Exercise 3.1: Cosine-square PDF and PDF with Dirac Functions

Cosine–square PDF (top), and
Dirac delta PDF (bottom)

The graph shows the probability density functions $\rm (PDF)$ of two random variables $x$ and $y$.

The PDF of the random variable $x$ in analytical form is:

$$f_x(x)=\left\{\begin{array}{*{4}{c}}A \cdot \cos^2({\pi}/{4}\cdot x) &\rm for\hspace{0.1cm} -2\le \it x\le \rm +2, \\0 & \rm else. \\\end{array}\right.$$

The PDF of the random variable $y$ consists of a total of five Dirac delta functions with the weights given in the graph.

If we consider these random variables as instantaneous values of two random signals $x(t)$ and $y(t)$,
it is obvious that both signals are "amplitude limited" to the range $\pm 2$. .

Hints:

The exercise belongs to the chapter Probability Density Function.
Reference is also made to the chapter From Random Experiment to Random Variable.
The following integral equation holds:

$$\int \cos^{\rm 2}( ax)\, {\rm d}x=\frac{x}{2}+\frac{1}{4 a}\cdot \sin(2 ax).$$

Questions

	The random variable $x$ is continuous in value.
	The random variable $y$ is discrete in value.
	The randomness of $y$ is also discrete in time.
	The PDF says nothing regarding "discrete-time/continuous-time."

$A \ = \ $

${\rm Pr}(x = 0)\ = \ $

${\rm Pr}(x > 0)\ = \ $

${\rm Pr}(y > 0)\ = \ $

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

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

Solution

For calculating PDF area

(1) Correct are statements 1, 2, and 4:

$x$ is continuous value.
$y$ is discrete value $(M = 5)$.
The PDF does not provide information about whether a random variable is discrete or continuous in time.

(2) The area under the PDF must yield $1$ .

By simple geometric reasoning, one arrives at the result $\underline{A=0.5}$.

(3) The probability that the continuous-valued random variable $x$ takes a fixed value $x_0$ is always negligibly small ⇒ $\underline{{\rm Pr}(x = 0) = 0}$.

On the other hand, for the discrete value random variable $y$ holds according to the specification: ${\rm Pr}(y = 0) = 0.4$ $($weight of the Dirac function at $y = 0)$.

(4) Because of ${{\rm Pr}(x = 0) = 0}$ and PDF symmetry, we get $\underline{{\rm Pr}(x > 0) = 0.5}$.

(5) Since $y$ is a discrete random variable, the probabilities for $y = 1$ and $y = 2$ add up:

$${\rm Pr}(y >0) = {\rm Pr}(y = 1) + {\rm Pr}( y = 2) \hspace{0.15cm}\underline {= 0.3}.$$

(6) The event $|\hspace{0.05cm} y \hspace{0.05cm} | < 1$ here is identical to $y = 0$. Thus we obtain:

$${\rm Pr}(|\hspace{0.05cm}y\hspace{0.05cm}| < 1) = {\rm Pr}( y = 0)\hspace{0.15cm}\underline { = 0.4}.$$

(7) The probability we are looking for is equal to the integral from $-1$ to $+1$ over the PDF of the continuous random variable $x$.

Taking into account the symmetry and the given equation, we obtain:

$${\rm Pr}(|\hspace{0.05cm} x\hspace{0.05cm}|<1)=2 \cdot \int_{0}^{1}{1}/{2}\cdot \cos^2({\pi}/{4}\cdot x)\hspace{0.1cm}{\rm d}x={x}/{2}+{1}/{\pi}\cdot \sin({\pi}/{2}\cdot x)\Big |_{\rm 0}^{\rm 1}=\rm{1}/{2} + {1}/{\pi} \hspace{0.15cm}\underline{ \approx 0.818}.$$