Moments do not characterize the distribution function

Question

Let $X \sim N(0,1)$ and $Y = e^X$. Consider another random variable $Z$ and its PDF is $$f_Z(t) = [1 + \sin(2\pi \log_e(t))] f_Y (t)\quad 1\{t \ge 0\}\quad (1)$$

a) Show that $(1)$ is a valid PDF, that is, $f_Z(t) \ge 0$ for any t and $\int^\infty_{-\infty} f_Z(t)\,dt = 1$.

b) Show that $E[Y^n] = E[Z^n]$, for all $n = 1,2,...$

c) Use your favorite computer program to plot $f_Y (t)$ and $f_Z(t)$ for the range $0 \le t \le 4$ and argue that the distribution function of $Y$ and $Z$ are different.

My attempt:

a) $$f_Y(y)=\frac{1}{y\sqrt{2\pi}}\exp\left(\frac{-(\ln y)^2}{2}\right)$$

$$f_Z(t) = [1+\sin(2\pi \log_e(t))]\frac{1}{t\sqrt{2\pi}}\exp\left(\frac{-(\ln t)^2}{2}\right)$$

exponential is always $\ge 0$, $\frac{1}{t}\ge0 \;\text{for} \;t\ge0$, $\sin x\in[-1,1]$ so $1+\sin x \ge 0$

thus it is proven that $f_Z(t) \ge 0$.

$$\int^\infty_{-\infty} f_Z(t)\,dt = \int^\infty_0f_Y(t)dt+\int^\infty_0\frac{\sin(2\pi \ln t)}{t\sqrt{2\pi}}\exp\left(\frac{-(\ln t)^2}{2}\right)\,dt$$

$$\int^\infty_0 f_Y(t)\,dt=F_Y(\infty)=1$$

But, I can't think of a way to solve the second part of integration.

Answer 1

Consider $$g(t) = \frac{\sin (2\pi \log t)}{t \sqrt{2\pi}} e^{-(\log t)^2/2}, \quad t > 0.$$ Then $$g(1/t) = \frac{t \sin (-2\pi \log t)}{\sqrt{2\pi}} e^{-(-\log t)^2/2} = -\frac{t \sin (2\pi \log t)}{\sqrt{2\pi}} e^{-(\log t)^2/2} = -t^2 g(t),\quad t > 0.$$ Therefore, with the substitution $t = 1/u$, $dt = -u^{-2} \, du$, we obtain $$\begin{align*} \int_{t=0}^\infty g(t) \, dt &= \int_{t=0}^1 g(t) \, dt + \int_{t=1}^\infty g(t) \, dt \\ &= \int_{t=0}^1 g(t) \, dt + \int_{u=1}^0 g(1/u) (-u^{-2}) \, du \\ &= \int_{t=0}^1 g(t) \, dt + \int_{u=1}^0 -u^2 g(u) (-u^{-2}) \, du \\ &= \int_{t=0}^1 g(t) \, dt + \int_{u=1}^0 g(u) \, du \\ &= 0. \end{align*}$$

Answer 2

(b)

\begin{align}\mathbb E(Z^n)&=\int_0^\infty t^n [1+sin(2\pi \ln(t))]\frac{1}{t\sqrt{2\pi}}exp(\frac{-(\ln t)^2}{2}) dt \\ &= \mathbb E(X^n)+ \int_0^\infty t^n sin(2\pi \ln t)\frac{1}{t\sqrt{2\pi}}exp(\frac{-(\ln t)^2}{2}) dt \end{align} It is enough to show $\int_0^\infty t^n sin(2\pi \ln t)\frac{1}{t\sqrt{2\pi}}exp(\frac{-(\ln t)^2}{2}) dt=0.$

$y=\ln t -n$

\begin{align} \int_0^\infty \frac{t^n sin(2\pi \ln t)}{t\sqrt{2\pi}}exp(\frac{-(\ln t)^2}{2}) dt &= \int_{-\infty}^{+\infty} \frac{e^{ny+n^2} sin(2\pi (y+n))}{\sqrt{2\pi}}e^{\frac{-(y+n)^2}{2}} dy \\ &= \frac{e^{n^2}}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} sin(2\pi y) e^{\frac{-y^2-n^2}{2}} dy=0 \end{align} because the function under integration is odd. Definite integral of an odd function is 0 (symmetric interval)

(c)

    #R code
   t<-seq(.01,4,len=100)
   fy<-exp(-.5*(log(t))^2)/(t*sqrt(2*pi))
   fz<-(1+sin(2*pi*log(t)))*fy
   plot(t,fz,type="l",lwd=1.6,col="blue")
   lines(t,fy,col="red",lwd=2)
   text(1,1,labels="fz",col="blue",cex=1.5)
   text(2.2,.3,labels="fy",col="red",cex=1.5)