Exercise 4.3Z: Dirac-shaped "2D-PDF"

Dirac-shaped 2D– PDF

The graph shows the two-dimensional probability density function $f_{xy}(x, y)$ of two discrete random variables $x$, $y$.

This 2D–PDF consists of eight Dirac points, marked by crosses.
The numerical values indicate the corresponding probabilities.
It can be seen that both $x$ and $y$ can take all integer values between the limits $-2$ and $+2$.
The variances of the two random variables are given as follows: $\sigma_x^2 = 2$, $\sigma_y^2 = 1.4$.

Hints:

The exercise belongs to the chapter Two-Dimensional Random Variables.
Reference is also made to the chapter Moments of a Discrete Random Variable

Questions

	The probabilities for $-2$, $-1$, $0$, $+1$ and $+2$ are equal.
	The random variable $x$ is mean-free $(m_x = 0)$.
	The probability ${\rm Pr}(x \le 1)=0.9$.

	The probabilities for $-2$, $-1$, $0$, $+1$ and $+2$ are equal.
	The random variable $y$ is mean-free $(m_y = 0)$.
	The probability ${\rm Pr}(y \le 1)=0.9$.

$F_{xy}(+1, +1) \ = \ $

${\rm Pr}(x ≤ 1\hspace{0.05cm} | \hspace{0.05cm}y ≤ 1)\ = \ $

$m_{xy}\ = \ $

$\rho_{xy}\ = \ $

$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x}\ = \ $

$\ \rm degrees$

	The random variables $x$ and $y$ are statistically independent.
	It can already be seen from the given 2D–PDF that $x$ and $y$ are statistically dependent on each other.
	From the calculated correlation coefficient $\rho_{xy}$ one can conclude the statistical dependence between $x$ and $y$ .

Solution

(1) Correct are the first two answers:

The marginal probability density function $f_{x}(x)$ is obtained from the 2D–PDF $f_{xy}(x, y)$ by integration over $y$.
For all possible values $ x \in \{-2, -1, \ 0, +1, +2\}$ the probabilities are equal $0.2$.
It holds ${\rm Pr}(x \le 1)= 0.8$. The mean is $m_x = 0$.

Discrete marginal PDF $f_{y}(y)$

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

By integration over $x$ one obtains the PDF $f_{y}(y)$ sketched on the right.
Due to symmetry, the mean value $m_y = 0$ is obtained.
The probability we are looking for is ${\rm Pr}(y \le 1)= 0.9$.

(3) By definition:

$$F_{xy}(r_x, r_y) = {\rm Pr} \big [(x \le r_x)\cap(y\le r_y)\big ].$$

For $r_x = r_y = 1$ it follows:

$$F_{xy}(+1, +1) = {\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ].$$

As can be seen from the 2D–PDF on the information page, this probability is ${\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ]\hspace{0.15cm}\underline{=0.8}$.

(4) For this, Bayes' theorem can also be used to write:

$$ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = \frac{ \rm Pr\big [(\it x \le \rm 1)\cap(\it y\le \rm 1)\big ]}{ \rm Pr(\it y\le \rm 1)} = \it \frac{F_{xy} \rm (1, \rm 1)}{F_{y}\rm (1)}.$$

With the results from (2) and (3) it follows $ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = 0.8/0.9 = 8/9 \hspace{0.15cm}\underline{=0.889}$.

(5) According to the definition, the common moment is:

$$m_{xy} = {\rm E}\big[x\cdot y \big] = \sum\limits_{i} {\rm Pr}( x_i \cap y_i)\cdot x_i\cdot y_i. $$

There remain five Dirac delta functions with $x_i \cdot y_i \ne 0$:

$$m_{xy} = \rm 0.1\cdot (-2) (-1) + 0.2\cdot(-1) (-1)+ 0.2\cdot 1\cdot 1 + 0.1\cdot 2\cdot 1+ 0.1\cdot 2\cdot 2\hspace{0.15cm}\underline{=\rm 1.2}.$$

2D–PDF and correlation line $y = K(x)$

(6) For the correlation coefficient:

$$\rho_{xy} = \frac{\mu_{xy}}{\sigma_x\cdot \sigma_y} = \frac{1.2}{\sqrt{2}\cdot\sqrt{1.4}}\hspace{0.15cm}\underline{=0.717}.$$

This takes into account that because $m_x = m_y = 0$ the covariance $\mu_{xy}$ is equal to the moment $m_{xy}$ .

The equation of the correlation line is:

$$y=\frac{\sigma_y}{\sigma_x}\cdot \rho_{xy}\cdot x = \frac{\mu_{xy}}{\sigma_x^{\rm 2}}\cdot x = \rm 0.6\cdot \it x.$$

See sketch on the right. The angle between the correlation straight line $K(x)$ and the $x$-axis is

$$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x} = \arctan(0.6) \hspace{0.15cm}\underline{=31^\circ}.$$

(7) The correct solutions are solutions 2 and 3:

If statistically independent, $f_{xy}(x, y) = f_{x}(x) \cdot f_{y}(y)$ should hold, which is not done here.
From correlatedness $($follows from $\rho_{xy} \ne 0)$ it is possible to directly infer statistical dependence,
because correlation means a special form of statistical dependence, namely linear statistical dependence.