Aufgaben:Exercise 4.8: Diamond-shaped Joint PDF

• Return to book

Diamond–shaped joint PDF

We consider a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  whose components arise as linear combinations of two random variables  $u$  and  $v$:

$$x=2u-2v+1,$$

$$y=u+3v.$$

Further,  note:

• The two statistically independent random variables  $u$  and  $v$  are each uniformly distributed between  $0$  and  $1$.
• In the figure you can see the joint PDF.  Within the parallelogram drawn in blue holds:

$$f_{xy}(x,\hspace{0.08cm} y) = H = {\rm const.}$$

• Outside the parallelogram no values are possible:  $f_{xy}(x,\hspace{0.08cm} y) = 0$.

Hints:

• The exercise belongs to the chapter  Linear Combinations of Random Variables.
• Reference is also made to the page  Regression line.
• We also refer here to the interactive applet  Correlation coefficient and regression line
• Assume - if possible - the given equations.  Use the information of the above sketch mainly only to check your results.

Questions

1

What is the height  $H$  of the joint PDF within the parallelogram?

$H \ = \ $

2

What values of  $u$  and  $v$  underlie the corner point  $(-1, 3)$?

$u \ = \ $

$v \ = \ $

3

Calculate the correlation coefficient  $\rho_{xy}$.

$\rho_{xy}\ = \ $

4

What is the regression line  $\rm (RL)$?  At what point  $y_0$  does it intersect the  $y$–axis?

$y_0\ = \ $

5

Calculate the marginal probability density function $f_x(x)$.  What is the probability that the random variable  $x$  is negative?

${\rm Pr}(x < 0)\ = \ $

6

Calculate the marginal probability density function $f_y(y)$.  What is the probability that the random variablee  $y >3$?

${\rm Pr}(y > 3)\ = \ $

Solution

Solution

(1)  The area of the parallelogram can be composed of two triangles of equal size.

• The area of the triangle $(1,0)\ (1,4)\ (-1,3)$  gives  $0.5 · 4 · 2 = 4$.
• The total area is double:   $F = 8$.
• Since the PDF volume is always  $1$ , then  $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.

(2)  The minimum value of  $x$  is obtained for  $\underline{ u=0}$  and  $\underline{ v=1}$.

• From the above equations,  the results  $x= -1$  and  $y= +3$  follow.

(3)  The equation given in the theory section is valid in general, i.e., for any PDF of the two statistically independent variables  $u$  and  $v$,  as long as they have equal standard deviations  $(\sigma_u = \sigma_v)$.

• With  $A = 2$,  $B = -2$,  $D = 1$  and  $E = 3$  we obtain:

$$\rho_{xy } = \frac {\it A \cdot D + B \cdot E}{\sqrt{(\it A^{\rm 2}+\it B^{\rm 2})(\it D^{\rm 2}+\it E^{\rm 2})}} =\frac {2 \cdot 1 -2 \cdot 3}{\sqrt{(4 +4)(1+9)}} = \frac {-4}{\sqrt{80}} = \frac {-1}{\sqrt{5}}\hspace{0.15cm}\underline{ = -0.447}. $$

(4)  The regression line is generally:

$$y=\frac{\sigma_y}{\sigma_x}\cdot\rho_{xy}\cdot(x-m_x)+m_y.$$

• From the linear means  $m_u = m_v = 0.5$  and the equations given in the problem statement,  we obtain  $m_x = 1$  and  $m_y = 2$.

• The variances of  $u$  and  $v$  are respectively  $\sigma_u^2 = \sigma_v^2 =1/12$.  It follows:

$$\sigma_x^2 = 4 \cdot \sigma_u^2 + 4 \cdot \sigma_v^2 = 2/3,$$

$$\sigma_y^2 = \sigma_u^2 + 9\cdot \sigma_v^2 = 5/6.$$

• Substituting these values into the equation of the regression line,  we get:

$$y=\frac{\sqrt{5/6}}{\sqrt{2/3}}\cdot (\frac{-1}{\sqrt{5}})\cdot(x-1)+2= - x/{2} + 2.5.$$

• From this follows with  $x=0$  the value  $y_0=\hspace{0.15cm}\underline{ = 2.5}$

(5)  With the auxiliary quantities   $q= 2u$,   $r= -2v$   and   $s= x-1$:  $s= q+r$.

Triangular PDF $f_x(x)$

• Since  $u$  and  $v$  are each uniformly distributed between  $0$  and  $1$,  $q$  has a uniform distribution in the range from  $0$  to  $2$  and  $r$  is uniformly distributed between  $-2$  and  $0$.
• In addition,  since  $q$  and  $r$  are not statistically dependent on each other, the PDF of the sum is:

$$f_s(s) = f_q(q) \star f_r(r).$$

• The addition  $x = s+1$  leads to a shift of the triangular–PDF by  $1$  to the right.
• For the sought probability  (highlighted in green in the graphic)  therefore holds:  

$${\rm Pr}(x < 0)\hspace{0.15cm}\underline{ = 0.125}.$$

Trapezoidal PDF $f_y(y)$

(6)  Analogous to the solution for the subtask  (5)  holds with  $t = 3v$:

$$f_y(y) = f_u(u) \star f_t(t).$$

• The convolution between two rectangles of different widths results in a trapezoid.  For the probability we are looking for,  we get:

$${\rm Pr}(y>3) =1/6\hspace{0.15cm}\underline{ \approx 0.167}.$$

• This probability is highlighted in green in the right sketch.