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Let $X_1, X_2, \cdots$ be iid. If $X_i \sim Beta(1,\beta)$, find the value of $\nu$ so that $n^\nu (1-X_{(n)})$ converges in distribution.

My thoughts:

Since $X_{(n)} \to 1$ in probability, I was thinking I can use this information. However, I don't really see how this can be important.

I would like to learn how to approach these convergence problems. When you look at this problem, what are your initial thoughts? Can you please take me through the thought process?

Work in Progress

First step: Find CDF. This is the first step because we can use either CDFs or MGFs to show convergence; however, MGFs are not as powerful for location-scale transformations.

Let $Y=n^\nu(1-X_{(n)})$. Then, $F_Y(y)=P(n^\nu(1-X_{(n)} \le y) = 1 - P(X_{(n)} < 1- \frac{y}{n^\nu})=1-F_{X_{(n)}}(1-\frac{y}{n^\nu})$ $=1-F_x(1-\frac{y}{n^\nu})^n.$

Now, we know that $X \sim Beta(1, \beta)$. So, we can find this last statement by differentiating both sides.

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    $\begingroup$ First step is to find the CDF of $n^\nu (1-X_{(n)})$. $\endgroup$ Nov 28, 2019 at 21:50
  • $\begingroup$ Dose $X_{(n)}$ reprsent $nth$ order statistic? You can refer to convergence in distribution and some examples by this link: statlect.com/asymptotic-theory/convergence-in-distribution $\endgroup$
    – G.fei
    Nov 29, 2019 at 1:05
  • $\begingroup$ @StubbornAtom I added some work using the first step. However, I can't get a nice form by differentiating both sides of the equality. What do you think? $\endgroup$
    – Ron Snow
    Nov 29, 2019 at 3:38
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    $\begingroup$ No need to differentiate to get the pdf. Try taking the limit $n\to\infty$ from the CDF itself. This is where the value of $\nu$ comes into play. $\endgroup$ Nov 29, 2019 at 8:40
  • $\begingroup$ @StubbornAtom Such as $F_Y(y)=1-F_X(1-\frac{y}{n^\nu})^n \to e^{-y}$ if $\nu =1$? $\endgroup$
    – Ron Snow
    Dec 2, 2019 at 21:23

1 Answer 1

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The CDF of $X_1$ is

$$P(X_1\le y)=\begin{cases}0&,\text{ if }y<0 \\ 1-(1-y)^\beta &,\text{ if }0<y<1 \\1&,\text{ if }y>1\end{cases}$$

Therefore verify that the CDF of $n^\nu(1-X_{(n)})$ is

\begin{align} P(n^\nu(1-X_{(n)})\le y)&=P\left(X_{(n)}\ge 1-\frac{y}{n^\nu}\right) \\&=1-P\left(X_{(n)}\le 1-\frac{y}{n^\nu}\right) \\&=1-\left\{P\left(X_1\le 1-\frac{y}{n^\nu}\right)\right\}^n \\&=\begin{cases}1&,\text{ if }y>n^\nu \\ 1-\left(1-\frac{y^\beta}{n^{\nu\beta}}\right)^n &,\text{ if }0<y<n^\nu \\0&,\text{ if }y<0 \end{cases} \end{align}

You are to take the limit of this function as $n\to\infty$.

Careful consideration is required, so first find out what happens if $\nu=1$ (say). Do you get a distribution function in the limit? Then proceed to argue for other values of $\nu$.

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    $\begingroup$ Would the answer be $\nu=\frac{1}{\beta}$ since $(1+\frac{-y^\beta}{n^{\nu\beta}})^n \to e^{-y^\beta}$ if $\nu\beta=1$? $\endgroup$
    – Ron Snow
    Dec 5, 2019 at 4:25

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