# Convergence in distribution for $Y_n=\min(X_1,X_2,\ldots,X_n)$ with i.i.d. Uniform$(0,1)$ random variables

I have this exercise where I want to check my solutions:

Let $$X_1 , X_2 , X_3 , \ldots$$ be a sequence of i.i.d. Uniform $$(0,1)$$ random variables. Define the sequence $$Y_n$$ as $$Y_n=\min(X_1,X_2,⋯,X_n)$$.

Prove the following convergence results independently (i.e, do not conclude the weaker convergence modes from the stronger ones).

$$Y_n \overset{d}{\rightarrow} 0$$

My solutions:

We can write

\begin{align} F_{Y_n}(y)&=P(Y_n\le y) \\&=1−P(Y_n>y) \\&=1−P(X_1>y,X_2>y,⋯,X_n>y) \\&=1−P(X_1>y)P(X_2>y)⋯P(X_n>y) \end{align}

(since $$X_i$$'s are independent)

$$=1−(1−F_{X_1}(y))(1−F_{X_2}(y))⋯(1−F_{X_n}(y))=1−(1−y)^n$$.

Therefore, we conclude $$\lim_{n\to\infty}F_{Y_n}(y)=\left\{\begin{matrix} 0 & y \leq 0 \\ 1& y>0 \end{matrix} \right.$$.

Here I cannot see the convergence in distribution since $$\lim_{n\to \infty}F_{Y_n}(y)$$ could also take value 1.

Where is my error?

• Your work is correct. What do mean $\lim F_{Y_n}(y)$ it could also take the value $1$? It is supposed to be piecewise constant function as you found. Jun 25, 2023 at 10:24
• what it means is $Y_n \rightarrow 0$ in distribution. Jun 25, 2023 at 10:26
• I cannot see the convergence to $0$ here. if $y>0$ then our function will converge to 1,right? @athalhaids Jun 25, 2023 at 10:27
• I think you are confusing what convergence in distribution means. It means the cdfs $F_{Y_n}$ converge to the cdf $F_0$ of the constant variable $0$. It doesn't mean $Y_n$ themselves converge to $0$; that is a much stronger form of convergence which we are not considering here. Jun 25, 2023 at 10:28
• Thank you very much, I understand now my error @athalhaids Jun 25, 2023 at 10:35

Your solution is almost correct, because the equality $$\mathbb P(X_1>y)=1-y$$ is valid only for $$y\in [0,1]$$. However, the cases $$y<0$$ and $$y>1$$ are trivial.
Let $$F$$ be the cumulative distribution function of the constant random variable equal to $$0$$ (thus, $$F(t)=0$$ if $$t<0$$ and $$F(t)=1$$ if $$t\geqslant 0$$). You showed that $$F_n(y)$$ converges to $$F(y)$$ at each continuity point of $$F$$, that is, for $$y\neq 0$$ since the only discontinuity point of $$F$$ is at $$0$$. By the way, the convergence fails for $$y=0$$ as you noticed.