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# Cumulative Distributive Function | ISI M.Stat 2019 PSB Problem B6

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 $x$.

## 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.

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 $x$.

## 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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