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Chapter 4: Problem 2

Let \(Y_{1}\) denote the minimum of a random sample of size \(n\) from a distribution that has pdf \(f(x)=e^{-(x-\theta)}, \theta

Short Answer

Expert verified
The limiting distribution of \(Z_n\) tends to an exponential distribution with parameter 1.

Step by step solution

01

Embark on Obtaining CDF of \(Y_1\)

First, calculate the cumulative distribution function (CDF) of \(Y_{1}\). Given that \(Y_{1}\) is the minimum of n IID random variables, the CDF of \(Y_{1}\) is \(F_{Y_1}(y)=1-\left(1-F(y)\right)^n\). Substituting given pdf, we get \(F_{Y_1}(y) = 1-\left[1-(1-e^{-(y-\theta)})\right]^n\) for \(\theta< y<\infty\) and zero elsewhere.
02

Determine the pdf of \(Y_1\)

Differentiate this cumulative distribution function of \(Y_1\) to identify the pdf. Using chain rule for differentiation, the pdf \(f_{Y1}(y) = n e^{-(y-\theta)} [\left(1-e^{-(y-\theta)}\right)]^{n-1}\), again for \(\theta<y<\infty\) and zero elsewhere.
03

Derive the distribution of \(Z_n\)

Now derive the distribution of \(Z_n\) which is \(n(Y_1 - \theta)\). Find cumulative distribution function (CDF) of \(Z_n\). Let \(z = n(y - \theta)\), then \(y = z/n + \theta\). By substituting this we can get \(F_{Z_n}(z) = 1-[1-(1-e^{-z})]^{n}\), for \(0<z<\infty\) and zero elsewhere.
04

Investigate the Limiting Distribution of \(Z_n\)

Calculate limit as \(n\) tends to infinity for \(F_{Z_n}(z)\). It can be shown using properties of limits along with applying the exponential limit, that \(\lim_{n\to\infty} F_{Z_n}(z) = 1-e^{-z}\) , for \(0<z<\infty\) and zero elsewhere. Hence the limiting distribution of \(Z_n\) is exponential with parameter 1.

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