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 $x$ .

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.

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

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.

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

Problem

Prerequisites

Solution

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

Related

Problem

Prerequisites

Solution

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

Related

Leave a ReplyCancel reply

Knowledge Partner

Cheenta Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

COMMUNITY

IN FOCUS

MORE