This problem based on Cumulative Distributive Function gives a detailed solution to ISI M.Stat 2019 PSB Problem 6, with a tinge of simulation and code.

## Problem

Suppose \(X_{1}, X_{2}, \ldots, X_{n} \) is a random sample from Uniform\((0, \theta)\) for some unknown \(\theta>0\). Let \(Y_{n}\) be the minimum of \(X_{1}, X_{2}, \ldots, X_{n}\).

(a) Suppose \(F_{n}\) is the cumulative distribution function (c.d.f.) of \(n Y_{n}\).

Show that for any real \(x, F_{n}(x)\) converges to \(F(x)\), where \(F\) is the c.d.f. of an exponential distribution with mean \(\theta\).

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

### Prerequisites

- Order Statistics
- Convergence in Distribution

## Solution

#### Distribution of \(Y_{n} = X_{(1), n}\)

\(F_{Y_n}(x) = 1 – (1-\frac{x}{\theta})^{n} \) from the results of the order statistics

Now, let’s compute \(F_n(x)\).

\(F_{n}(x) = \text{Pr}\left(nY_{n} \leq x\right) = \text{Pr}\left(Y_{n} \leq \frac{x}{n}\right) = 1 – (1-\frac{x}{ \theta n})^{n} \).

Observe that \(F_{n}(0) = 0\). We will need it in the second part.

So, \( \lim_{n \rightarrow \infty}F_{n}(x) = 1 – \lim_{n \rightarrow \infty} (1-\frac{x}{\theta n})^{n} = 1 – e^{- \frac{x}{\theta}} = F(x) = F_{Y}(x) \), where \(Y\) ~ exp(mean \( \theta \)).

Let’s add a computing dimension to it, we will verify the result using simulation.

Let’s take \( \theta = 2\).

- Generate \(X_{1}, X_{2}, \ldots, X_{n} \) is a random sample from Uniform\((0, \theta)\). n = 100000.
- Select \(Y_n = X_{(1)}\). Calculate \( nY_n\).
- Repeat this process k = 1000 times.
- Plot the histogram.
- Plot and match with exp(mean \( \theta \)).

```
v = NULL
for (i in 1:100000) {
r = runif(100, 0, 2)
m = 100*min(r)
v = c(v,m)
}
hist(v, freq = FALSE)
x = seq(0, 10, 0.0001)
curve(dexp(x, 0.5), from = 0, col = "red", add = TRUE)
require(vcd)
require(MASS)
#Fitting the data with exponential distribution
fit <- fitdistr(v, "exponential")
fit
# rate = 0.504180128 ( which is close to 0.5 = the actual rate)
```

The following is the diagram.

We need to compute this \(\lim_{n \rightarrow \infty} \mathrm{P}\left(n\left[Y_{n}\right]=k\right)\)

So, observe that for any fixed \( k \in \mathbb{N},\) , \( \forall n > k, [Y_n] = k/n \notin \mathbb{Z} \).

Hence, \(\mathrm{P}\left(n\left[Y_{n}\right]=k\right) = 0 \) . Hence, for k > 0, \(\lim_{n \rightarrow \infty} \mathrm{P}\left(n\left[Y_{n}\right]=k\right) = 0\).

Let’s compute separately for k = 0.

\(\mathrm{P}\left(n\left[Y_{n}\right]=0 \right) = \mathrm{P}([Y_{n}]=0) = \mathrm{P}( 0 \leq Y_{n} < 1) = \mathrm{P}( 0 \leq nY_{n} < n) = F_n(n)\).

\(\lim_{n \rightarrow \infty} \mathrm{P}\left(n\left[Y_{n}\right]=0 \right) = \lim_{n \rightarrow \infty} F_n(n) = F(\infty) = 1 \) .

Thus, this problem is just an application of the first part.

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