Difference between revisions of "Aufgaben:Exercise 3.1: Cosine-square PDF and PDF with Dirac Functions"
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | + | '''(1)''' Correct are <u>statements 1, 2, and 4</u>: | |
− | '''(1)''' Correct are <u>statements 1, 2, and 4</u>: | + | * $x$ is continuous in value. |
− | * $x$ is continuous value. | + | * $y$ is discrete in value $(M = 5)$. |
− | * $y$ is discrete value $(M = 5)$. | ||
*The PDF does not provide information about whether a random variable is discrete or continuous in time. | *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 $1$ | + | [[File:P_ID174__Sto_A_3_1_b.png|right|frame|For calculating the PDF area]] |
− | *By simple geometric | + | '''(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 | + | '''(3)''' The probability that the continuous-valued random variable $x$ takes a fixed value $x_0$ is always negligibly small: |
− | *On the other hand, for the discrete value random variable $y$ holds according to the specification: ${\rm Pr}(y = 0) = 0.4$ | + | :$$\underline{{\rm Pr}(x = 0) = 0}.$$ |
+ | *On the other hand, for the discrete value random variable $y$ holds according to the specification: | ||
+ | :$${\rm Pr}(y = 0) = 0.4,$$ | ||
+ | :because the given weight of the Dirac delta function at $y = 0$ is $0.4$. | ||
− | '''(4)''' Because of ${{\rm Pr}(x = 0) = 0}$ and PDF symmetry, we get $\underline{{\rm Pr}(x > 0) = 0.5}$. | + | '''(4)''' Because of ${{\rm Pr}(x = 0) = 0}$ and the 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: | + | '''(5)''' Since $y$ is a discrete random variable, the probabilities for $y = 1$ and $y = 2$ add up: |
:$${\rm Pr}(y >0) = {\rm Pr}(y = 1) + {\rm Pr}( y = 2) \hspace{0.15cm}\underline {= 0.3}.$$ | :$${\rm Pr}(y >0) = {\rm Pr}(y = 1) + {\rm Pr}( y = 2) \hspace{0.15cm}\underline {= 0.3}.$$ | ||
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'''(7)''' The probability we are looking for is equal to the integral from $-1$ to $+1$ over the PDF of the continuous random variable $x$. | '''(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: | + | *Taking into account the symmetry and the given equation, we obtain: |
:$${\rm Pr}(|\hspace{0.05cm} x\hspace{0.05cm}|<1)=2 \cdot \int_{0}^{1}{1}/{2}\cdot \cos^2({\pi}/{4}\cdot x)\hspace{0.1cm}{\rm d}x={x}/{2}+{1}/{\pi}\cdot \sin({\pi}/{2}\cdot x)\Big |_{\rm 0}^{\rm 1}=\rm{1}/{2} + {1}/{\pi} | :$${\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{ | \hspace{0.15cm}\underline{ |
Revision as of 18:44, 31 December 2021
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
Solution
- $x$ is continuous in value.
- $y$ is discrete in 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:
- $${\rm Pr}(y = 0) = 0.4,$$
- because the given weight of the Dirac delta function at $y = 0$ is $0.4$.
(4) Because of ${{\rm Pr}(x = 0) = 0}$ and the 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:
- $${\rm Pr}(|\hspace{0.05cm} x\hspace{0.05cm}|<1)=2 \cdot \int_{0}^{1}{1}/{2}\cdot \cos^2({\pi}/{4}\cdot x)\hspace{0.1cm}{\rm d}x={x}/{2}+{1}/{\pi}\cdot \sin({\pi}/{2}\cdot x)\Big |_{\rm 0}^{\rm 1}=\rm{1}/{2} + {1}/{\pi} \hspace{0.15cm}\underline{ \approx 0.818}.$$