In Casella and Berger (2002) I found an example for non-unique moments (example 2.3.10 on page 64). They are providing the following 2 pdfs:

$f_1(x) = \frac 1{\sqrt{2\pi}x}e^{-(\log x)^2/2}$, where $0 \leqslant x \leqslant\infty$


$f_1(x) = f_1(x) [1 + \sin(2\pi\log x)]$, where $0 \leqslant x \leqslant\infty$$

The former is lognormal with $\mu = 0$ and $\sigma^2 = 1$. It can be shown that if $x_1 \sim f_1(x)$, then $E(x_1^r) = e^{r^2/2}$. Hence, $x_1$ has all the moments. However,

$E(x_2^r) = \int_0^{\infty}x^rf1(x)[1+\sin(2\pi\log x)]dx = E(x_1^r)+\int_0^{\infty}x^rf1(x)[\sin(2\pi\log x)]dx.$

Since the last integral is equal to zero, we can show that $x_1$ and $x_2$ have distinct pfds but the same moments.

My question is about how to show that the last integral is equal to zero. In exercise 2.35, Casella and Berger (2002, p. 81) show how to prove this (also see here on page 2-11).

$\int_0^{\infty}x^rf1(x)[\sin(2\pi\log x)]dx = \int_0^{\infty}x^r\frac 1{\sqrt{2\pi}x}e^{-(\log x)^2/2}[\sin(2\pi\log x)]dx$

Now, Cassella and Berger propose to substitute $y = \log x$ so that $dy = (1/x)dx$. Hence,

$\int_0^{\infty}e^{(y+r)r}\frac 1{\sqrt{2\pi}x}e^{-(y+r)^2/2}[\sin(2\pi y + 2\pi r)]dx$

$\int_{-\infty}^{\infty}\frac 1{\sqrt{2\pi}x}e^{(r^2-y^2)/2}[\sin(2\pi y + 2\pi r)]dx$

Hence, the the integrand is an odd function. Thus, the negative integral cancels the positive one. This all makes sense.

However, when I try to understand the substitution performed, I wonder if Casella and Berger meant to substitute using $y + r = \log x$. My question may be trivial but would this substitution be allowed/ common? If so, then I would never have seen such an integration by substitution using 2 summands.


It looks like they have indeed substituted $y + r = \log x$. And this is perfectly allowed, because it just means they substituted $\color{blue}{y = \log x - r}$ (remember, $r$ is just some constant).

The reason they included the $r$ is to get the integrand to become an odd function.


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