# Exercise 3.9Z: Sine Transformation

Input PDF, characteristic curve

In this task,  we consider a random variable  $x$  with sine-square shaped  $\rm PDF$  in the range between  $x= 0$  and  $x= 2$:

$$f_x(x)= \sin^2({\rm\pi}/{\rm 2}\cdot x) \hspace{1cm}\rm for\hspace{0.25cm}{\rm 0\le \it x \le \rm 2} .$$

Outside of this,  the PDF is identically zero.

The mean and the standard deviation 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:

• 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 range  $0 \le y \le 1$  . $y$  is limited to the 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 the  standard deviation of the random variable  $y$.

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

(4)  Due to the symmetry of the PDF  $f_x(x)$  and the 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 get the searched PDF in the range  $0 \le y \le 1$:
$$f_y(y)= 2\cdot\frac{\sin^{ 2}({ \pi}/{ 2}\cdot x)}{{ \pi}/{ 2}\cdot \cos({ \pi}/{ 2}\cdot x)}\Big|_{\, x={ 2}/{ \pi}\cdot \arcsin( y)}.$$
PDF after transformation
• Outside of this range:  $f_y(y) \equiv 0$.  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 quantity,  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 delta function.
• Or more casually expressed:   An infinity point in the PDF is  "less"  than a Dirac delta function.