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Chapter 1: Problem 19

Find the 25 th percentile of the distribution having pdf \(f(x)=|x| / 4\), where \(-2

Short Answer

Expert verified
The 25th percentile does not exist for this distribution.

Step by step solution

01

Understanding the Distribution Function

The distribution function given is \(f(x) = \frac{|x|}{4}\) for \(-2 < x < 2\), and 0 elsewhere. This is an absolute value function, symmetrical about the y-axis. Thus, the function is equivalent to \(f(x) = \frac{x}{4}\) for \(0 < x < 2\) and \(f(x) = \frac{-x}{4}\) for \(-2 < x < 0\).
02

Finding the cumulative distribution function (CDF)

The cumulative distribution function F(x) is given by the integral of f(x) from \(-\infty\) to \(x\). Since the pdf is 0 for \(x < -2\) and \(x > 2\), these limits can be replaced by -2 and 2, respectively. For \(-2 < x < 0\), we have \(F(x) = \int_{-2}^{x} \frac{-t}{4} dt = \frac{-x^2}{8} + \frac{1}{2}\). For \(0 < x < 2\), we have \(F(x) = \int_{0}^{x} \frac{t}{4} dt = \frac{x^2}{8} + \frac{1}{2}\). Combining these two expressions, we have \(F(x) = \frac{|x^2|}{8} + \frac{1}{2}\) for \(-2 < x < 2\).
03

Finding the 25th Percentile

We find the 25th percentile by setting \(F(x) = 0.25\), which gives us \(|x^2| = 2(0.25 - 0.5) = -0.5\). However, absolute expressions can never equal a negative number, therefore, there are no solutions for this equation within the given range \(-2 < x < 2\). This implies that the 25th percentile does not exist for this distribution.

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Most popular questions from this chapter

Show that the moment generating function of the random variable \(X\) having the pdf \(f(x)=\frac{1}{3},-1

Let \(X\) denote a random variable for which \(E\left[(X-a)^{2}\right]\) exists. Give an example of a distribution of a discrete type such that this expectation is zero. Such a distribution is called a degenerate distribution.

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$$ \begin{aligned} &\text {Let } X \text { have a pmf } p(x)=\frac{1}{3}, x=1,2,3, \text { zero elsewhere. Find the pmf of }\\\ &Y=2 X+1 \end{aligned} $$
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