Exercise 4.1: PDF, CDF and Probability

$\rm (CDF)$ (top), $\rm (PDF)$ (bottom)

To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with

the probability density function $\rm (PDF)$,
the cumulative distribution function $\rm (CDF)$.

The upper plot shows the cumulative distribution function $F_X(x)$ of a value-discrete random variable $X$. The corresponding probability density function $f_X(x)$ has to be determined in subtask (1).

The equation

$$ {\rm Pr}(A < X \le B) = F_X(B) - F_X(A) = \lim_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} \int_{A+\varepsilon}^{B+\varepsilon} \hspace{-0.15cm} f_X(x) \hspace{0.1cm}{\rm d}x $$

represents two ways to calculate the probability for the event "The random variable $X$ lies in a given interval" from the CDF and the PDF, respectively.

The lower graph shows the probability density function

$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\ \end{array} \right.\quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}l} | y| \le 2, \\ y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$

of a continuous random variable $Y$, which is restricted to the range $|Y| \le 2$ . In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable $Y$ as for a value-discrete random variable. Nevertheless, you will notice some differences in details.

For example, for the continuous random variable $Y$, the boundary transition can be omitted in the above equation, and we obtain simplified:

$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm} f_Y(y) \hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$

Hints:

The exercise belongs to the chapter Differential Entropy.
Useful hints for solving this problem and further information on continuous random variables can be found in the third chapter "Continuous Random Variables" of the book Theory of Stochastic Signals.

Given also is the following indefinite integral:

$$\int \hspace{0.1cm} \cos^2(A \eta) \hspace{0.1cm}{\rm d}\eta = \frac{\eta}{2} + \frac{1}{4A} \cdot \sin(2A \eta).$$

Questions

	The PDF is composed of five Dirac functions.
	${\rm Pr}(X= 0) = 0.4$ and ${\rm Pr}(X= 1) = 0.2$ are true.
	${\rm Pr}(X= 2) = 0.4$ is true.

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

${\rm Pr}(|X| ≤ 1) \ = \ $

$F_Y(y = 0) \ = \ $

$F_Y(y = 1) \ = \ $

$F_Y(y = 2) \ = \ $

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

	The result $Y = 0$ is impossible.
	The result $Y = 3$ is impossible.

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

${\rm Pr}(|Y| ≤ 1) \ = \ $

Solution

PDF and CDF of the value-discrete random variable $X$

(1) Proposed solutions 1 and 2 are correct:

The cumulative distribution function $F_X(x)$ is obtained from the probability density function $f_X(x)$ by integration over the (renamed) random variable in the range from $- \infty$ to $x$.
The inverse is: Given the CDF, obtain the PDF by differentiation.
The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:

$$f_X(x) = 0.1 \cdot {\rm \delta}( x+2) + 0.2 \cdot {\rm \delta}( x+1) + 0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1) + 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$

The Dirac weights give the occurrence probabilities of the random variable $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ , e.g.:

$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) = 0.7 - 0.3 = 0.4\hspace{0.05cm}.$$

Accordingly, the other probabilities are:

$${\rm Pr}(X = +1) = {\rm Pr}(X = -1) = 0.2\hspace{0.05cm},\hspace{0.3cm} {\rm Pr}(X = +2) = {\rm Pr}(X = -2) = 0.1\hspace{0.05cm}.$$

(2) From the PDF just calculated, we obtain:

$${\rm Pr}(X >0) = {\rm Pr}(X = +1) + {\rm Pr}(X = +2) \hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$

$${\rm Pr}(|X| \le 1) ={\rm Pr}(X = -1) + {\rm Pr}(X = 0) + {\rm Pr}(X = +1) = 0.2 + 0.4 +0.2 \hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$

The same result is obtained using the CDF. Here the general equation, which is equally valid for value-discrete and continuous random variables, is:

$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$

Thus, with $A= 0$ and $B = +2$ we obtain:

$${\rm Pr}(0 < X \le +2) = {\rm Pr}(X >0)= F_X(+2) - F_X(0) = 1 - 0.7 \hspace{0.15cm}\underline {= 0.3} \hspace{0.05cm}.$$

Setting $A=-2$ and $B = +1$, we get:

$${\rm Pr}(-2 < X \le +1) = {\rm Pr}(|X| \le 1)= F_X(+1) - F_X(-2) = 0.9 - 0.1 \hspace{0.15cm}\underline {= 0.8} \hspace{0.05cm}.$$

PDF and CDF of the continuous random variable $Y$

(3) The cumulative distribution function $F_Y(y)$ is obtained from the (renamed) WDF $f_Y(\eta)$ by integrating $- \infty$ to $x$. Due to symmetry, this can be written in the range $0 \le y \le +2$:

$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$

$$\Rightarrow \hspace{0.3cm}F_Y(y) = \frac{1}{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{0.1cm}\frac{1}{2} \cdot \cos^2({\pi}/{4} \cdot \eta) \hspace{0.1cm}{\rm d}\eta = \frac{1}{2}+\frac{y}{4} + \frac{1}{2\pi} \cdot \sin({\pi}/{2} \cdot y).$$

The equation holds in the entire range $0 \le y \le +2$. The CDF values we are looking for are thus:

$F_Y(y=0)\hspace{0.15cm}\underline{= 0.5}$ (integral over half the PDF),
$F_Y(y=1)= 3/4 + 1/(2 \pi)\hspace{0.15cm}\underline{= 0.909}$ (area in red background in the PDF),
$F_Y(y=2)\hspace{0.15cm}\underline{= 1}$ (integral over the entire PDF).

(4) The probability that the continuous random variable $Y$ lies in the range from $-\varepsilon$ to $+\varepsilon$ can be calculated using the given equation as follows:

$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$

It was taken into account that for the random variable $Y$ the "<"sign can be replaced by the "≤" sign without distortion.
With the boundary transition $\varepsilon \to 0$, the probability we are looking for is obtained:

$${\rm Pr}(Y = 0) =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) = F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$

Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.

In general: The probability ${\rm Pr}(Y = y_0)$ that a continuous random variable $Y$ takes a fixed value $y_0$, is always zero.

(5) Proposed solution 2 is correct:

Based on the PDF at hand, the result $Y=3$ can be excluded.
The result $Y=0$ on the other hand is quite possible, although ${\rm Pr}(Y = 0) = 0$ .
For example, if one performs a random experiment $N \to \infty$ times and obtains the result $Y= 0$ for $N_0$ times, then with finite $N_0$ according to the classical definition of probability:

$${\rm Pr}(Y = 0) = \lim_{N\hspace{0.05cm}\rightarrow\hspace{0.05cm}\infty}\hspace{0.1cm}{N_0}/{N} = 0\hspace{0.05cm}.$$

(6) We again assume the equation $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$ valid for the continuous random quantity $Y$:

With $A = 0$ and $B \to \infty$ $($or $B = 2)$ we obtain:

$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty) = {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0) \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$

Thus, for the symmetric continuous random variable $Y$ holds indeed as expected: ${\rm Pr}( Y > 0) = 1/2$.
Although the value-discrete random variable $X$ is also symmetrical about $x= 0$ ⇒ ${\rm Pr}( X > 0) = 0.3$ was determined in subtask (3), on the other hand.
Further, with $A = -1$ and $B = +1$, one obtains because of $F_Y(-1) = 1- F_Y(+1)$:

$${\rm Pr}( |Y| \le 1) = {\rm Pr}(-1 \le Y \le +1) = F_Y(+1) - F_Y(-1) = 2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$