Translation examples
Thomas Bayes-isä, Joshua Bayes, oli yksi kuuden ensimmäisen Nonconformist ministereitä on vihitty Englanti.
Prezentare Thomas Bayes' father, Joshua Bayes, was one of the first six Nonconformist ministers to be ordained in England.
Naiivi Bayes-valitsin on termi Bayes tilastot käsitellessään yksinkertainen todennäköisyyspohjaisen valitsimeen perust
A naive Bayes classifier is a term in Bayesian statistics dealing with a simple probabilistic classifier based on applying Bayes' theorem with strong (naive) independence assumptions.
Monia käytännön sovelluksia parametrin estimointi naiivi Bayes malleja käyttää menetelmää suurin todennäköisyys; Toisin sanoen voi työskennellä naiivi Bayes-mallin ilman uskoa Bayes-todennäköisyys tai Bayes-men
In many practical applications, parameter estimation for naive Bayes models uses the method of maximum likelihood; in other words, one can work with the naive Bayes model without believing in Bayesian probability or using any Bayesian methods.
Monet Bayes' s paperit olivat sen vuoksi hintaa.
Several of Bayes 's papers were therefore given to Price.
Tämä vaikuttaa ovat syntyneet kautta hänen ystävyyttä Bayes.
This seems to have come about through his friendship with Bayes .
Bayesin teoreema (myös Bayesin sääntö tai Bayesin laki) on ehdolliseen todennäköisyyteen liittyvä matemaattinen teoreema.
In probability theory and statistics, Bayes' theorem (alternatively Bayes' law or Bayes' rule) describes the probability of an event, based on prior knowledge of conditions that might be related to the event.
Bayesilainen roskapostisuodatus soveltaa Bayesin teoreemaa.
A central rule of Bayesian inference is Bayes' theorem.
Tätä vaikutusta kutsutaan posteriori-todennäköisyydeksi ja se lasketaan Bayesin teoreemaa soveltaen.
This contribution is called the posterior probability and is computed using Bayes' theorem.
Bayesin tiedetään julkaisseen kaksi teosta: Divine Benevolence, or an Attempt to Prove That the Principal End of the Divine Providence and Government is the Happiness of His Creatures (1731) ja An Introduction to the Doctrine of Fluxions, and a Defence of the Mathematicians Against the Objections of the Author of the Analyst (1736), jossa hän puolustaa Isaac Newtonin laskentaoppia George Berkeleyn hyökkäystä vastaan.
He is known to have published two works in his lifetime, one theological and one mathematical: Divine Benevolence, or an Attempt to Prove That the Principal End of the Divine Providence and Government is the Happiness of His Creatures (1731) An Introduction to the Doctrine of Fluxions, and a Defence of the Mathematicians Against the Objections of the Author of The Analyst (published anonymously in 1736), in which he defended the logical foundation of Isaac Newton's calculus ("fluxions") against the criticism of George Berkeley, author of The Analyst It is speculated that Bayes was elected as a Fellow of the Royal Society in 1742 on the strength of the Introduction to the Doctrine of Fluxions, as he is not known to have published any other mathematical works during his lifetime.
Tällöin Zi:n ehdollinen jakauma voidaan kirjoittaa todennäköisyytenä Bayesin kaavan mukaisesti: T j , i ( t ) := P ( Z i = j | X i = x i ; θ ( t ) ) = τ j ( t ) f ( x i ; μ j ( t ) , Σ j ( t ) ) τ 1 ( t ) f ( x i ; μ 1 ( t ) , Σ 1 ( t ) ) + τ 2 ( t ) f ( x i ; μ 2 ( t ) , Σ 2 ( t ) ) {\displaystyle T_{j,i}^{(t)}:=\operatorname {P} (Z_{i}=j|X_{i}=\mathbf {x} _{i};\theta ^{(t)})={\frac {\tau _{j}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{j}^{(t)},\Sigma _{j}^{(t)})}{\tau _{1}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{1}^{(t)},\Sigma _{1}^{(t)})+\tau _{2}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{2}^{(t)},\Sigma _{2}^{(t)})}}} .
Given our current estimate of the parameters θ(t), the conditional distribution of the Zi is determined by Bayes theorem to be the proportional height of the normal density weighted by τ: T j , i ( t ) := P ( Z i = j | X i = x i ; θ ( t ) ) = τ j ( t ) f ( x i ; μ j ( t ) , Σ j ( t ) ) τ 1 ( t ) f ( x i ; μ 1 ( t ) , Σ 1 ( t ) ) + τ 2 ( t ) f ( x i ; μ 2 ( t ) , Σ 2 ( t ) ) {\displaystyle T_{j,i}^{(t)}:=\operatorname {P} (Z_{i}=j|X_{i}=\mathbf {x} _{i};\theta ^{(t)})={\frac {\tau _{j}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{j}^{(t)},\Sigma _{j}^{(t)})}{\tau _{1}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{1}^{(t)},\Sigma _{1}^{(t)})+\tau _{2}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{2}^{(t)},\Sigma _{2}^{(t)})}}} .
Tapahtumat A = ”vanki A armahdetaan” B = ”vanki B armahdetaan” C = ”vanki C armahdetaan” b = ”vartija nimeää vangin B joutuvan teloitettavaksi (ei armahdettu)” Nyt Bayesin teoreemaa käyttäen saadaan, että vangin A mahdollisuus tulla armahdetuksi on: P ( A | b ) = P ( b | A ) P ( A ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(A|b)={\frac {P(b|A)P(A)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 2 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 1 3 . {\displaystyle ={\frac {{\tfrac {1}{2}}\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {1}{3}}.} Jokaisella vangilla on 1/3 mahdollisuus tulla armahdetuksi.
Call A {\displaystyle A} , B {\displaystyle B} and C {\displaystyle C} the events that the corresponding prisoner will be pardoned, and b {\displaystyle b} the event that the warden tells A that prisoner B is to be executed, then, using Bayes' theorem, the posterior probability of A being pardoned, is: P ( A | b ) = P ( b | A ) P ( A ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(A|b)={\frac {P(b|A)P(A)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 2 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 1 3 . {\displaystyle ={\frac {{\tfrac {1}{2}}\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {1}{3}}.} The probability of C being pardoned, on the other hand, is: P ( C | b ) = P ( b | C ) P ( C ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(C|b)={\frac {P(b|C)P(C)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 2 3 . {\displaystyle ={\frac {1\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {2}{3}}.} The crucial difference making A and C unequal is that P ( b | A ) = 1 2 {\displaystyle P(b|A)={\tfrac {1}{2}}} but P ( b | C ) = 1 {\displaystyle P(b|C)=1} .
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