vault backup: 2023-03-24 12:29:32
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@@ -43,7 +43,7 @@ $$H(x)=-\sum\limits_{i=1}^{3}P(x_{i})ld\ P(x_{i})= P(x_{1})=P(x_1y_1)+P(x_1y_{2)
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$$P(y_{1})=P(x1y1)+p(x2y1)+p(x3y1)=0.6 max ld 1$$
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Średnia entropia dwóch źródeł - max ld 6
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H(x,y)= $-\sum_{i}\sum_{j}P(xiyj)ldP(xiyj)$
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H(x,y)= $-\sum_{i}\sum_{j}P(x_iy_j)ldP(x_iy_j)$
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H(x/y) x pod warunkiem y , a posteriori=$-\sum_{i}\sum_{j}P(xiyj)ldP(xi/yj)$
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H(x/y)=-01ld0.1-0.25ld0.25-0.2ld0.2-0ld0-0.3ld0.3-0.15ld0.15 = max ld 6
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@@ -63,4 +63,6 @@ hahahahahahahahahahahahahahahahahahahahahahahahhahahahhahahashdambgbvakabs
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"
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:I just read a book about Stockholm syndrome. It was pretty bad at first, but by the end I liked it."
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\end{gathered}$$
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\end{gathered}$$
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prawdopodobieństwo warunkowe: $P(X/Y)=\frac{P(X,Y)}{P(Y)}$
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18
TIiK/Wykład/3. Łańcuchy markowa.md
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18
TIiK/Wykład/3. Łańcuchy markowa.md
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@@ -0,0 +1,18 @@
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0->0 (p(0/0)=1/8)
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0->1 (p(1/0)=/8)
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1->1(p(1/1)=1/4)
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1->0(p(0/1)=3/4)
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Źródło nieergodyczne - uśrednianie po czasie
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o->0 = 0.999
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0-1 = 0.001
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1-1 = 1
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$$
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\begin{bmatrix}
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0.8 & 0.2 & 0 & 0 \\0 & 0 & 0.5 & 0.5 \\ 0.5 & 0.5 & 0 & 0 \\ 0 & 0 & 0.2 & 0.8 \\
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\end{bmatrix}
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$$
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$Hz(x)$
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0
TIiK/Ćwiczenia/2. Markow.md
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0
TIiK/Ćwiczenia/2. Markow.md
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