# Limiting Distribution of $n\left[Y_n\right]$ where $Y_n$ is the minimum of a sample of size n from Uniform$\left(0,\theta\right)$ distribution

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.

• Since $[Y_n]$ converges to $1,$ which makes (b) uninteresting, did you perhaps intend to write "$[nY_n]$" instead of "$n[Y_n]$"? – whuber Jun 9 at 19:07
• 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. – Soumya Mukherjee Jun 9 at 20:52
• 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. – whuber Jun 9 at 20:56
• Could you provide me a proof of what you stated? It would be very helpful. – Soumya Mukherjee Jun 9 at 21:06
• Then, if you solved the problem, you can answer it yourself! – kjetil b halvorsen Jun 10 at 10:16