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Proof. We need to prove that if you add a burst of length to a codeword (i.e. to a polynomial that is divisible by ()), then the result is not going to be a codeword (i.e. the corresponding polynomial is not divisible by ()).
where is the instance, [] the expectation value, is a class into which an instance is classified, (|) is the conditional probability of label for instance , and () is the 0–1 loss function: L ( x , y ) = 1 − δ x , y = { 0 if x = y 1 if x ≠ y {\displaystyle L(x,y)=1-\delta _{x,y}={\begin{cases}0&{\text{if }}x=y\\1&{\text{if }}x\neq y\end ...
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The BER is the likelihood of a bit misinterpretation due to electrical noise ().Considering a bipolar NRZ transmission, we have = + for a "1" and () = + for a "0".Each of () and () has a period of .
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x erf x 1 − erf x; 0: 0: 1: 0.02: 0.022 564 575: 0.977 435 425: 0.04: 0.045 111 106: 0.954 888 894: 0.06: 0.067 621 594: 0.932 378 406: 0.08: 0.090 078 126: 0.909 ...
The "68–95–99.7 rule" is often used to quickly get a rough probability estimate of something, given its standard deviation, if the population is assumed to be normal. It is also used as a simple test for outliers if the population is assumed normal, and as a normality test if the population is potentially not normal.