Question

Question: Suppose that and are the events that an incoming mail \({E_1}\)message contains the words \({w_1}\) and \({w_2}\), respectively. Assuming that \({E_1}\) and \({E_2}\) are independent events and that \({E_1}\left| S \right.\) and \({E_2}\left| S \right.\) are independent events, where S is the event that an incoming message is spam, and that we have no prior knowledge regarding whether or not the message is spam, show that

\(p(S|{E_1} \cap {E_2}) = \frac{{p({E_1}|S)p({E_2}|S)}}{{p({E_1}|S)p({E_2}|S) + p({E_1}|\bar S)p({E_2}|\bar S)}}\)

Short answer

Answer:

We have proved that \(p(S|{E_1} \cap {E_2}) = \frac{{p({E_1}|S)p({E_2}|S)}}{{p({E_1}|S)p({E_2}|S) + p({E_1}|\bar S)p({E_2}|\bar S)}}\)

Step-by-step solution

Given

\({E_1}\)and \({E_2}\) are the events that an incoming mail message contains the words \({w_1}\) and \({w_2}\), respectively

To Show: \(p(S|{E_1} \cap {E_2}) = \frac{{p({E_1}|S)p({E_2}|S)}}{{p({E_1}|S)p({E_2}|S) + p({E_1}|\bar S)p({E_2}|\bar S)}}\)

Formula

Generalized Bayes’ Theorem:

\(P({F_j}/E) = \frac{{P(E/{F_j})P(F)}}{{\sum\nolimits_{j = 1}^n {P(E/{F_j})P({F_j})} }}\)

Proof

By applying Bayes’ Theorem

\(p(S/{E_1} \cap {E_2}) = \frac{{p({E_1} \cap {E_2}/S)p(S)}}{{p({E_1} \cap {E_2}/S)p(S) + p({E_1} \cap {E_2}/\bar S)p(\bar S)}}\)

We are assuming that there is no prior knowledge whether a message is or is not a spam,

\(p(S) = p(\bar S) = 0.5\)

So, the above equation becomes

\(p(S/{E_1} \cap {E_2}) = \frac{{p({E_1} \cap {E_2}/S)}}{{p({E_1} \cap {E_2}/S) + p({E_1} \cap {E_2}/\bar S)}}\)

We have assumed the independence of \({E_1}\) and \({E_2}\)and\(S\)

So, we have

\(p({E_1} \cap {E_2}/S) = p({E_1}|S).p({E_2}|S)\)

And similarly for the S

Final Answer

Hence, we showed that \(p(S|{E_1} \cap {E_2}) = \frac{{p({E_1}|S)p({E_2}|S)}}{{p({E_1}|S)p({E_2}|S) + p({E_1}|\bar S)p({E_2}|\bar S)}}\)