Exercise 4.4: Two-dimensional Gaussian PDF

Table: Gaussian error functions

We consider two-dimensional random variables, where both components are always assumed to be mean-free.

The 2D–PDF of the random variable $(u, v)$ is:

$$f_{uv}(u, v)={1}/{\pi} \cdot {\rm e}^{-(2u^{\rm 2} \hspace{0.05cm}+ \hspace{0.05cm}v^{\rm 2}\hspace{-0.05cm}/\rm 2)}.$$

The following parameters are known from another two-dimensional random variable $(x, y)$ which is also Gaussian:

$$\sigma_x= 0.5, \hspace{0.5cm}\sigma_y = 1,\hspace{0.5cm}\rho_{xy} = 1. $$

The values of the Gaussian error integral ${\rm \phi}(x)$ and the complementary function ${\rm Q}(x) = 1- {\rm \phi}(x)$ can be found in the adjacent table.

Hints:

The exercise belongs to the chapter Two-dimensional Gaussian Random Variables.
Reference is also made to the chapter Gaussian distributed random variables.
More information on this topic is provided in the (German language) learning video "Gaußsche 2D-Zufallsgrößen":

Part 1: Gaussian random variables without statistical bindings,

Part 2: Gaussian random variables with statistical bindings.

Questions

	The random variables $u$ and $v$ are uncorrelated.
	The random variables $u$ and $v$ are statistically independent.

$\sigma_u/\sigma_v \ = \ $

${\rm Pr}(u < 1)\ = \ $

${\rm Pr}\big[(u < 1) ∩ (υ > 1)\big]\ = \ $

	The joint PDF $f_{xy}(x, y)$ is always zero outside the straight line $y = 2x$.
	For all pairs of values on the straight line $y = 2x$ holds: $f_{xy}(x, y)= 0.5$.
	With respect to the edge PDFs: $f_{x}(x) = f_{u}(u)$ and $f_{y}(y) = f_{v}(v)$ holds.

${\rm Pr}(x < 1)\ = \ $

${\rm Pr}\big[(x < 1) ∩ (y > 1)\big]\ = \ $

Solution

(1) Both statements are true:

Comparing the given 2D PDF with the general 2D PDF.

$$f_{uv}(u,v) = \frac{\rm 1}{{\rm 2}\it\pi \cdot \sigma_u \cdot \sigma_v \cdot \sqrt{{\rm 1}-\it \rho_{\it uv}^{\rm 2}}} \cdot \rm exp\left[\frac{\rm 1}{2\cdot (\rm 1-\it \rho_{uv}^{\rm 2}{\rm )}}(\frac{\it u^{\rm 2}}{\it\sigma_u^{\rm 2}} + \frac{\it v^{\rm 2}}{\it\sigma_v^{\rm 2}} - \rm 2\it\rho_{uv}\frac{\it u\cdot \it v}{\sigma_u\cdot \sigma_v}\rm )\right],$$

so it can be seen that no term with $u \cdot v$ occurs in the exponent, which is only possible with $\rho_{uv} = 0$ mögible.

But this means that $u$ and $v$ are uncorrelated.
For Gaussian random variables, however, statistical independence always follows from uncorrelatedness.

(2) With statistical independence holds:

$$f_{uv}(u, v) = f_u(u)\cdot f_v(v), \hspace{0.5cm} f_u(u)=\frac{{\rm e}^{-{\it u^{\rm 2}}/{(2\sigma_u^{\rm 2})}}}{\sqrt{\rm 2\pi}\cdot\sigma_u} , \hspace{0.5cm} \it f_v{\rm (}v{\rm )}=\frac{{\rm e}^{-{\it v^{\rm 2}}/{{\rm (}{\rm 2}\sigma_v^{\rm 2}{\rm )}}}}{\sqrt{\rm 2\pi}\cdot\sigma_v}.$$

By comparing coefficients, we get $\sigma_u = 0.5$ and $\sigma_v = 1$.
The quotient is thus $\sigma_u/\sigma_v\hspace{0.15cm}\underline{=0.5}$.

