Aufgaben:Exercise 3.9Z: Sine Transformation

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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

1

Which of the following statements are true?

	$y$  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  $m_x$.

2

Calculate the mean of the random variable  $y$.

$m_y \ = \ $

3

Calculate the root mean square of  $y$  and the rms $\sigma_y$ .

$\sigma_y \ = \ $

4

Calculate the PDF $f_y(y)$.  Note the symmetry properties.  What PDF–value results for  $y = 0.6$ ?

$f_y(y=0.6) \ = \ $

5

What is the PDF value for  $y = 1$?  Interpret the result.  What is the probability that  $y$  is exactly equal  $1$ ?

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

Solution

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.