Convergence of independent $\mathcal U {(n,n^2)}$ random variables?

Question

What does this sequence of random variable's distribution converge to? The random variables are given as follows $$Y_n=\frac{X_n-n}{n^2}, \quad n=1,2,3, \dots$$ and $X_n\sim\mathcal{U(n,n^2)}$- a sequence of independent variables. The answer is given by: $$P(Y_n<y)=P\left(\frac{X_n-n}{n^2}<y\right)=\begin{cases}0, &y\leq 0 \\[0.2cm] \dfrac{ny}{n-1}, &0<y \leq 1- \frac{1}{n} \\[0.2cm] 1, &y>1-\frac{1}{n} \end{cases}:n\to \infty =\mathcal U(0,1)$$

The interval that I do not understand is $$0<y \leq 1- \frac{1}{n}$$ Can someone explain this?

Answer 1

I assume you understand how $F_{Y_n}(y)=P(Y_n\le y)$ was obtained $$P(Y_n<y)=P\left(\frac{X_n-n}{n^2}<y\right)=\begin{cases}0, &y\leq 0 \\[0.2cm] \dfrac{ny}{n-1}, &0<y \leq 1- \frac{1}{n} \\[0.2cm] 1, &y>1-\frac{1}{n} \end{cases}$$ but you have a problem when you have to le $n \to \infty$. There are three parts, so let's see what happens to each part:

1. The first part $y\le 0$ is independent of $n$, so remains the same when we take the limit.
2. The third part $F_{Y_n}(y)=1$ for $y\ge 1-\frac1n$ becomes $$\lim_{n\to \infty} 1-\frac1n=1-0=1$$ So $F_Y(y)=1$ for $y>1$.
3. The second part, where your problem is: The domain is $0\le y \le 1-\frac1n$ so when you let $n\to \infty$ it becomes $0\le y \le 1$. And for the function (in this domain) you have $$\lim_{n\to \infty}F_{Y_n}(y)=\lim_{n\to \infty}\frac{ny}{n-1}=\lim_{n\to \infty}\frac{y}{1-\frac1n}\cdot \frac{\not n}{\not n}=y$$ So, in total you have that $$F_Y(y)=\lim_{n\to \infty}F_{Y_n}(y)=\begin{cases}0,& y\le 0\\ y, & 0\le y\ \le 1 \\ 1, & 1\le y\end{cases}$$ which is the cdf of $\mathcal U(0,1)$.

Answer 2

You're asking two different questions, one in your title about what $Y_n$ converges to and one about the behavior of $Y_n<y$ for $0<y\leq 1-1/n$. For the latter,

$$P(Y_n<y)=P(X_n< n^2y+n),$$

and since $X_n$ is uniform in $[n,n^2]$, you'll need to figure out when $n^2y+n=n^2$, achieving the upper bound of $X_n$, which gives $y=1-1/n$. So for $y\geq 1-1/n$, $P(Y_n<y)=1$ and otherwise:

$$P(X_n<n^2y+n)=\frac{(n^2y+n)-n}{n^2-n}=\frac{ny}{n-1}$$

Answer 3

$X_n$ lies in $[n,n^2]$, so $Y_n$ lies in $[0,1-1/n]$. When $y$ lies in this latter interval, we have [from the CDF of $\mathcal{U}(n,n^2)$] the following. $$P\left(\frac{X_n-n}{n^2} < y\right) = P(X_n < n^2 y + n) = \frac{(n^2 y + n)-n}{n^2-n} = \frac{ny}{n-1}.$$

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Regarding the convergence in distribution, just take $n \to \infty$ in your computation of $P(Y_n<y)$ to obtain the result stated in your question.