Probability: $\rm Pr\big[(\it u < \rm 1) \cap (\it v > \rm 1)\big]$

(3) Since $u$ is a continuous random variable, holds:

$$\rm Pr(\it u < \rm 1) = \rm Pr(\it u \le \rm 1) =\it F_u\rm (1). $$

With the mean $m_u = 0$ and the standard deviation $\sigma_u = 0.5$ we get:

$$\rm Pr(\it u < \rm 1) = \rm \phi({\rm 1}/{\it\sigma_u})= \rm \phi(\rm 2) \hspace{0.15cm}\underline{=\rm 0.9772}. $$

(4) Due to the statistical independence between $u$ and $v$ holds:

$$\rm Pr\big[(\it u < \rm 1) \cap (\it v > \rm 1)\big] = \rm Pr(\it u < \rm 1)\cdot \rm Pr(\it v > \rm 1).$$

The probability ${\rm Pr}(u < 1) =0.9772$ has already been calculated.
For the second probability ${\rm Pr}(v > 1)$ holds for reasons of symmetry:

$$\rm Pr(\it v > \rm 1) = \rm Pr(\it v \le \rm (-1) = \it F_v\rm (-1) = \rm \phi(\frac{\rm -1}{\it\sigma_v}) = \rm Q(1) =0.1587$$

$$\Rightarrow \hspace{0.3cm} \rm Pr\big[(\it u < \rm 1) \cap (\it v > \rm 1)\big] = \rm 0.9772\cdot \rm 0.1587 \hspace{0.15cm}\underline{ = \rm 0.1551}.$$

The sketch illustrates the given constellation:

The PDF (blue) height lines are stretched ellipses due to $\sigma_v > \sigma_u$ in vertical direction.
Drawn in red shading is the area whose probability should be calculated in this subtask.

2D Dirac "wall" on the correlation line

(5) Correct are the first and the third suggested solutions:

Because $\rho_{xy} = 1$ there is a deterministic correlation between $x$ and $y$

⇒ All values lie on the straight line $y =K(x) \cdot x$.

Because of the standard deviations $\sigma_x = 0.5$ and $\sigma_y = 1$ it holds $K = 2$.
On this straight line $y = 2x$ all PDF values are infinitely large.
This means: The 2D PDF is here a "Dirac wall".
As you can see from the sketch, the PDF values are distributed evenly on the straight line $y = 2x$&nbsp.
The straight line $y = 2x$ also represents the correlation line.
The two marginal probability densities are also Gaussian functions, each with zero mean.
Because $\sigma_x = \sigma_u$ and $\sigma_y = \sigma_v$ also holds:

$$f_x(x) = f_u(u), \hspace{0.5cm}f_y(y) = f_v(v).$$

Probability calculation for the Dirac wall

(6) Since the PDF of the random variable isö&aerospace;e $x$ identical to the PDF $f_u(u)$, it also results in exactly the same probability as calculated in the subtask (3) :

$$\rm Pr(\it x < \rm 1) \hspace{0.15cm}\underline{ = \rm 0.9772}.$$

(7) The random event $y > 1$ is identical to the event $x > 0.5$.

Thus, the wanted probability is equal to:

$$\rm Pr \big[(\it x > \rm 0.5) \cap (\it x < \rm 1)\big] = \it F_x \rm( 1) - \it F_x\rm (0.5). $$

With the standard deviation $\sigma_x = 0.5$ follows further:

$$\rm Pr \big[(\it x > \rm 0.5) \cap (\it x < \rm 1)\big] = \rm \phi(\rm 2) - \phi(1)=\rm 0.9772- \rm 0.8413\hspace{0.15cm}\underline{=\rm 0.1359}.$$