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Chapter 7: Problem 16

Show that \(X\) and \(-X\) have the same distribution if and only if \(\phi_{X}\) is a purely real-valued function.

Short Answer

Expert verified
\(X\) and \(-X\) have the same distribution if and only if \(\phi_X(t)\) is purely real-valued, as shown by the symmetric property \(\phi_X(t) = \phi_X(-t)\).

Step by step solution

01

Understanding Characteristic Functions

The characteristic function \(\phi_X(t)\) of a random variable \(X\) is defined as \(\phi_X(t) = E[e^{itX}]\). It transforms the random variable into a function that captures its distribution in the frequency domain.
02

Formulating the Condition for Symmetric Distribution

Two random variables \(X\) and \(-X\) have the same distribution if for all \(t\), the characteristic functions are equal: \(\phi_X(t) = \phi_{-X}(t)\). Since \(\phi_{-X}(t) = E[e^{it(-X)}] = E[e^{-itX}] = \phi_X(-t)\), we require \(\phi_X(t) = \phi_X(-t)\).
03

Real-Valued Characteristic Function

A purely real-valued characteristic function means \(\phi_X(t)\) has no imaginary component. If \(X\) and \(-X\) have the same distribution, then \(\phi_X(t) = \phi_X(-t)\) implies that \(\phi_X(t)\) is symmetric and thus purely real-valued.
04

If and Only If Proof

To show an 'if and only if' condition, prove both implications. - **If** \(\phi_X(t) = \phi_X(-t)\), then \(\phi_X(t)\) is real-valued, indicating symmetry in \(X\) and \(-X\).- **If** \(\phi_X(t)\) is real-valued, it has no imaginary parts, and thus \(\phi_X(t) = \overline{\phi_X(t)} = \phi_X(-t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Symmetric distribution
The concept of a symmetric distribution is central to understanding the relationship between a random variable and its negative counterpart. When we say that a distribution is symmetric, we mean that it is mirror-like and balanced around a central point.

For a random variable \(X\), having the same distribution as \(-X\) implies that the distribution does not change if we flip it around its mean or origin. This symmetry is mathematically expressed through its characteristic function, where if \( \phi_X(t) = \phi_X(-t) \), the distribution is deemed symmetric.

Symmetry in distributions is a vital concept because it often suggests that the mean of the distribution is its central point, around which data points are equally dispersed in both directions. This type of distribution can make data analysis and interpretation simpler and more intuitive.
Real-valued function
A real-valued function is a function that returns real numbers as its values. In the context of characteristic functions of random variables, being real-valued has specific implications.

The characteristic function \( \phi_X(t) \) is defined as \(E[e^{itX}]\), and for it to be real-valued, it must have no imaginary part. This happens exactly when the characteristic function of the random variable is symmetric, meaning \( \phi_X(t) = \phi_X(-t) \).

Real-valued characteristic functions are crucial because they simplify the verification of symmetric distributions. By examining whether the characteristic function returns purely real numbers, it becomes clear whether or not a distribution is symmetric. This connection reflects deeper properties of the random variable concerning its balance and spread across the real number line.
Random variable distribution
Random variable distribution describes how the probabilities are spread over the possible outcomes of the random variable. It can be visualized as a pattern or shape drawn on a graph, indicating where values of a random variable are most likely to occur.

In our scenario, the distribution of \(X\) compared to \(-X\) is crucial. If both have the same distribution, then for all values, the characteristic functions are equal: \( \phi_X(t) = \phi_X(-t) \). This indicates that the probabilities of \(X\) and \(-X\) are mirrored and hence, balanced around a central point or zero.

Understanding these distributions can aid in identifying whether a set of data exhibits neutrality or bias. Symmetry implies balance, while asymmetry might indicate skewness, guiding further statistical decision-making and interpretation of data trends.

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Most popular questions from this chapter

Let \(p \geq 1\). By Jensen's inequality or otherwise, find the smallest value of the constant \(c_{p}\) such that \((a+b)^{p} \leq c_{p}\left(a^{p}+b^{p}\right)\) for all \(a, b \geq 0\). (Cambridge 2006)

Prove that if \(X=X_{1}+\cdots+X_{n}\) and \(Y=Y_{1}+\cdots+Y_{n}\), where \(X_{i}\) and \(Y_{j}\) are independent whenever \(i \neq j\), then \(\operatorname{cov}(X, Y)=\sum_{i=1}^{n} \operatorname{cov}\left(X_{i}, Y_{i}\right)\). (Assume that all series involved are absolutely convergent.) Two players A and B play a series of independent games. The probability that A wins any particular game is \(p^{2}\), that \(\mathrm{B}\) wins is \(q^{2}\), and that the game is a draw is \(2 p q\), where \(p+q=1\). The winner of a game scores 2 points, the loser none; if a game is drawn, each player scores 1 point. Let \(X\) and \(Y\) be the number of points scored by A and B, respectively, in a series of \(n\) games. Prove that \(\operatorname{cov}(X, Y)=-2 n p q .\) (Oxford \(1982 \mathrm{M})\) )

Let \(X\) and \(Y\) be random variables with equal variance. Show that \(U=X-Y\) and \(V=X+Y\) are uncorrelated. Give an example to show that \(U\) and \(V\) need not be independent even if, further, \(X\) and \(Y\) are independent.

Let \(X\) have moment generating function \(M(t)\). (a) Show that \(M(t) M(-t)\) is the moment generating function of \(X-Y\), where \(Y\) is independent of \(X\) but has the same distribution. (b) In a similar way, describe random variables which have moment generating functions $$ \frac{1}{2-M(t)}, \quad \int_{0}^{\infty} M(u t) e^{-u} d u $$

Show that every distribution function has only a countable set of points of discontinuity.
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