The lifetime of a type-A bulb is exponentially distributed with parameter $2$. The lifetime of a type-B bulb is exponentially distributed with parameter $3$. You have a box full of lightbulbs of the same type, with an a prior probability of $1/3$ that the box contains type-B lightbulbs.Based on observing $T_1=2$, Find the LMS estimate of $T_2$, the lifetime of another lightbulb from the same box.
Because LMS is conditional expectation, I tried $$E[T_2|T_1] = E[A]P(A|T_1)+E[B]P(B|T_1)$$ But my answer is wrong, where did I make the mistake?
Best Answer
Close. What you want is:
$$\begin{align} & \mathsf E(T_2\mid T_1=2) \\[1ex] = & \qquad(\text{by Law of Iterated Expectation}) \\[1ex] & \mathsf E(T_2\mid T_1=2, \mathrm A)\;\mathsf P(\mathrm A\mid T_1=2)+\mathsf E(T_2\mid T_1=2,\mathrm B)\;\mathsf P(\mathrm B\mid T_1=2) & \\[1ex] = & \qquad(\text{since expected life of bulbs in the same box are conditionally independent}) \\[1ex] &\mathsf E(T_2\mid \mathrm A)\;\mathsf P(\mathrm A\mid T_1=2)+\mathsf E(T_2\mid \mathrm B)\;\mathsf P(\mathrm B\mid T_1=2) \\[1ex] = & \qquad(\text{Using the Means of Exponential Distributions}) \\[1ex] & \tfrac 1 2\;\mathsf P(\mathrm A\mid T_1=2)+\tfrac 1 3\;\mathsf P(\mathrm B\mid T_1=2) \end{align}$$
Now, can you evaluate those posterior probabilities?