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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.

Short Answer

Expert verified
A pdf or pmf is symmetric about 0 iff its mgf is symmetric about 0, as shown above.

Step by step solution

01

Assume the pdf/pmf is symmetric

Let's first assume that \(f(x)\), our pdf (probability density function) or pmf (probability mass function), is symmetric about 0. This means that for any value of \(x\), it is true that \(f(x) = f(-x)\).
02

Verify that the mgf is symmetric

If \(f(x)\) is symmetric, our goal is to show that the moment generating function (mgf) \(M(t)\) of \(f(x)\) is also symmetric about 0. By the definition, the mgf for a random variable \(X\) is \(M(t) = E(e^{tX})\). By substituting \(X\) with \(-X\) and using the fact that \(f(x) = f(-x)\), you will see that \(M(-t) = E(e^{-tX}) = M(t)\). Thus showing that the mgf \(M(t)\) is symmetric about \(t = 0\).
03

Assume that the mgf is symmetric

Now let's assume that the moment generating function \(M(t)\) is symmetric about 0, i.e. \(M(-t) = M(t)\). This means that, \(E(e^{-tX}) = E(e^{tX})\).
04

Verify that the pdf/pmf is symmetric

To show that the pdf/pmf \(f(x)\) is symmetric around 0, we can apply Fourier's theorem. By applying Fourier’s theorem, which states that the characteristic functions determine the distribution, we then know that if \(M(-t) = M(t)\) for all \(t\), then \(f(x) = f(-x)\), so \(f(x)\) is symmetric.

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Key Concepts

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

Probability Density Function (PDF)
The probability density function (PDF) is a concept used to represent the likelihood of various outcomes of continuous random variables. Unlike a probability mass function, which is used for discrete random variables, the PDF provides probabilities over a range of values rather than exact points. A key property of a PDF is that the area under the curve and between any two points on the x-axis represents the probability of the random variable falling within that range. To be a valid PDF, a function must be non-negative for all values and the total area under the curve must equal 1, signifying the certainty of the occurrence of some outcome within the given variable's range.

A PDF is said to be symmetric about 0 when the probability is evenly distributed on both sides of the origin, which means for any value of \( x \), the function satisfies \( f(x) = f(-x) \). In the context of symmetry, the symmetry axis does not have to be at zero; it can be any vertical line. However, our exercise focuses specifically on symmetry about 0.
Probability Mass Function (PMF)
The probability mass function (PMF) is similar to a pdf but is used for discrete random variables, which can only take on specific, isolated values. It gives the probability that a discrete random variable is exactly equal to some value. Formally, for a discrete random variable \( X \), the pmf \( p(x) \) is defined as \( p(x) = P(X = x) \), where \( P \) represents the probability. Much like a PDF, for a PMF to be valid, the sum of all probabilities for all possible outcomes must be 1. This ensures that the probabilities encompass all potential events for the given variable's domain of definition.

A PMF is symmetric about a point \( c \) if \( p(x) = p(2c - x) \) for any value in the support of \( X \). When \( c = 0 \), this reduces to \( p(x) = p(-x) \), which is the condition discussed in the original exercise.
Moment Generating Function (MGF)
The moment generating function (MGF) of a random variable is a tool that provides a series of moments (expected values of powers of the random variable), encapsulated in one function. It is defined as \( M(t) = E(e^{tX}) \), where \( E \) denotes the expected value and \( X \) is a random variable. If the MGF for a distribution exists, it uniquely determines the distribution. MGFs are useful for mathematical convenience, especially when trying to derive the moments of a distribution since the nth moment can be found by taking the nth derivative of the MGF and evaluating it at zero. The MGF's symmetry property connects directly to the PDF or PMF's symmetry, as seen in the original exercise. If the MGF of a distribution is symmetric about zero, it indicates that the distribution itself is symmetric about zero.

This property is crucial as it can also be used to prove symmetry of the underlying distribution because if \( M(t) = M(-t) \) for all \( t \), the random variable's distribution is symmetric about the origin.
Characteristic Function
The characteristic function of a random variable is another tool similar to the MGF and is crucial in probability theory. It is defined using complex numbers: for a given random variable \( X \), the characteristic function \( \[\phi(t) \] = E(e^{itX}) \), with \( i \) being the imaginary unit. This function uniquely determines the distribution of \( X \) and has the property that it always exists, unlike the MGF, which may not exist in some cases.

The symmetry of the characteristic function can also be used to demonstrate the symmetry of the underlying probability distribution. This is where Fourier's theorem comes into play, which states that the inversion of a characteristic function—essentially a type of Fourier transform—yields the probability distribution. The theorem assures that if the characteristic function is symmetric, so is the probability distribution related to the random variable. The characteristic function's properties are important in the realms of signal processing and quantum mechanics, demonstrating the interdisciplinary nature of these mathematical concepts.
Fourier's Theorem
Fourier's theorem is a fundamental principle within the field of mathematics, especially in harmonic analysis and signal processing. It states that any reasonably smooth or integrable function can be decomposed into a series of sine and cosine functions that oscillate at different frequencies—essentially a frequency spectrum of the original function. In the context of probability theory, Fourier's theorem is associated with the characteristic function's ability to determine the probability distribution.

