3
$\begingroup$

Suppose $X_1,X_2,\dots,X_n$ is a random sample from Uniform$(0,\theta)$ for some unknown $\theta > 0$. Let $Y_n$ be the minimum of $X_1,X_2,\dots,X_n$.

(a) Suppose $F_n$ is the CDF of $nY_n$. Show that for any real x, $F_n(x)$ converges to $F(x)$, where F is the CDF of an exponential distribution with mean $\theta$.

(b) Find $\lim_{n\rightarrow\infty} P(n[Y_n]=k)$ for $k = 0,1,2,\dots$, where $[x]$ denotes the largest integer less than or equal to x.

I have been able to solve part (a) but unable to solve part (b).Could someone help solve this question?

My attempt at solving part (b):

$P(n[Y_n]=k) = P([Y_n]=\frac{k}{n}) = P(\frac{k}{n}\leq Y_n<\frac{k}{n}+1) = (\frac{\theta-\frac{k}{n}}{\theta})^n-\{\frac{\theta-(\frac{k}{n}+1)}{\theta}\}^n$

The first terms tends to $e^\frac{-k}{\theta}$ as $n\rightarrow\infty$ but I am unable to find the limit of the second term.

It has not been mentioned whether part (a) has to be used to solve part (b) but it is very much likely that it should be the case.

$\endgroup$
8
  • $\begingroup$ Since $[Y_n]$ converges to $1,$ which makes (b) uninteresting, did you perhaps intend to write "$[nY_n]$" instead of "$n[Y_n]$"? $\endgroup$
    – whuber
    Jun 9, 2019 at 19:07
  • $\begingroup$ No the question is as stated and could you clarify why $[Y_n]$ converges to 1? I do notice that if the question involved $[nY_n]$ then the limiting distribution is a geometric distribution. However that is unfortunately not what the question states. $\endgroup$ Jun 9, 2019 at 20:52
  • $\begingroup$ When the sample size is large, the minimum $Y_n$ will be close to zero and is unlikely to be $1$ or greater, making it highly probable that $[Y_n]=0.$ (I mistyped the value as "$1$" in my first comment because I had misread your question.) Thus, it will become highly probable that $n[Y_n]=0$ as $n$ grows large. $\endgroup$
    – whuber
    Jun 9, 2019 at 20:56
  • $\begingroup$ Could you provide me a proof of what you stated? It would be very helpful. $\endgroup$ Jun 9, 2019 at 21:06
  • 1
    $\begingroup$ Then, if you solved the problem, you can answer it yourself! $\endgroup$ Jun 10, 2019 at 10:16

1 Answer 1

3
$\begingroup$

(a) We assume for simplicity that $\theta=1$. Let $X_1, X_2, \dotsc, X_n$ be iid $\mathcal{U}(0,1)$ and $Y_n$ their minimum. Then $$ \DeclareMathOperator{\P}{\mathbb{P}} \P( Y_n \le y) = 1- \P(Y_n > y) = 1-\P(X_1>y, X_2>y, \dotsc, X_n>y)=\\ 1-\P(X_1 > y)^n = 1- (1-y)^n $$ then $$ \P(nY_n \le y) = \P(Y_n \le y/n) = 1 - (1-y/n)^n$$ and it is a standard result that this converges to $1-e^{-y}$ when $n$ increases without bounds. That is the cdf of a standard exponential distribution.

(b) As the OP found in a comment, $\P(n [Y_n] =0) = \P([Y_n]=0) = \P(Y_n < 1) = \P(Y_n \le 1)=1 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.