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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?

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  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Oct 20, 2023 at 18:46
  • $\begingroup$ 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. $\endgroup$
    – user_42
    Oct 20, 2023 at 19:36
  • $\begingroup$ 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, $a_n = 1$ $\endgroup$
    – BGM
    Oct 21, 2023 at 16:26

1 Answer 1

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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.

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