Exercise 4.7

From the statement of the exercise, we know that

\begin{align} \ Pr(X \mid Y = yes) & = f_{yes}(x) = N(\mu = 10, \sigma^2 = 36),\ Pr(X \mid Y = no) & = f_{no}(x) = N(\mu = 0, \sigma^2 =36),\ Pr(Y = yes) & = \pi_{yes} = 0.8,\ Pr(Y = no)& = \pi_{no} = 0.2.\ \end{align}

We want to calculate \(Pr(Y = "Yes \mid X = 4)\). Using Bayes’ theorem and substituting the expressions above:

\begin{align} \ \pi_{yes}(x) & = \frac{\pi_{yes} f_{yes}(x) }{ \pi_{yes} f_{yes}(x) + \pi_{no} f_{no}(x)},\ \ \pi_{yes}(4) & = \frac{\pi_{yes} f_{yes}(4) }{ \pi_{yes} f_{yes}(4) + \pi_{no} f_{no}(4)} = \ & = \frac{0.8 e^{\left( -\frac{1}{2\times 36}(4-10)^2 \right)} }{ 0.8 e^{\left( -\frac{1}{2\times 36}(4-10)^2 \right)} + 0.2 e^{\left( -\frac{1}{2\times 36}(4-0)^2 \right)}} \ & = 0.7571.\ \end{align}