Aufgaben:Exercise 4.4Z: Contour Lines of the "2D-PDF"

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Gaussian 2D PDF:   contour lines

Given a two-dimensional Gaussian random variable  $(x, y)$  with mean  $(0, 0)$  and the 2D PDF.

$$f_{xy}(x, y) = C\cdot{\rm e}^{-(x^{\rm 2} + y^{\rm 2} +\sqrt{\rm 2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm} y)}.$$

It is further known that the two standard deviations  $\sigma_x$  and  $\sigma_y$  are respectively equal  $1$ .

Entered in the sketch are:

• a height line of this PDF for  $f_{xy}(x,y) =0.2$,
• the (dark blue) ellipse major axis  $\rm (EA)$, and
• the (red) correlation line  $y=K(x)$.

Hints:

• The exercise belongs to the chapter  Two-dimensional Gaussian Random Variables.

• 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

How large is the correlation coefficient $\rho_{xy}$?

$\rho_{xy} \ = \ $

2

What is the maximum value  $C = f_{xy}(0, 0)$  of the PDF?

$C \ = \ $

3

What is the angle  $\alpha$  between ellipse major axis  $\rm (EA)$  and  $x$–axis?

$\alpha\ = \ $ $ \ \rm degrees$

4

At what values  $x_0$  and  $y_0$  respectively, does the hemline  $f_{xy}(x,y) = 0.2$  intersect the ellipse major axis?  What is the relationship between  $x_0$  and  $y_0$?

$x_0/y_0 \ = \ $

5

Which statements are true regarding the correlation line  $K(x)$ ?

	The correlation line is steeper than the ellipse major axis.
	The angle of  $K(x)$  with respect to  $x$–axis is about  $-35^\circ$.
	The correlation line intersects all hemlines where a vertical tangent can be applied to the ellipse.

Solution

Solution

(1)  Even without specifying  $\sigma_x = \sigma_y = 1$  one could see that the standard deviations  $\sigma_x$  and  $\sigma_y$  are equal,

• since in the exponent of $f_{xy}(x, y)$  the coefficients at  $x^2$  and  $y^2$  are equal.
• By comparing coefficients, we thus obtain:

$$\frac{- 2 \rho_{xy}}{\sigma_x\cdot\sigma_y} = \sqrt{2}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} \rho_{xy}=\frac{-1}{\sqrt{2}} \hspace{0.15cm}\underline{\approx -0.707}.$$

(2)  Using the numerical values calculated in point  (1)  we also obtain:

$$C=\frac{\rm 1}{\rm 2\it\pi\cdot\sigma_x\cdot\sigma_y\cdot\sqrt{\rm 1 - \rho_{xy}^{\rm 2}}} =\frac{\rm 1}{\rm 2\pi\cdot\rm 1\cdot 1\cdot\sqrt{0.5}}=\frac{\rm 1}{\sqrt{\rm 2}\cdot \pi}\hspace{0.15cm}\underline{\approx \rm 0.225}.$$

(3)  The general equation is:

$$\alpha = {\rm 1}/{\rm 2}\cdot \rm arctan \ (\rm 2 \cdot\it \rho_{xy}\cdot \frac{\sigma_x\cdot\sigma_y}{\sigma_x^{\rm 2} - \sigma_y^{\rm 2}}{\rm ).}.$$

• Applies  $\sigma_x = \sigma_y$  and  $\rho_{xy} \ne 0$,  then the angle is always  $\alpha = \pm 45^\circ$, where the sign is equal to the sign of  $\rho_{xy}$  .
• In the present case  $\alpha\hspace{0.15cm}\underline{ = -45^\circ}$ holds.

