Exercise 1.16Z: Bounds for the Gaussian Error Function

Function ${\rm Q}(x)$ and approximations;
it holds: ${\rm Q_u}(x)\le{\rm Q}(x)\le{\rm Q_o}(x)$

The probability that a zero-mean Gaussian random variable $n$ with standard deviation $\sigma$ ⇒ variance $\sigma^2$ is greater in amount than a given value $A$ is equal to

$${\rm Pr}(n > A) = {\rm Pr}(n < -A) ={\rm Q}(A/\sigma) \hspace{0.05cm}.$$

Here is used one of the most important functions for Communications Engineering (drawn in red in the diagram):
the "complementary Gaussian error function"

$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u \hspace{0.05cm}.$$

${\rm Q}(x)$ is a monotonically decreasing function with ${\rm Q}(0) = 0.5$. For very large values of $x$ ⇒ ${\rm Q}(x)$ tends $\to 0$.

The integral of the ${\rm Q}$–function is not analytically solvable and is usually given in tabular form. From the literature, however, manageable approximations or bounds for positive $x$ values are known:

the "upper bound" ⇒ upper $($German: "obere" ⇒ subscript: "o"$)$ blue curve in adjacent graph, valid for $x > 0$:

$$ {\rm Q_o}(x)=\frac{\rm 1}{\sqrt{\rm 2\pi}\cdot x}\cdot {\rm e}^{-x^{\rm 2}/\rm 2}\hspace{0.15cm} \ge \hspace{0.15cm} {\rm Q} (x) \hspace{0.05cm},$$

the "lower bound" ⇒ upper $($German: "untere" ⇒ subscript: "u"$)$ blue curve in adjacent graph, valid for $x > 1$:

$$ {\rm Q_u}(x)=\frac{\rm 1-{\rm 1}/{\it x^{\rm 2}}}{\sqrt{\rm 2\pi}\cdot x}\cdot \rm e^{-x^{\rm 2}/\rm 2} \hspace{0.15cm} \le \hspace{0.15cm} {\rm Q} (x) \hspace{0.05cm},$$

the "Chernoff-Rubin bound" $($green curve in the graph, drawn for $K = 1)$:

$${\rm Q_{CR}}(x)=K \cdot {\rm e}^{-x^{\rm 2}/\rm 2} \hspace{0.15cm} \ge \hspace{0.15cm} {\rm Q} (x) \hspace{0.05cm}.$$

In the exercise it is to be investigated to what extent these bounds can be used as approximations for ${\rm Q}(x)$ and what corruptions result.

Hints:

This exercise belongs to the chapter "Bounds for block error probability".

Reference is also made to the chapter "Gaussian distributed random variables" in the book "Stochastic Signal Theory".

The exercise provides some important hints for solving "Exercise 1.16", in which ${\rm Q}_{\rm CR}(x)$ is used to derive the "Bhattacharyya Bound" for the AWGN channel.

Further we refer to the interactive applet "Complementary Gaussian error functions".

Questions

${\rm Q_{o}}(x = 4) \ = \ $

$\ \cdot 10^{-5} $

${\rm Q_{u}}(x = 4) \ = \ $

$\ \cdot 10^{-5} $

	For $x ≥ 2$: Both bounds are usable.
	For $x < 1$: ${\rm Q_{u}}(x)$ is unusable $($because ${\rm Q_{u}}(x)< 0)$.
	For $x < 1$: ${\rm Q_{o}}(x)$ is unusable $($because ${\rm Q_{o}}(x)> 1)$.

${\rm Q}_{\rm CR}(x = 2)/{\rm Q_{o}}(x = 2 ) \ = \ $

${\rm Q}_{\rm CR}(x = 4)/{\rm Q_{o}}(x = 4 ) \ = \ $

${\rm Q}_{\rm CR}(x = 6)/{\rm Q_{o}}(x = 6 ) \ = \ $

$K \ = \ $

Solution

(1) The upper bound is:

$${\rm Q_o}(x)=\frac{1}{\sqrt{\rm 2\pi}\cdot x}\cdot {\rm e}^{-x^{\rm 2}/\rm 2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q_o}(4 )=\frac{1}{\sqrt{\rm 2\pi}\cdot 4}\cdot {\rm e}^{-8 }\hspace{0.15cm}\underline{\approx 3.346 \cdot 10^{-5}}\hspace{0.05cm}.$$

The lower bound can be converted as follows:

$${\rm Q_u}( x)=(1-1/x^2) \cdot {\rm Q_o}(x) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q_u}(4 ) \hspace{0.15cm}\underline{\approx 3.137 \cdot 10^{-5}} \hspace{0.05cm}.$$

The relative deviations from the "real" value ${\rm Q}(4) = 3.167 · 10^{–5}$ sind $+5\%$ bzw. $–1\%$.

(2) Correct are the solutions 1 and 2:

For $x = 2$, the actual function value ${\rm Q}(x) = 2.275 - 10^{-2}$ is bounded by ${\rm Q_{o}}(x) = 2.7 - 10^{-2}$ and ${\rm Q_u}(x) = 2.025 - 10^{-2}$, respectively.
The relative deviations are therefore $18.7\%$ and $-11\%,$ respectively.
The last statement is wrong: Only for $x < 0.37$ ${\rm Q_o}(x) > 1$ is valid.

(3) For the quotient of ${\rm Q}_{\rm CR}(x)$ and ${\rm Q_o}(x)$, according to the given equations:

$$q(x) = \frac{{\rm Q_{CR}}(x)}{{\rm Q_{o}}(x)} = \frac{{\rm exp}(-x^2/2)}{{\rm exp}(-x^2/2)/({\sqrt{2\pi} \cdot x})} = {\sqrt{2\pi} \cdot x}$$

$$\Rightarrow \hspace{0.3cm} q(x) \approx 2.5 \cdot x \hspace{0.3cm} \Rightarrow \hspace{0.3cm} q(x =2) \hspace{0.15cm}\underline{=5}\hspace{0.05cm}, \hspace{0.2cm}q(x =4)\hspace{0.15cm}\underline{=10}\hspace{0.05cm}, \hspace{0.2cm}q(x =6) \hspace{0.15cm}\underline{=15}\hspace{0.05cm}.$$

The larger the abscissa value $x$ is, the more inaccurately ${\rm Q}(x)$ is approximated by ${\rm Q}_{\rm CR}(x)$.
When looking at the graph on the information page, one has (I had) the impression that ${\rm Q}_{\rm CR}(x)$ results from ${\rm Q}(x)$ by shifting down or shifting up. But this is only an optical illusion and does not correspond to the facts.

(4) With $\underline{K = 0.5}$ the new bound $0.5 - {\rm Q}_{\rm CR}(x)$ for $x = 0$ agrees exactly with ${\rm Q}(x=0) = 0.500$.

For larger abscissa values, the corruption $q \approx 1.25 - x$ thus also becomes only half as large.