Find $a_n,b_n$ so that $b_n(X_n-a_n)$ converges to a non-degenerate limit

I have the following problem:

Let $$X_n$$ be the maximum of a random sample $$Y_1,...,Y_n$$ from the density $$f(x)=2(1-x), x\in [0,1]$$. Find constants $$a_n,b_n$$ so that $$b_n(X_n-a_n)$$ converges in distribution to a non-degenerate limit.

I did the following:

Calculate $$F(x)=x(2-x), x\in [0,1]$$ and find distribution of $$X_n$$: $$F_{X_n}(x) = \mathbb{P}[max{Y_i} \leq x] = \mathbb{P}[Y_1 \leq x,...,Y_n \leq x] = \mathbb{P}[Y_1 \leq x]^n = x^n(2-x)^n$$

So

$$\mathbb{P}[b_n(X_n-a_n) \leq x] = \mathbb{P}[X_n \leq \dfrac{x}{b_n}+a_n] = F_{X_n}(\dfrac{x}{b_n}+a_n)=(\dfrac{x}{b_n}+a_n)^n (2-\dfrac{x}{b_n}-a_n)^n$$

My intuition here is to use that $$(1+\dfrac{x}{n})^n \rightarrow e^x$$ and choose $$a_n=1 , b_n=n$$

In that case $$\mathbb{P}[b_n(X_n-a_n) \leq x] \rightarrow e^xe^{-x}=1$$ However if I understand correctly a "non-degenerate" limit means that it should not be a constant.

Can someone please correct me? What should $$a_n,b_n$$ be?

• your calculation is not correct. You derived a formula for $F(x)$, which you said is valid for $x\in[0,1]$. If you take $a_n=1,b_n=n$, then $x/b_n \notin a_n$ if $x\in[0,1]$. In any case, it may turn out that your calculations are OK for the purposes of completing this problem; just clarify what $x$ actually is first. Oct 20, 2023 at 18:46
• To be more clear regarding what $x$ is, I can write it like that in order not to use $x$ again: let $Z_n = b_n(X_n-a_n)$ then $F_{Z_n}(z)= \mathbb{P}[b_n(X_n-a_n) \leq z] =\left(\dfrac{z}{b_n}+a_n \right)^n \left(2-\dfrac{z}{b_n}-a_n\right)^n$ when $0 \leq \dfrac{z}{b_n}+a_n < 1$. I can see why the $a_n,b_n$ that I choose were wrong, but still this does not help me to find the answer. Oct 20, 2023 at 19:36
• As what you have calculated, $F_{X_n}(x) = [1 - (1-x)^2]^n$ which will converge to $0$ for $0 \leq x < 1$. So intuitively $X_n$ converge to $1$ in distribution. In order for the limit of $b_n(X_n - a_n)$ to be non-degenerate, $a_n = 1$
– BGM
Oct 21, 2023 at 16:26

As what you have calculated, $$F_{X_n}(x) = [1 - (1-x)^2]^n$$

which will converge to $$0$$ for $$0 \leq x < 1$$. So intuitively $$X_n$$ converge to $$1$$ in distribution. In order for the limit of $$b_n(X_n - a_n)$$ to be non-degenerate, $$a_n = 1$$.

Consider the distribution of $$b_n(X_n - 1)$$ with $$b_n > 0$$. As the support of $$X_n$$ is $$[0, 1]$$, the support of $$X_n - 1$$ is $$[-1, 0]$$ and thus the support of $$b_n(X_n - 1)$$ is $$[-b_n, 0]$$. Its CDF is

\begin{align} F_{b_n(X_n-1)}(x) &= \Pr\{b_n(X_n-1) \leq x\} \\ &= \Pr\left\{X_n \leq \frac {x} {b_n} + 1\right\} \\ &= \left[1 - \left(1- \frac {x} {b_n} - 1\right)^2\right]^n \\ &= \left(1 - \frac {x^2} {b_n^2}\right)^n \end{align}

So here we require $$b_n^2$$ to go to infinity in the same order of $$n$$, in order the function to converge to the exponential function $$e^{-x^2}$$. We can take $$b_n = \sqrt{n}$$

The result is $$\lim_{n\to\infty} F_{\sqrt{n}(X_n - 1)}(x) = \lim_{n\to\infty}\left(1 - \frac {x^2} {n}\right)^n = e^{-x^2}, x < 0$$

i.e. $$\sqrt{n}(X_n - 1)$$ converge in distribution, with the CDF

$$F(x) = \begin{cases} e^{-x^2} & \text{if} & x < 0 \\ 1 & \text{if} & x \geq 0\end{cases}$$

So this is the negative of a Weibull distribution with $$k = 2$$ and $$\lambda = 1$$, which is one of the three extreme value distribution. If you take $$b_n = -\sqrt{n}$$ you get the regular Weibull distribution.