[[File:P_ID2752__Inf_Z_3_2_neu.png|right|frame|$\rm PMF$ of the two-dimensional random variable $XY$]]
Wir betrachten die Zufallsgrößen
We consider the random variables $X = \{ 0,\ 1,\ 2,\ 3 \}$ and $Y = \{ 0,\ 1,\ 2 \}$, whose joint probability mass function $P_{XY}(X,\ Y)$ is given.
*From this two-dimensional probability mass function $\rm (PMF)$, the one-dimensional probability mass functions $P_X(X)$ and $P_Y(Y)$ are to be determined.
*Such a one-dimensional probability mass function is sometimes also called "marginal probability".
$X$ = { 0, 1, 2, 3 },
$Y$ = { 0, 1, 2 },
If $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables $X$ and $Y$ are statistically independent. Otherwise, there are statistical dependencies between them.
deren gemeinsame Wahrscheinlichkeitsfunktion $P_{X,Y}(X,Y)$ gegeben ist. Aus dieser 2D–Wahrscheinlichkeitsfunktion sollen die eindimensionalen Wahrscheinlichkeitsfunktionen $P_X(X)$ und $P_Y(Y)$ ermittelt werden. Man nennt eine solche manchmal auch Randwahrscheinlichkeit (englisch: Marginal Probability).
In the second part of the task we consider the random variables $U= \big \{ 0,\ 1 \big \}$ and $V= \big \{ 0,\ 1 \big \}$, which result from $X$ and $Y$ by modulo-2 operations:
Gilt $P_{X,Y}(X,Y)$ = $P_X(X)$ . $P_Y(Y)$, so sind die beiden Zufallsgrößen X und Y statistisch unabhängig. Andernfalls bestehen statistische Bindungen zwischen $X$ und $Y$.
:$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm} V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
Im zweiten Teil der Aufgabe betrachten wir die Zufallsgrößen
$U$ = { 0, 1 }, $V$ = { 0, 1 },
die sich aus $X$ und $Y$ durch Modulo–2–Operationen ergeben:
$U$ = $X$ mod 2, $V$ = $Y$ mod 2.
'''Hinweis:''' Die Aufgabe gehört zum Themengebiet von [http://http://en.lntwww.de/Informationstheorie_und_Quellencodierung/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen Kapitel 3.1]. Ausgegangen wird hier von der gleichen Konstellation wie in
<u>Hints:</u>
*The exercise belongs to the chapter [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on two-dimensional random variables]].
*The same constellation is assumed here as in [[Aufgaben:Aufgabe_3.2:_Erwartungswertberechnungen|Exercise 3.2]].
*There the random variables $Y = \{ 0,\ 1,\ 2,\ 3 \}$ were considered, but with the addition ${\rm Pr}(Y = 3) = 0$.
*The property $|X| = |Y|$ forced in this way was advantageous in the previous task for the formal calculation of the expected value.
===Questions===
<quiz display=simple>
{What is the probability mass function $P_X(X)$?
|type="{}"}
$P_X(0) \ = \ $ { 0.5 3% }
$P_X(1) \ = \ $ { 0.125 3% }
$P_X(2)\ = \ $ { 0. }
$P_X(3) \ = \ ${ 0.375 3% }
{What is the probability mass function $P_Y(Y)$?
|type="{}"}
$P_Y(0) \ = \ $ { 0.5 3% }
$P_Y(1) \ = \ $ { 0.25 3% }
$P_Y(2) \ = \ $ { 0.25 3% }
===Fragebogen===
{Are the random variables $X$ and $Y$ statistically independent?
<quiz display=simple>
|type="()"}
{Multiple-Choice Frage
- Yes,
|type="[]"}
+ No.