(4)  For the plotted contour line holds:

$$f_{xy}(x, y)=\frac{1}{\sqrt{2}\cdot \pi}\cdot {\rm e}^{(x^{2} + y^{2} + \sqrt{2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm}y)}=0.2\hspace{0.3cm} \rightarrow \hspace{0.3cm}{\rm e}^{-(x^{2} + y^{2} + \sqrt{2}\hspace{0.05cm} \cdot \hspace{0.05cm} x \hspace{0.05cm} \cdot \hspace{0.05cm}y)} = 0.8885 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} x^{\rm 2} + y^{\rm 2} + \sqrt{\rm 2}\cdot\hspace{0.05cm} x \hspace{0.05cm} \cdot \hspace{0.05cm}y = -{\rm ln(0.8885)} \approx\rm 0.118.$$

• The angle of the ellipse major axis is  $\alpha = -45^\circ$.  Therefore  $y_0 = - x_0$  must hold. It further follows:

$$x_{\rm 0}^{\rm 2} + (-x_{\rm 0})^{\rm 2} + \sqrt{\rm 2}\cdot x_{\rm 0}(-x_{\rm 0}) = 0.118$$

$$\Rightarrow \hspace{0.3cm}(\rm 2 - \sqrt{\rm 2})\cdot \it x_{\rm 0}^{\rm 2} = {\rm 0.118} \hspace{0.5cm}\rightarrow \hspace{0.5cm} x_{\rm 0}^{\rm 2} \approx \frac{\rm0.118}{\rm0.585}\approx\rm 0.202; \hspace{0.5cm} {\it x}_{\rm 0}\approx\pm\rm 0.450.$$

• The two intersections of the plotted contour lines with the ellipse major axis are thus at  $(+0.45, -0.45)$  and  $(-0.45, +0.45)$.
• The quotient in both cases is  $x_0/y_0 \hspace{0.15cm}\underline{ = -1}$.

(5)  Correct are the proposed solutions 2 and 3:

• With  $\sigma_x = \sigma_y$  and the result of the subtask  (1)  holds for the angle of the correlation line:

$$\theta_{y\hspace{0.05cm}\rightarrow \hspace{0.05cm}x} = \arctan (\rho_{\it xy})=\arctan(-{\rm 1}/{\sqrt{\rm 2}})\approx -\rm 35.3^{\circ}.$$

• This means:   The first statement is false and the second is true.

The following is the proof of the correctness of the last statement:

• Solving the elliptic equation  $($with  $z = 0. 118)$,  so  $x^{\rm 2}+ y^{\rm 2} +\sqrt{\rm 2}\cdot \it x\cdot \it y - \it z = \rm 0$,  to  $y$  we get after solving a quadratic equation:

$$y_{\rm 1, \ 2}={\sqrt{\rm 2}}/ {\rm 2} \cdot x\pm\sqrt{{x^{\rm 2}}/{\rm 2}-x^{\rm 2}+{\it z}} \hspace{0.5cm}\rightarrow \hspace{0.5cm} y_{\rm 1, \ 2}={\it x}/{\sqrt{\rm 2}}\pm \sqrt{z-{x^{\rm 2}}/{\rm 2}}.$$

• The vertical tangent results for the case that the two solutions  $y_{\rm 1, \rm 2}$  are identical.  That is:   The root expression must result in zero.
• The solution for positive  $x$  is then:   $x_{\rm T}=\sqrt{\rm 2\cdot \it z}=\rm \rm 0.485.$
• Inserted into the ellipse equation one obtains für the  $y$–value of the tangent point:

$$x_{\rm T}^{\rm 2} + y_{\rm T}^{\rm 2} + \sqrt{2} \cdot x_{\rm T} \cdot y_{\rm T} - z = 0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} 2 z + y_{\rm T}^{\rm 2} + 2\sqrt{ z}\cdot y_{\rm T} - z = 0$$

$$\Rightarrow \hspace{0.3cm}y_{\rm T}^{\rm 2} + 2\sqrt{ z}\cdot y_{\rm T} + z = 0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} (y_{\rm T} + \sqrt{ z}) = 0\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_{\rm T} = -\sqrt{ z} = -0.343.$$

• This gives  $y_{\rm T}=-{x_{\rm T}}/{\sqrt{\rm 2}}. $  But this also means:   The tangent point  $(x_{\rm T}, y_{\rm T})$  lies exactly on the correlation line  $y=K(x)=-{ x}/{\sqrt{\rm 2}}.$