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Chapter 10: Problem 6

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
If a pdf or pmf \(f(x)\) is symmetric about 0, this means \(f(x) = f(-x)\) for all \(x\). If this is true, it implies that its moment generating function \(M(t) = E[e^{tx}]\) is also symmetric around 0, so \(M(t) = M(-t)\). Conversely, if the mgf is symmetric around 0, then by taking the inverse Fourier transform we also find that the original pdf or pmf is symmetric around 0.

Step by step solution

01

Understand function symmetry

A function \(f(x)\) is symmetric about 0 if for any x, we have \(f(x) = f(-x)\). This means if you take any point \(x\) to the right of 0, there is a corresponding point \(-x\) to the left of 0 where the function values are equal. It's like folding the function over the y-axis, the two halves match exactly.
02

Define Moment Generating Function

The moment generating function (mgf) is defined as \(M(t) = E[e^{tx}]\) = \(\int e^{tx}f(x)dx\) for a continuous random variable, or \(M(t) = E[e^{tx}]\) = \(\sum e^{tx}f(x)\) for a discrete random variable, where \(E[]\) stands for expectation. If \(f(x)\) is symmetric about 0, it implies that \(f(x) = f(-x)\) for all \(x\). Therefore, the mgf becomes \(M(t) = E[e^{tx}] = E[e^{-tx}]\).
03

Prove symmetry of mgf

We have represented the mgf, in case if our random variable is symmetric about 0, as \(M(t)=E[e^{tx}] = E[e^{-tx}]\). Now, to complete the if-and-only-if statement, we need to show that if \(E[e^{tx}] = E[e^{-tx}]\) for all \(t\), then the distribution is symmetric. We could do this by showing that the characteristic function, \(E[e^{itx}]\), is real for all \(t\), which implies symmetry.
04

Prove symmetry of pdf or pmf

If the mgf is symmetric about 0, i.e. \(M(t) = E[e^{tx}] = E[e^{-tx}]\), then by taking the inverse Fourier transform we get the pdf or pmf back and because Fourier transform preserves the symmetry about origin, the pdf or pmf will also be symmetric about 0. The proof of this statement involves complex analysis and advanced mathematics.

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Most popular questions from this chapter

Show that the test given by (10.2.6) has asymptotically level \(\alpha\); that is, show that under \(H_{0}\) $$ \left[S\left(\theta_{0}\right)-(n / 2)\right] /(\sqrt{n} / 2) \stackrel{D}{\rightarrow} Z $$ where \(Z\) has a \(N(0,1)\) distribution.

(a) For \(n=3\), expand the mgf (10.3.5) to show that the distribution of the signed-rank Wilcoxon is given by $$ \begin{array}{|l|ccccccc|} \hline j & -6 & -4 & -2 & 0 & 2 & 4 & 6 \\ \hline P(T=j) & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{2}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$ (b) Obtain the distribution of the signed-rank Wilcoxon for \(n=4\).

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

Show that the power function of the sign test is nonincreasing for the hypotheses $$ H_{0}: \theta=\theta_{0} \text { versus } H_{1}: \theta<\theta_{0} $$

Consider the hypotheses (10.4.2). 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 $$ n \doteq\left(\frac{\left(z_{\alpha}-z_{\gamma^{*}}\right) 2 \tau_{\varphi}}{\Delta^{*}}\right)^{2} $$
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