Find $\lim_{n \rightarrow \infty} P(n[Y_n]=k)$

Question

Suppose $X_1,X_2,...,X_n$ is a random sample from $U(0,\theta)$ for some unknown $\theta>0$.Let $Y_n$ be the minimum of $X_1,X_2,..,X_n$. Find $\lim_{n \rightarrow \infty} P(n[Y_n]=k)$ for $k=0,1,2,...$ where $[x]$ denotes the largest integer less than or equal to $x$.

My approach:

We first write down the pdf of $Y_{n}$:

$f(y)=\frac{n}{\theta}(1-\frac{y}{\theta})^{n-1}$

Now, let $Z=[Y_n]$

$P(Z=z)=P([Y_n]=z)=P(z \le Y_n < z+1)=\int_{z}^{z+1} \frac{n}{\theta}(1-\frac{y}{\theta})^{n-1} dy=(1-\frac{z}{\theta})^n-(1-\frac{z+1}{\theta})^n$

Now, plugging in $\frac{k}{n}$ for $z$,we have $P(Z=\frac{k}{n})=(1-\frac{k}{n \theta})^n-(1-\frac{k+n}{n \theta})^n$ , but I cannot find the limit . Perhaps, I have made a mistake somewhere. Please help!!

Answer 1

Consider $k>0$, if $n>k$, then $\frac{k}{n} \notin \mathbb{Z}$, hence

$$P\left( n[Y_n]=k\right)=P\left( [Y_n]=\frac{k}{n}\right)=0$$

Hence if $k>0$, $\lim_{n \to \infty}P(n[Y_n]=k)=0.$

Now consider the case where $k=0$.

Let's first consider the CDF of $Y_n$. Let $y \in (0, \theta),$

\begin{align} P(Y_n \le y) &= 1-P(Y_n > y) \\ &= 1-P\left( \min(X_1, \ldots, X_n) > y\right) \\ &= 1- \prod_{i=1}^n P(X_i > y) \\ &= 1- \left( \frac{\theta-y}{\theta}\right)^n \end{align}

That is

$$P(Y_n \le y ) =\begin{cases} 0 & , y \le 0 \\1-\left( \frac{\theta-y}{\theta}\right)^n & ,0 < y < \theta\\ 1 &, y \ge \theta\end{cases}$$

Then

\begin{align} P (n [Y_n] = 0) &= P([Y_n]=0) \\ &=P(0 \le Y_n < 1)\\ &= P(Y_n \le 1) \\ &= \begin{cases} 1-\left( \frac{\theta-1}{\theta}\right)^n & \theta > 1\\ 1 & \theta \le 1\end{cases} \end{align}

Hence $$\lim_{n \to \infty}P (n [Y_n] = 0) = 1$$