Theory of Stochastic Signals/Linear Combinations of Random Variables

1. Theory of Stochastic Signals
2. Random Variables with Statistical Dependence
3. Linear Combinations of Random Variables

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Prerequisites and mean values

Throughout the chapter  "Linear Combinations of Random Variables"  we make the following assumptions:

• The random variables  $u$  and  $v$  are zero mean each   ⇒   $m_u = m_v = 0$  and also statistically independent of each other   ⇒   $ρ_{uv} = 0$.
• The two random variables  $u$  and  $v$  each have equal standard deviation  $σ$.  No statement is made about the nature of the distribution.
• Let the two random variables  $x$  and  $y$  be linear combinations of  $u$  and  $v$,  where:

$$x=A \cdot u + B \cdot v + C,$$

$$y=D \cdot u + E \cdot v + F.$$

Thus,  for the  (linear)  mean values of the new random variables  $x$  and  $y$  we obtain according to the general rules of calculation for expected values:

$$m_x =A \cdot m_u + B \cdot m_v + C =C,$$

$$m_y =D \cdot m_u + E \cdot m_v + F =F.$$

Thus,  the coefficients  $C$  and  $F$  give only the mean values of  $x$  and  $y$.  Both are always set to zero in the following sections.

Resulting correlation coefficient

Let us now consider the  »variances«  of the random variables according to the linear combinations.

• For the random variable  $x$  holds independently of the parameter  $C$:

$$\sigma _x ^2 = {\rm E}\big[x ^{\rm 2}\big] = A^{\rm 2} \cdot {\rm E}\big[u^{\rm 2}\big] + B^{\rm 2} \cdot {\rm E}\big[v^{\rm 2}\big] + {\rm 2} \cdot A \cdot B \cdot {\rm E}\big[u \cdot v\big].$$

• The expected values of  $u^2$  and  $v^2$  are by definition equal to  $σ^2$,  because  $u$  and  $v$  are zero mean.
• Since  $u$  and  $v$  are moreover assumed to be statistically independent,  one can write for the expected value of the product:

$${\rm E}\big[u \cdot v\big] = {\rm E}\big[u\big] \cdot {\rm E}\big[v\big] = m_u \cdot m_v = \rm 0.$$

• Thus,  for the variances of the random variables formed by linear combinations,  we obtain:

$$\sigma _x ^2 =(A^2 + B^2) \cdot \sigma ^2,$$

$$\sigma _y ^2 =(D^2 + E^2) \cdot \sigma ^2.$$

The  »covariance«  $μ_{xy}$  is identical to the joint moment  $m_{xy}$ for zero mean random variables  $x$  and  $y$   ⇒   $C = F = 0$:

$$\mu_{xy } = m_{xy } = {\rm E}\big[x \cdot y\big] = {\rm E}\big[(A \cdot u + B \cdot v)\cdot (D \cdot u + E \cdot v)\big].$$

Note here that  ${\rm E}\big[ \text{...} \big]$  denotes an expected value,  while  $E$  describes a coefficient.

$\text{Conclusion:}$ After evaluating this equation in an analogous manner to above,  it follows:

$$\mu_{xy } = (A \cdot D + B \cdot E) \cdot \sigma^{\rm 2 } \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \rho_{xy } = \frac{\rho_{xy } }{\sigma_x \cdot \sigma_y} = \frac {A \cdot D + B \cdot E}{\sqrt{(A^{\rm 2}+B^{\rm 2})(D^{\rm 2}+E^{\rm 2} ) } }. $$

We now exclude two special cases:

• $A = B = 0$  ⇒   $x ≡ 0$,
• $D = E = 0$  ⇒   $y ≡ 0$.

Then the above equation always yields unique values for the correlation coefficient in the range  $-1 ≤ ρ_{xy} ≤ +1$.

$\text{Example 1:}$  If we set  $A = E = 0$,  we get the correlation coefficient  $ρ_{xy} = 0$.  This result is insightful:

• Now  $x$  depends only on  $v$  and  $y$  depends exclusively on  $u$.
• But since  $u$  and  $v$  were assumed to be statistically independent,  there are also no relationships between  $x$  and  $y$.
• Similarly,   $ρ_{xy} = 0$  results for the combination  $B = D = 0$.

$\text{Example 2:}$  The constellation  $B = E = 0$  leads to the fact that both  $x$  and  $y$  depend only on  $u$.  Then the following correlation coefficient is obtained:

$$\rho_{xy } = \frac {A \cdot D }{\sqrt{A^{\rm 2}\cdot D^{\rm 2} } } = \frac {A \cdot D }{\vert A\vert \cdot D\vert } =\pm 1. $$

• If  $A$  and  $D$  have the same sign,  then  $ρ_{xy} = +1$.
• For different signs,  the correlation coefficient is  $-1$.
• Also for  $A = D = 0:$  The coefficient  $ρ_{xy} = ±1$,  if  $B \ne 0$  and  $E \ne 0$  holds.

Generation of correlated random variables

The  $\text{previously used equations}$  can be used to generate a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  with given characteristics  $σ_x$,  $σ_y$  and  $ρ_{xy}$.

• If no further preconditions are met other than these three nominal values, one of the four coefficients  $A, \ B, \ D$  and  $E$  is arbitrary.
• In the following, we always arbitrarily set  $E = 0$ .
• With the further specification that the statistically independent random variables  $u$  and  $v$  each have standard deviation  $σ =1$  we obtain:

$$D = \sigma_y, \hspace{0.5cm} A = \sigma_x \cdot \rho_{xy}, \hspace{0.5cm} B = \sigma_x \cdot \sqrt {1-\rho_{xy}^2}.$$

• For  $σ ≠ 1$  these values must still be divided by  $σ$  in each case.

$\text{Example 3:}$  We always assume zero mean Gaussian random variables  $u$  and  $v$.  Both have variance  $σ^2 = 1$.

$(1)$   To generate a two-dimensional random variable with desired characteristics   $σ_x =1$,  $σ_y = 1.55$  and  $ρ_{xy} = -0.8$   the parameter set is suitable, for example:

Two-dimensional random variables generated by linear combination

$$A = -0.8, \; B = 0.6, \; D = 1.55, \; E = 0.$$

• This parameter set underlies the left graph.
• The regression line  $(\rm RL)$  is shown in red.
• It runs at an angle of about  $-50^\circ$.
• Drawn in purple is the ellipse major axis, which lies slightly above the regression line.

$(2)$   The parameter set for the right graph is:

$$A = -0.625,\; B = 0.781,\; D = 1.501,\; E = -0.390.$$

• In a statistical sense,  the same result is obtained even though the two point clouds differ in detail.
• With respect to regression line  $(\rm RL)$  and ellipse major axis  $(\rm EA)$,  there is no difference with respect to the parameter set  $(1)$.

Exercises for the chapter

Exercise 4.7: Weighted Sum and Difference

Exercise 4.7Z: Generation of a Joint PDF

Exercise 4.8: Diamond-shaped Joint PDF

Exercise 4.8Z: AWGN Channel