[[File:P_ID146__Sto_Z1_1.png|right|framed|Sum $S$ of two <br>ternary signals $X$ and $Y$]]
[[File:P_ID146__Sto_Z1_1.png|right|framed|Sum $S$ of two <br>ternary signals $X$ and $Y$]]
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*At the signal source $X$, the values $-1$, $0$ and $+1$ occur with equal probability.
*At the signal source $X$, the values $-1$, $0$ and $+1$ occur with equal probability.
*For source $Y$, the signal value $0$ is twice as likely as the other two values $-1$ and $+1$, respectively.
*For source $Y$, the signal value $0$ is twice as likely as the other two values $-1$ and $+1$, respectively.
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[[File:EN_Sto_Z1_1_c_neu.png|right|frame|400px|Sum and difference of ternary random variables]]
'''(2)''' $S$ can take a total of $\underline {I =5}$ values, namely $0$, $\pm 1$ and $\pm 2$.
'''(2)''' $S$ can take a total of $\underline {I =5}$ values, namely $0$, $\pm 1$ and $\pm 2$.
[[File:EN_Sto_Z1_1_c.png|right|frame|400px|Sum and difference of ternary random variables]]
'''(3)''' Since $Y$ is not equally distributed, one cannot (actually) apply the "Classical Definition of Probability" here.
'''(3)''' Since $Y$ is not equally distributed, one cannot (actually) apply the "Classical Definition of Probability" here.
*However, if we divide $Y$ into four ranges according to the graph, assigning two of the ranges to the event $Y = 0$ , we can still proceed according to the classical definition.
*However, if we divide $Y$ into four ranges according to the graph, assigning two of the ranges to the event $Y = 0$, we can still proceed according to the classical definition.
*One then obtains:
*One then obtains:
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'''(4)''' It is also evident from the graph that the difference signal $D$ and the sum signal $S$ take the same values with equal probabilities.
'''(4)''' It is also evident from the graph that the difference signal $D$ and the sum signal $S$ take the same values with equal probabilities.
*This was to be expected, since ${\rm Pr}(Y = +1) ={\rm Pr}(Y = -1)$ is given ⇒ <u>Proposed solution 1</u>.
*This was to be expected, since ${\rm Pr}(Y = +1) ={\rm Pr}(Y = -1)$ is given ⇒ <u>Proposed solution 1</u>.
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[[Category:Theory of Stochastic Signals: Exercises|^1.1 Some Basic Definitions
[[Category:Theory of Stochastic Signals: Exercises|^1.1 Some Basic Definitions
Let two three-stage message sources $X$ and $Y$ be given, whose output signals can only assume the values $-1$, $0$ and $+1$ respectively. The signal sources are statistically independent of each other.
A simple circuit now forms the sum signal $S = X + Y$.
At the signal source $X$, the values $-1$, $0$ and $+1$ occur with equal probability.
For source $Y$, the signal value $0$ is twice as likely as the other two values $-1$ and $+1$, respectively.
Solve the subtasks (3) and (4) according to the classical definition.
Nevertheless, consider the different occurrence frequencies of the signal $Y$.
The topic of this section is illustrated with examples in the (German language) learning video Klassische Definition der Wahrscheinlichkeit $\Rightarrow$ "Classical definition of probability".
(2) $S$ can take a total of $\underline {I =5}$ values, namely $0$, $\pm 1$ and $\pm 2$.
(3) Since $Y$ is not equally distributed, one cannot (actually) apply the "Classical Definition of Probability" here.
However, if we divide $Y$ into four ranges according to the graph, assigning two of the ranges to the event $Y = 0$, we can still proceed according to the classical definition.
