Aufgaben:Exercise 3.9: Characteristic Curve for Cosine PDF

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Rectangular and cosine PDF

We are looking for a continuous,  monotonically increasing nonlinear characteristic  $y =g(x)$,  which generates a new random variable  with  "cosine"  PDF from a between  $-1$  and  $+1$  uniformly distributed random variable  $x$:

$$f_y(y)=A\cdot\cos({\pi}/{2}\cdot y).$$

• The random variable  $y$  can also only take values between  $-1$  and  $+1$.
• The two density functions  $f_x(x)$  and  $f_y(y)$  are sketched on the right.

Hints:

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

Question

1

Which of the following statements are true?

	Outside the range  $-1 \le x \le +1$  ⇒   $g(x)$  can be arbitrary.
	The characteristic curve must be symmetrical about  $x= 0$  :   $g(-x) = g(x)$.
	The random variable  $y$  has a smaller variance than  $x$.

2

Calculate the  $f_y(y)$ value at  $y = 0$:   $A = f_y(0)$.

$A \ = \ $

3

Determine the slope  $h\hspace{0.05cm}'(y)$  of the inverse function  $x = h(y)$,  where for  $|y| \le 1$  always  $h\hspace{0.05cm}'(y) > 0$  should hold?  What slope holds for  $y = 0$ ?

$h'(y = 0) \ = \ $

4

Compute with the result from  (3)  the function  $x = h(y)$  under the constraint  $h(0) = 0$.  What value results für  $y = 1$ ?

$h(y=1) \ = \ $

5

Determine the function  $y = g(x)$  of the characteristic we are looking for.  What is the function value at the point  $x = 1$ ?

$g(x = 1) \ = \ $

Solution

Solution

(1)  Correct are statements 1 and 3:

• Since  $x$  can only take values between  $\pm 1$ , the course of the characteristic curve outside of this range is irrelevant for the random variable  $y$ .
• The condition  $g(-x) = g(x)$  does not have to be met.  There are any number of characteristic curves that can generate the desired PDF.
• For example, the characteristic curve calculated in point  (5)  is point symmetric:   $g(-x) = -g(x)$.
• The graphical representations of the two density functions already show that  $\sigma_y^2 < \sigma_x^2$  is.

(2)  The integral over the PDF must always equal  $1$ . It follows that:

$$\int_{-\rm 1}^{\rm 1}A\cdot \cos({\pi}/{\rm 2}\cdot y)\, {\rm d} y=\frac{A\cdot \rm 4}{\pi}\hspace{0.3cm} \rightarrow\hspace{0.3cm} A=\frac{\pi}{\rm 4} \hspace{0.15cm}\underline{= \rm 0.785}.$$

(3)  The transformation formula can be transformed as follows:

$$f_y(y)=\frac{f_x(x)}{| g'(x)|}\Big|_{\, x=h(y)}=f_x(x)\cdot |h'(y)| \Big|_{\, x=h(y)}.$$

• The inverse function  $x = h(y)$  of a monotonically increasing characteristic  $y = g(x)$  also increases monotonically.
• Therefore one does not need to make use of the absolute value and subsequently obtains:

$$h\hspace{0.05cm}'(y)=\frac{f_y(y)}{f_x(x)\Big|_{\, x=h(y)}}={\pi}/{\rm 2}\cdot \cos({\pi}/{2}\cdot y).$$

• At the point  $y = 0$  the slope has the value  $h\hspace{0.05cm}'(y= 0)=π/2\hspace{0.15cm}\underline{\approx 1.571}$.

(4)  One obtains by (indefinite) integration:

$$h(y)=\int h\hspace{0.05cm}'(y)\, {\rm d} y + C = \frac{\pi}{2}\cdot \frac{2}{\pi}\cdot \sin(\frac{\pi}{ 2}\cdot y) + C.$$

• The constraint  $h(y= 0) = 0$  leads to the constant  $C = 0$  and thus to the result:

$$h(y) = \sin({\pi}/{2}\cdot y) \hspace{0.5cm} \rightarrow\hspace{0.5cm} h(y = 1) \hspace{0.15cm}\underline{= +1}.$$

(5)  The inverse function of the function determined in subtask  (4)  is  $x = h(y)$  :

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

• This characteristic increases monotonically in the range  $-1 \le x \le +1$  from  $y = -1$  to  $y = +1$ .
• So the value we are looking for is  $g(x= 1) \hspace{0.15cm}\underline{= +1}$.