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Chapter 10: Problem 2

Let \(X\) be a continuous random variable with pdf \(f(x)\). Suppose \(f(x)\) is symmetric about \(a\); i.e., \(f(x-a)=f(-(x-a))\). Show that the random variables \(X-a\) and \(-(X-a)\) have the same pdf.

Short Answer

Expert verified
Given the symmetry of the density function about \(a\), we found that the pdf's of the random variables \(X - a\) and \(- (X - a)\) are indeed the same.

Step by step solution

01

Define the new random variables

Let \(Y = X - a\) and \(Z = - (X - a) = a - X\). The goal is to derive the pdf for each of these variables and to show they are the same.
02

Derive the pdf of \(Y\)

The pdf of \(Y\) is obtained by replacing \(X\) with \(Y + a\) in \(f(x)\). That is, \(f_Y(y) = f(y + a)\).
03

Derive the pdf of \(Z\)

To find the pdf of \(Z\), we first note that \(Z = -Y\). It is a known result that if \(Y\) is a random variable with pdf \(f_Y(y)\), the pdf of \(-Y\) is given by \(f_{-Y}(y) = f_Y(-y)\). Therefore, \(f_Z(z) = f_Y(-z) = f(-z + a)\).
04

Use symmetry property

It is given that \(f(x - a) = f(-(x - a))\). Therefore, by replacing \(x\) with \(-z\) in this equation, we get \(f(-z + a) = f(z - a)\). Hence, we have \(f_Z(z) = f(-z + a) = f(z - a) = f_Y(y)\).

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Most popular questions from this chapter

In Exercise \(10.9 .5\), the influence function of the variance functional was derived directly. Assuming that the mean of \(X\) is 0, note that the variance functional, \(V\left(F_{X}\right)\), also solves the equation $$ 0=\int_{-\infty}^{\infty}\left[t^{2}-V\left(F_{X}\right)\right] f_{X}(t) d t $$ (a) Determine the natural estimator of the variance by writing the defining equation at the empirical cdf \(F_{n}(t)\), for \(X_{1}-\bar{X}, \ldots, X_{n}-\bar{X}\) iid with cdf \(F_{X}(t)\), and solving for \(V\left(F_{n}\right)\). (b) As in Exercise \(10.9 .6\), write the defining equation for the variance functional at the contaminated \(\operatorname{cdf} F_{x, \epsilon}(t)\). (c) Then derive the influence function by implicit differentiation of the defining equation in part (b).

Consider the rank correlation coefficient given by \(r_{q c}\) in part (c) of Exercise \(10.8 .5 .\) Let \(Q_{2 X}\) and \(Q_{2 Y}\) denote the medians of the samples \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{n}\), respectively. Now consider the four quadrants: $$ \begin{aligned} I &=\left\\{(x, y): x>Q_{2 X}, y>Q_{2 Y}\right\\} \\ I I &=\left\\{(x, y): xQ_{2 Y}\right\\} \\ I I I &=\left\\{(x, y): xQ_{2 X}, y

Show that Kendall's \(\tau\) satisfies the inequality \(-1 \leq \tau \leq 1\).

Spearman's rho is a rank correlation coefficient based on Wilcoxon scores. In this exercise we consider a rank correlation coefficient based on a general score function. Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate continuous cdf \(F(x, y) .\) Let \(a(i)=\varphi(i /(n+1))\), where \(\sum_{i=1}^{n} a(i)=0 .\) In particular, \(\bar{a}=0 .\) As in expression \((10.5 .6)\), let \(s_{a}^{2}=\sum_{i=1}^{n} a^{2}(i)\). Consider the rank correlation coefficient, $$ r_{a}=\frac{1}{s_{a}^{2}} \sum_{i=1}^{n} a\left(R\left(X_{i}\right)\right) a\left(R\left(Y_{i}\right)\right) $$ (a) Show that \(r_{a}\) is a correlation coefficient on the items $$ \left\\{\left(a\left[R\left(X_{1}\right)\right], a\left[R\left(Y_{1}\right)\right]\right),\left(a\left[R\left(X_{2}\right)\right], a\left[R\left(Y_{2}\right)\right]\right), \ldots,\left(a\left[R\left(X_{n}\right)\right], a\left[R\left(Y_{n}\right)\right]\right)\right\\} $$ (b) For the score function \(\varphi(u)=\sqrt{12}(u-(1 / 2))\), show that \(r_{a}=r_{S}\), Spearman's rho. (c) Obtain \(r_{a}\) for the sign score function \(\varphi(u)=\operatorname{sgn}(u-(1 / 2)) .\) Call this rank correlation coefficient \(r_{q c}\). (The subscript \(q c\) is obvious from Exercise \(10.8 .7\).)

Prove that a pdf (or pmf) \(f(x)\) is symmetric about 0 if and only if its mgf is symmetric about 0 , provided the mgf exists.
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