Question

Question:Suppose that a Bayesian spam filter is trained on a set of\({\bf{500}}\)spam messages and\({\bf{200}}\)messages that are not spam. The word “exciting” appears in\({\bf{40}}\)spam messages and in\({\bf{25}}\)messages that are not spam. Would an incoming message be rejected as spam if it contains the word “exciting” and the threshold for rejecting spam is\({\bf{0}}.{\bf{9}}\)?

Short answer

Answer

Hence, the incoming message containing the word ‘exciting’ is not rejected as spam.

Step-by-step solution

Given Data

• A Bayesian spam filter is trained on a set of \(500\) spam messages and \(200\) messages that are not spam. The word “exciting” appears in \(40\) spam messages and in \(25\) messages that are not spam.

Formula to be used

According to Bayes’ theorem, \(P\left( {F\left| E \right.} \right) = \frac{{P\left( {E\left| F \right.} \right)P\left( F \right)}}{{P\left( {E\left| F \right.} \right)P\left( F \right) + P\left( {E\left| {\overline F } \right.} \right)P\left( {\overline F } \right)}}\)

find the message category

Assume the following events-

\(E\)is a received message contains the word ‘exciting’.

\({F_1}\)is that an incoming message is spam and\({F_2}\)is that the message received is not a spam.

Thus, the probability for a message received containing the word ‘exciting’ to be a spam is-

\(\begin{array}{l}P\left( {{F_1}\left| E \right.} \right) = \frac{{P\left( {E\left| {{F_1}} \right.} \right)P\left( {{F_1}} \right)}}{{P\left( {E\left| {{F_1}} \right.} \right)P\left( {{F_1}} \right) + P\left( {E\left| {{F_2}} \right.} \right)P\left( {{F_2}} \right)}}\\ = \frac{{\left( {\frac{{40}}{{500}}} \right)\left( {\frac{{500}}{{700}}} \right)}}{{\left( {\frac{{40}}{{500}}} \right)\left( {\frac{{500}}{{700}}} \right) + \left( {\frac{{25}}{{200}}} \right)\left( {\frac{{200}}{{700}}} \right)}}\\ = \frac{8}{{13}}\\ = 0.61538 < 0.9\end{array}\)

So, the incoming message is not rejected as spam.