Applying Fourier's theorem to probability, the inverse of the characteristic function, can tell us the nature of the probability distribution of a random variable. Consequently, in the original exercise, if the moment generating function (which is closely related to the characteristic function through a simple transformation) of a random variable is symmetric about zero, Fourier's theorem implies that the probability distribution itself must be symmetric about zero. In essence, Fourier’s theorem constructs the bridge between symmetry in the frequency domain and symmetry in the spatial domain, a concept that has profound implications across various scientific disciplines.

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Most popular questions from this chapter

Let \(X\) be a random variable with cdf \(F(x)\) and let \(T(F)\) be a functional. We say that \(T(F)\) is a scale functional if it satisfies the three properties $$ \text { (i) } T\left(F_{a X}\right)=a T\left(F_{X}\right), \text { for } a>0 $$ (ii) \(T\left(F_{X+b}\right)=T\left(F_{X}\right), \quad\) for all \(b\) $$ \text { (iii) } T\left(F_{-X}\right)=T\left(F_{X}\right) \text { . } $$ Show that the following functionals are scale functionals. (a) The standard deviation, \(T\left(F_{X}\right)=(\operatorname{Var}(X))^{1 / 2}\). (b) The interquartile range, \(T\left(F_{X}\right)=F_{X}^{-1}(3 / 4)-F_{X}^{-1}(1 / 4)\).

In this section, as discussed above expression \((10.5 .2)\), the scores \(a_{\varphi}(i)\) are generated by the standardized score function \(\varphi(u) ;\) that is, \(\int_{0}^{1} \varphi(u) d u=0\) and \(\int_{0}^{1} \varphi^{2}(u) d u=1\). Suppose that \(\psi(u)\) is a square-integrable function defined on the interval \((0,1)\). Consider the score function defined by $$ \varphi(u)=\frac{\psi(u)-\bar{\psi}}{\int_{0}^{1}[\psi(v)-\bar{\psi}]^{2} d v}, $$ where \(\bar{\psi}=\int_{0}^{1} \psi(v) d v\). Show that \(\varphi(u)\) is a standardized score function.

Consider the hypotheses (10.4.4). Suppose we select the score function \(\varphi(u)\) and the corresponding test based on \(W_{\varphi} .\) Suppose we want to determine the sample size \(n=n_{1}+n_{2}\) for this test of significance level \(\alpha\) to detect the alternative \(\Delta^{*}\) with approximate power \(\gamma^{*}\). Assuming that the sample sizes \(n_{1}\) and \(n_{2}\) are the same, show that $$ n \approx\left(\frac{\left(z_{\alpha}-z_{\gamma^{*}}\right) 2 \tau_{\varphi}}{\Delta^{*}}\right)^{2} $$

For any \(n \times 1\) vector \(\mathbf{v}\), define the function \(\|\mathbf{v}\|_{W}\) by $$ \|\mathbf{v}\|_{W}=\sum_{i=1}^{n} a_{W}\left(R\left(v_{i}\right)\right) v_{i} $$ where \(R\left(v_{i}\right)\) denotes the rank of \(v_{i}\) among \(v_{1}, \ldots, v_{n}\) and the Wilcoxon scores are given by \(a_{W}(i)=\varphi_{W}[i /(n+1)]\) for \(\varphi_{W}(u)=\sqrt{12}[u-(1 / 2)] .\) By using the correspondence between order statistics and ranks, show that $$ \|\mathbf{v}\|_{W}=\sum_{i=1}^{n} a(i) v_{(i)}, $$ where \(v_{(1)} \leq \cdots \leq v_{(n)}\) are the ordered values of \(v_{1}, \ldots, v_{n} .\) Then, by establishing the following properties, show that the function \((10.9 .53)\) is a pseudo-norm on \(R^{n} .\) (a) \(\|\mathbf{v}\|_{W} \geq 0\) and \(\|\mathbf{v}\|_{W}=0\) if and only if \(v_{1}=v_{2}=\cdots=v_{n}\). Hint: First, because the scores \(a(i)\) sum to 0, show that $$ \sum_{i=1}^{n} a(i) v_{(i)}=\sum_{ij} a(i)\left[v_{(i)}-v_{(j)}\right] $$ where \(j\) is the largest integer in the set \(\\{1,2, \ldots, n\\}\) such that \(a(j)<0\). (b) \(\|c \mathbf{v}\|_{W}=|c|\|\mathbf{v}\|_{W}\), for all \(c \in R\). (c) \(\|\mathbf{v}+\mathbf{w}\|_{W} \leq\|\mathbf{v}\|_{W}+\|\mathbf{w}\|_{W}\), for all \(\mathbf{v}, \mathbf{w} \in R^{n}\) Hint: Determine the permutations, say, \(i_{k}\) and \(j_{k}\) of the integers \(\\{1,2, \ldots, n\\}\), which maximize \(\sum_{k=1}^{n} c_{i_{k}} d_{j_{k}}\) for the two sets of numbers \(\left\\{c_{1}, \ldots, c_{n}\right\\}\) and \(\left\\{d_{1}, \ldots, d_{n}\right\\} .\)

Suppose the random variable \(e\) has cdf \(F(t)\). Let \(\varphi(u)=\sqrt{12}[u-(1 / 2)]\), \(0

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