- Falsch
+ Richtig
{Input-Box Frage
{Determine the probabilities $P_{UV}( U,\ V)$.
|type="{}"}
|type="{}"}
$\alpha$ = { 0.3 }
$P_{UV}( U = 0,\ V = 0) \ = \ $ { 0.375 3% }
$P_{UV}( U = 0,\ V = 1) \ = \ $ { 0.375 3% }
$P_{UV}( U = 1,\ V = 0) \ = \ $ { 0.125 3% }
$P_{UV}( U =1,\ V = 1) \ = \ $ { 0.125 3% }
{Are the random variables $U$ and $V$ statistically independent?
|type="()"}
+ Yes,
- No.
Line 47:
Line 70:
</quiz>
</quiz>
===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''1.'''
'''(1)''' You get from $P_{XY}(X,\ Y)$ to the one-dimensional probability mass function $P_X(X)$ by summing up all $Y$ probabilities:
'''(3)''' With statistical independence, $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$ should be.
*This does not apply here: answer <u>'''NO'''</u>.
'''(4)''' Starting from the left-hand table ⇒ $P_{XY}(X,Y)$, we arrive at the middle table ⇒ $P_{UY}(U,Y)$, <br>by combining certain probabilities according to $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.
If one also takes into account $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, one obtains the probabilities sought according to the right-hand table:
[[File:P_ID2753__Inf_Z_3_2d_neu.png|right|frame|Different probability functions]]
:$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
:$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
:$$P_{UV}( U = 1, V = 0) = 1/8 \hspace{0.15cm}\underline{= 0.125},$$
:$$P_{UV}( U = 1, V = 1) = 1/8 \hspace{0.15cm}\underline{= 0.125}.$$
'''(5)''' The correct answer is <u>'''YES'''</u>:
*The corresponding one-dimensional probability mass functions are:
$\rm PMF$ of the two-dimensional random variable $XY$
We consider the random variables $X = \{ 0,\ 1,\ 2,\ 3 \}$ and $Y = \{ 0,\ 1,\ 2 \}$, whose joint probability mass function $P_{XY}(X,\ Y)$ is given.
From this two-dimensional probability mass function $\rm (PMF)$, the one-dimensional probability mass functions $P_X(X)$ and $P_Y(Y)$ are to be determined.
Such a one-dimensional probability mass function is sometimes also called "marginal probability".
If $P_{XY}(X,\ Y) = P_X(X) \cdot P_Y(Y)$, the two random variables $X$ and $Y$ are statistically independent. Otherwise, there are statistical dependencies between them.
In the second part of the task we consider the random variables $U= \big \{ 0,\ 1 \big \}$ and $V= \big \{ 0,\ 1 \big \}$, which result from $X$ and $Y$ by modulo-2 operations:
$$U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2, \hspace{0.3cm} V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2.$$
(3) With statistical independence, $P_{XY}(X,Y)= P_X(X) \cdot P_Y(Y)$ should be.
This does not apply here: answer NO.
(4) Starting from the left-hand table ⇒ $P_{XY}(X,Y)$, we arrive at the middle table ⇒ $P_{UY}(U,Y)$, by combining certain probabilities according to $U = X \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$.
If one also takes into account $V = Y \hspace{0.1cm}\text{mod} \hspace{0.1cm} 2$, one obtains the probabilities sought according to the right-hand table:
Different probability functions
$$P_{UV}( U = 0, V = 0) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
$$P_{UV}( U = 0, V = 1) = 3/8 \hspace{0.15cm}\underline{= 0.375},$$
$$P_{UV}( U = 1, V = 0) = 1/8 \hspace{0.15cm}\underline{= 0.125},$$
$$P_{UV}( U = 1, V = 1) = 1/8 \hspace{0.15cm}\underline{= 0.125}.$$
(5) The correct answer is YES:
The corresponding one-dimensional probability mass functions are:
$$P_U(U) = \big [1/2 , \ 1/2 \big ],$$
$$P_V(V)=\big [3/4, \ 1/4 \big ].$$
Thus: $P_{UV}(U,V) = P_U(U) \cdot P_V(V)$ ⇒ $U$ and $V$ are statistically independent.
