Aufgaben:Exercise 4.6: Coordinate Rotation

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Coordinate rotation of a joint PDF

In the exercise we consider a two-dimensional Gaussian random variable  $(x, y)$  with statistically independent components  Let the standard deviations of the two components be  $\sigma_x = 1$  and  $\sigma_y = 2$.

We want to calculate the probability that the two-dimensional random variable $(x, y)$  lies within the shaded area:

$$-C \le x + y \le +C.$$

Perform a coordinate transformation to solve:

$$\xi = \hspace{0.4cm} x +y,$$

$$\eta= -x +y .$$

This corresponds to a rotation of the coordinate system by  $45^\circ$.

• From  $x+y= \pm C$  it thus follows  $\xi=\pm C$.
• The two two-dimensional density functions are then:

$$f_{xy} (x,y) = \frac{1}{4 \pi} \cdot \exp \left [ - ( x^2\hspace {-0.1cm} /2 + y^2\hspace {-0.1cm} /8) \right ] ,$$

$$f_{\xi\eta} (\xi, \eta) = \frac{1}{2 \pi \cdot \sigma_\xi \cdot \sigma_\eta \cdot \sqrt{1 - \rho_{\xi\eta}^2}} \cdot \exp \left [ - \frac{1}{2 \cdot (1 - \rho_{\xi\eta}^2)} \cdot ( \frac {\xi^2}{\sigma_\xi^2} + \frac {\eta^2}{\sigma_\eta^2 }- 2 \rho_{\xi\eta}\cdot \frac {\xi \cdot \eta}{\sigma_\xi \cdot \sigma_\eta}) \right ] .$$

Hints:

• The exercise belongs to the chapter  Two-dimensional Gaussian Random Variables.
• Reference is also made to the chapter  Rotation of the coordinate system.

• Given are the approximations  ${\rm Q}(2.3) \approx 0.01$  and  ${\rm Q}(2.6) \approx 0.005$  for the complementary Gaussian error integral.
• More information on this topic is provided in the learning video  Gaussian 2D random variables:

Part 1:   Gaussian random variables without statistical bindings,

Part 2:   Gaussian random variables with statistical bindings.

Questions

1

Determine the ratio of the two standard deviations of the new random variables by comparing coefficients $(\xi, \eta)$.

$\sigma_\xi/\sigma_\eta \ = \ $

2

Calculate the standard deviation  $\sigma_\xi$  and correlation coefficient  $\rho_{\xi\eta}$  between the new random variables  $\xi$  and  $\eta$.

$\sigma_\xi \ = \ $

$\rho_{\xi\eta} \ = \ $

3

Calculate the probability that  $ |\hspace{0.05cm}x+y\hspace{0.05cm}| \le C$  holds.  How large should  $C$  be chosen so that  $99\%$  of all sizes are in the shaded area?

$C_{99\%} \ = \ $

Solution

Solution

(1)  From  $\xi = x + y$  and  $\eta = -x + y$  it follows directly:

$$x = {1}/{2} \cdot ( \xi - \eta ) ,\hspace{0.5cm}y = {1}/{2}\cdot ( \xi +\eta ) .$$

• Substituting these values for the negative exponent, we get:

$$\frac{x^2}{2} + \frac{y^2}{8} = \frac{1}{8} \cdot ( \xi - \eta )^2 + \frac{1}{32} \cdot ( \xi + \eta )^2.$$

• Multiplied out, this gives:

$$\frac{5}{32} \cdot \xi^2 + \frac{5}{32} \cdot \eta^2 - \frac{3}{16} \cdot \xi \cdot \eta .$$

• Since the coefficients on  $\xi^2$  and  $\eta^2$  are equal, then  $\sigma_\xi = \sigma_\eta$.    The quotient we are looking for is therefore 1.

(2)  By comparing coefficients, we obtain for  $\sigma_\xi = \sigma_\eta$  the system of equations:

$$2 \cdot \sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)= \frac{32}{5},$$

$$\frac{\sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)}{\rho_{\xi\eta}}= \frac{16}{3}.$$

• Substituting the first equation into the second, we get  $\rho_{\xi\eta}\hspace{0.15cm}\underline {= 0.6}$  and  $\sigma_{\xi} = \sqrt{5}\hspace{0.15cm}\underline {= 2.236}$.

(3)  After coordinate transformation, we can write für this probability:

$${\rm Pr} ( | x + y | \le C ) = {\rm Pr} ( | \xi | \le C ) = 1 - 2 \cdot {\rm Pr} ( \xi >C ).$$

• With the complementary Gaussian error integral, it further follows:

$${\rm Pr} ( | x + y | \le C ) = 1 - 2 \cdot {\rm Q} ( {C}/{\sigma_\xi}) = 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q} ( {C}/{\sigma_\xi}) = 0.005.$$

• With the given value  ${\rm Q}(2.6) \approx 0.005$  we thus obtain the result:

$$C \approx 2.6 \cdot \sigma_{\xi}\hspace{0.15cm}\underline {= 5.814}.$$