Alexander Brock

$e^{\ln {x}}=x,\forall x\in \mathbb {R} ^{+}$

$\ln({a\cdot b})=\ln a+\ln b$

$\ln {a \over b}=\ln a-\ln b$

$P(w_{i}|C)={N_{C,i} \over N_{C}}$

$P(A|B)={{P(B|A)\cdot P(A)} \over {P(B)}}$

$P(w_{1},w_{2},...,w_{n}|C)=P(w_{1}|C)\cdot P(w_{2}|C)\cdot ...\cdot P(w_{n}|C)={\prod _{k=1}^{i}{N_{C,k} \over N_{C}}}$

$P(Spam)={{\color {Brown}N_{Spam}}\prod _{k=1}^{i}{{\color {Red}N_{Spam,k}} \over {\color {Brown}N_{Spam}}} \over {{\color {Blue}N_{Ham}}\prod _{k=1}^{i}{{\color {OliveGreen}N_{Ham,k}} \over {\color {Blue}N_{Ham}}}}}$

$P(Spam)={{{\color {Brown}N_{Spam}} \over {\color {Blue}N_{Ham}}}{\prod _{k=1}^{i}{{\color {Red}N_{Spam,k}} \over {\color {OliveGreen}N_{Ham,k}}}}{\prod _{k=1}^{i}{{\color {Blue}N_{Ham}} \over {\color {Brown}N_{Spam}}}}}|\ln(...)$

$\ln(P(Spam))={\ln({\color {Brown}N_{Spam}})-\ln({\color {Blue}N_{Ham}})+\sum _{k=1}^{i}{\ln \left({{\color {Red}N_{Spam,k}} \over {\color {OliveGreen}N_{Ham,k}}}\right)}+\sum _{k=1}^{i}{\ln \left({{\color {Blue}N_{Ham}} \over {\color {Brown}N_{Spam}}}\right)}}$

$\ln(P(Spam))={{\Bigl (}i-1{\Bigr )}{{\Bigl (}\ln({\color {Blue}N_{Ham}})-\ln({\color {Brown}N_{Spam}}){\Bigr )}}+\sum _{k=1}^{i}{{\Bigl (}\ln({\color {Red}N_{Spam,k}})-\ln({\color {OliveGreen}N_{Ham,k}}){\Bigr )}}}$

$P(Spam)=\exp {\Biggl (}{{\Bigl (}i-1{\Bigr )}{{\Bigl (}\ln({\color {Blue}N_{Ham}})-\ln({\color {Brown}N_{Spam}}){\Bigr )}}+\sum _{k=1}^{i}{{\Bigl (}\ln({\color {Red}N_{Spam,k}})-\ln({\color {OliveGreen}N_{Ham,k}}){\Bigr )}}{\Biggr )}}$

$P(Spam)=\exp {\bigl (}{sum(ln)+(i-1)({\color {Blue}ham}-{\color {Brown}spam}){\bigr )}}$

$(i-1)(\ln({\color {Blue}N_{Ham}})-\ln({\color {Brown}N_{Spam}}))$

$10^{-1500}$

$\ln {N_{Spam,i} \over N_{Ham,i}}$

$spam=spam+\Delta spam$

$\ln(spam+\Delta spam)-\ln {e^{(}\ln(spam)-ln)},spam=spam+n$