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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 is a random sample from Uniform for some unknown . Let be the minimum of .
(a) Suppose is the cumulative distribution function (c.d.f.) of .
Show that for any real converges to , where is the c.d.f. of an exponential distribution with mean .
(b) Find for where denotes the largest integer less than or equal to .

## Solution

#### Distribution of   from the results of the order statistics

Now, let's compute . .

Observe that . We will need it in the second part.

So, , where ~ exp(mean ).

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

Let's take .

• Generate is a random sample from Uniform . n = 100000.
• Select . Calculate .
• Repeat this process k = 1000 times.
• Plot the histogram.
• Plot and match with exp(mean ).
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 So, observe that for any fixed , .

Hence, . Hence, for k > 0, .

Let's compute separately for k = 0. . .

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 is a random sample from Uniform for some unknown . Let be the minimum of .
(a) Suppose is the cumulative distribution function (c.d.f.) of .
Show that for any real converges to , where is the c.d.f. of an exponential distribution with mean .
(b) Find for where denotes the largest integer less than or equal to .

## Solution

#### Distribution of   from the results of the order statistics

Now, let's compute . .

Observe that . We will need it in the second part.

So, , where ~ exp(mean ).

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

Let's take .

• Generate is a random sample from Uniform . n = 100000.
• Select . Calculate .
• Repeat this process k = 1000 times.
• Plot the histogram.
• Plot and match with exp(mean ).
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 So, observe that for any fixed , .

Hence, . Hence, for k > 0, .

Let's compute separately for k = 0. . .

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

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