$ then $F_{Y_n}(t) = 1$ because $0<F_k(t)<1$ thus $\prod^{n}_{k=1}(1-F_k(t))$ converges to $0$.

My problem appears when I choose $t = 0$. As I have no information(well at least no direct information) about $F_k(0)$.Based on the result I should found, I should have $F_k(0)=0$. Now I need to prove this. I know that is increasing, and right-continuous. Does this allow me to state that $F_k(0)=0$?

Answer 1

In order to show that $Y_n \to Y=0$ in law you have to prove that $F_{Y_n}(t) \to F_Y(t)$ for all continuity points $t$ of $F_Y$. Since $0$ is not a continuity point of $F_Y$, there is nothing to show for $t=0$.

A remark concerning your proof. Instead of

Now if $t>0$ then $F_{Y_n}(t)=1$ [...]

you probably wanted to write

Now if $t>0$ then $F_{Y_n}(t) \to 1$ [...]