Question

Question: Suppose that we have prior information concerning whether a random incoming message is spam. In particular, suppose that over a time period, we find that s spam messages arrive and h messages arrive that are not spam.

a) Use this information to estimate p(S), the probability that an incoming message is spam, and\(p(\bar S)\), the probability an incoming message is not spam.

b) Use Bayes’ theorem and part (a) to estimate the probability that an incoming message containing the word w is spam, where p(w)is the probability that w occurs in a spam message and q(w) is the probability that w occurs in a message that is not spam.

Short answer

Answer:

(a)\(p(S) = \frac{s}{{s + h}}\) and \(p(\bar S) = \frac{h}{{s + h}}\)

(b) The probability that an incoming message containing the word w is spam is \(\frac{{p(w).s}}{{p(w).s + q(w).h}}\)

Step-by-step solution

Given

We have prior information concerning whether a random incoming message is spam. In particular, suppose that over a time period, we find that s spam messages arrive and h messages arrive that are no spam.

Formula

Bayes’ Formula:

\(P(F/E) = \frac{{P(E/F)P(F)}}{{P(E/F)P(F) + P(E/\bar F)P(\bar F)}}\)

Calculation for part(a)

We have\(p(S) = \frac{s}{{s + h}}\)and

\(\begin{array}{l}p(\bar S) = 1 - p(S)\\ = 1 - \frac{s}{{s + h}}\\ = \frac{{s + h - s}}{{s + h}}\\ = \frac{h}{{s + h}}\end{array}\)

Final Answer for part(a)

Hence,

\(p(S) = \frac{s}{{s + h}}\)and \(p(\bar S) = \frac{h}{{s + h}}\)

Calculation for part(b)

The probability that an incoming message containing the word w is spam is

\(\frac{{p(E/S).p(S)}}{{p(E/S).p(S) + p(E/\bar S).p(\bar S)}} = \frac{{p(w).s}}{{p(w).s + q(w).h}}\)

Here p(w)is the probability that w occurs in a spam message and q(w) is the probability that w occurs in a message that is not spam.

Final Answer for part(b)

The probability that an incoming message containing the word w is spam is\(\frac{{p(w).s}}{{p(w).s + q(w).h}}\)