The lifetime of a type-A bulb is exponentially distributed with parameter λ. The lifetime of a type-B bulb is exponentially distributed with parameter μ, where μ>λ>0. You have a box full of lightbulbs of the same type, and you would like to know whether they are of type A or B. Assume an a priori probability of 1/4 that the box contains type-B lightbulbs.
Assume that λ=3 and μ=4. Find the LMS estimate of T2, the lifetime of another lightbulb from the same box, based on observing T1=2. Assume that conditioned on the bulb type, bulb lifetimes are independent.
The LMS estimate of T2 given T1=2 can be calculated using the conditional expectation formula. Let X be the type of bulb (A or B), and let T be the lifetime of a bulb. We are interested in finding E[T2 | T1=2].
By the law of total expectation:
E[T2 | T1=2] = E[T2 | X=A]P(X=A | T1=2) + E[T2 | X=B]P(X=B | T1=2]
Since P(X=A | T1=2) = 3/4 and P(X=B | T1=2) = 1/4, we have:
E[T2 | T1=2] = E[T2 | X=A] * 3/4 + E[T2 | X=B] * 1/4
Given that T1=2 and X=A, T2 follows an exponential distribution with parameter λ=3. Therefore, E[T2 | X=A] = 1/λ = 1/3.
Given that T1=2 and X=B, T2 follows an exponential distribution with parameter μ=4. Therefore, E[T2 | X=B] = 1/μ = 1/4.
Plugging in the values, we get:
E[T2 | T1=2] = (1/3) * 3/4 + (1/4) * 1/4 = 1/4 + 1/16 = 5/16
Therefore, the LMS estimate of T2 given T1=2 is 5/16.