Question

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. (For this part, you will need a calculator. Provide an answer with an accuracy of three decimal places.)

LMS estimate of T2 = ?

Answer 1

It is mentioned that conditioned on the bulb type, bulb lifetimes are independent.

That means, lifetimes on type A bulbs are independent and lifetimes on type B bulbs are independent.

Again, we can assume that the lifetime of one type A bulb and lifetime of one type B bulbs- both are independent.

So, the lifetime of another lightbulb from the same box does not depend on the lifetime of the previous light bulb/(s) irrespective of it is of which type. LMS estimate of T2 given T1=2 and LMS estimate of T2 are same as T2 does not depend on T1.

LMS(Least mean squared) estimate of T2 = E(T2/T1= 2) = E(T2)

= E(lifetime of type A bulb/type A bulb is chosen)

+ E(lifetime of type B bulb/type B bulb is chosen)

Conditional probability density function

f(lifetime of type A bulb/type A bulb is chosen)

= f(lifetime of type A bulb, type A bulb is chosen) * f(type A bulb is chosen)

= (1/3)Exp(-x/3) * (3/4)

[[priori probability that the box contains type-B lightbulbs is 1/4. So, priori probability that the box contains type-B lightbulbs is 1-(1/4) = 3/4. here X denotes the lifetime of type A bulb]]

Similarly,

Conditional probability density function

f(lifetime of type B bulb/type B bulb is chosen)

= f(lifetime of type B bulb, type B bulb is chosen) * f(type B bulb is chosen)

= (1/4)Exp(-y/4) * (1/4)

[[priori probability that the box contains type-B lightbulbs is 1/4. here Y denotes the lifetime of type A bulb]]

E(lifetime of type A bulb/type A bulb is chosen)

= x*f(lifetime of type A bulb/type A bulb is chosen) dx = (3/4) E(X) = (3/4)*3 = 9/4

Similarly,

E(lifetime of type B bulb/type B bulb is chosen)

= x*f(lifetime of type B bulb/type B bulb is chosen) dx = (1/4) E(Y) = (1/4)*4 = 1

LMS(Least mean squared) estimate of T2 = E(T2/T1= 2) = E(T2) = (9/4) + 1 = 3.25