Exercise 3.9Z: Sine Transformation

Input–PDF and characteristic curve

In this task, we consider a random variable $x$ with $\sin^2$& shaped PDF in the range between $x= 0$ and $x= 2$:

$$f_x(x)= \sin^2({\rm\pi}/{\rm 2}\cdot x) \hspace{1cm}\rm for\hspace{0.15cm}{\rm 0\le \it x \le \rm 2} .$$

Outside of this, the PDF is identically zero.

The mean and rms of this random variable $x$ have already been determined in the Exercise 3.3 :

$$m_x = 1,\hspace{0.2cm}\sigma_x = 0.361.$$

Another random variable is obtained by transformation using the nonlinear characteristic curve

$$y= g(x) =\sin({\rm\pi}/{\rm 2}\cdot x).$$

The figure shows in each case in the range $0 \le x \le 2$:

above the PDF $f_x(x)$,
below the nonlinear characteristic $y = g(x)$.

Hints:

The exercise belongs to the chapter Exponentially Distributed Random Variables.
In particular, reference is made to the page Transformation of random variables and the chapter Expected Values and Moments.

Given are the two indefinite integrals:

$$\int \sin^{\rm 3}( ax)\,{\rm d}x = \frac{\rm 1}{ 3 a} \cdot \cos^{\rm 3}( ax)-\frac{\rm 1}{ a}\cdot \cos(ax),$$

$$\int \sin^{\rm 4}(ax)\,{\rm d}x =\frac{\rm 3}{\rm 8}\cdot x-\frac{\rm 1}{\rm 4 a} \cdot \sin(2 ax)+\frac{\rm 1}{32 a}\cdot \sin(4 ax).$$

Questions

	$y$ is limited to the value range $0 \le y \le 1$ .
	$y$ is limited to the value range $0 < y \le 1$ .
	The mean $m_y$ is less than the mean $m_x$.

$m_y \ = \ $

$\sigma_y \ = \ $

$f_y(y=0.6) \ = \ $

${\rm Pr}(y=1) \ = \ $

Solution

(1) Correct are the second and the third suggested solutions:

Because of the range of values of $x$ and the given characteristic curve, $y$ cannot take values smaller than $0$ or larger than $1$ respectively.
The value $y = 0$ cannot occur either, however, since neither $x = 0$ nor $x = 2$ are possible.
With these properties, the result is surely $m_y < 1$, i.e., a smaller value than $m_x = 1$ (see specification).

(2) To solve this task, one could, for example, first determine the PDF $f_y(y)$ and calculate $m_y$ from it in the usual way.

The direct way leads to the same result:

$$m_y={\rm E}\big[y\big]={\rm E}\big[g(x)\big]=\int_{-\infty}^{+\infty}g(x)\cdot f_x(x)\,{\rm d}x.$$

With the current functions $g(x)$ and $f_x(x)$ we obtain:

$$m_y=\int_{\rm 0}^{\rm 2}\hspace{-0.1cm}\sin^{\rm 3}({\pi}/{ 2}\cdot x)\,{\rm d}x=\frac{\rm 2}{\rm 3\cdot \pi}\cdot \cos^{\rm 3}({\pi}/{ 2}\cdot x)-\frac{\rm 2}{\rm \pi} \cdot \cos({3 \rm \pi}/{\rm 2}\cdot x)\Big|_{\rm 0}^{\rm 2}=\frac{\rm 8}{\rm 3\cdot \pi} \hspace{0.15cm}\underline{=\rm 0.849}.$$

(3) By analogy with point (2) holds:

$$m_{2 y}={\rm E}[y^{\rm 2}]={\rm E}[g^{\rm 2}( x)]=\int_{-\infty}^{+\infty}\hspace{-0.35cm}g^{2}( x)\cdot f_x(x)\,{\rm d}x.$$

This leads to the result:

$$ m_{ 2 y}=\int_{\rm 0}^{\rm 2}\hspace{-0. 15cm}\sin^{\rm 4}({\rm \pi}/{\rm 2}\cdot x)\, {\rm d} x= \frac{\rm 3}{\rm 8}\cdot x-\frac{\rm 1}{\rm 2\cdot\pi}\cdot \sin(\rm \pi\cdot{\it x})+\frac{\rm 1}{\rm 16\cdot\pi}\cdot \sin(\rm 2 \pi\cdot {\it x})\Big|_{\rm 0}^{\rm 2} \hspace{0.15cm}{= \rm 0.75}.$$

With the result from (2) it thus follows for the rms:

$$ \sigma_{y}=\sqrt{\frac{\rm 3}{\rm 4}-\Big(\frac{\rm 8}{\rm 3\cdot\pi}\Big)^{\rm 2}} \hspace{0.15cm}\underline{\approx \rm 0.172}.$$

(4) Due to the symmetry of PDF $f_x(x)$ and characteristic curve $y =g(x)$ um $x = 1$ the two domains yield.

$0 \le x \le 1$ and
$1 \le x \le 2$

each give the same contribution for $f_y(y)$.

In the first domain, the derivative of the characteristic curve is positive: $g\hspace{0.05cm}'(x)={\rm \pi}/{\rm 2}\cdot \cos({\rm \pi}/{\rm 2}\cdot x).$

The inverse function is: $ x=h(y)={\rm 2}/{\rm \pi}\cdot \arcsin( y).$

Taking into account the second contribution by the factor $2$ we getär the searched PDF in the range $0 \le y \le 1$ ):

$$f_y(y)= 2\cdot\frac{\sin^{ 2}({ \pi}/{ 2}\cdot x)}{{ \pi}/{ 2}\cdot \cos({ \pi}/{ 2}\cdot x)}\Big|_{\, x={ 2}/{ \pi}\cdot \arcsin( y)}.$$

File:P ID138 Sto Z 3 9 e new.png

Sought PDF

outside $f_y(y) is \equiv 0$. This leads to the intermediate result

$$f_y(y)=\frac{4}{\pi}\cdot \frac{\sin^{2}(\arcsin( y ))}{\sqrt{\rm 1-\sin^{ 2}(\arcsin( y \rm ))}}.$$

And because of $\sin\big (\arcsin(y)\big) = y$:

$$f_y(y)=\frac{ 4}{\pi}\cdot \frac{ y^{2}}{\sqrt{1- y^{\rm 2}}.$$

At the point $y = 0.6$ one obtains the value $f_y(y= 0.6)\hspace{0.15cm}\underline{=0.573}$.
On the right, this PDF $f_y(y)$ is shown graphically.

(5) The PDF is infinitely large at the point $y = 1$ .

This is due to the fact that at this point the derivative $g\hspace{0.05cm}'(x)$ of the characteristic curve runs horizontally.
However, since $y$ is a continuous random größe, nevertheless ${\rm Pr}(y = 1) \hspace{0.15cm}\underline{= 0}$ holds.

This means:

An infinity point in the PDF is not identical to a Dirac function.
Or more casually expressed: An infinity point in the PDF is "less" than a Dirac function.