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 μλ0. 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 1/3 that the box contains type-B lightbulbs.

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 1/3 that the box contains type-B lightbulbs. You observe the value t1 of the lifetime, T1, of a lightbulb. A MAP decision rule implies that the lightbulb is of type A if and only if t1≥α. Find α. Express your answer in terms of μ and λ. Use ‘mu', ‘lambda' and ‘ln' to denote μ, λ, and the natural logarithm function, respectively. For example, ln2μλ should be entered as ‘ln((2*mu)/lambda)'. α=0 - incorrect 0 What is the probability of error of the MAP estimator? 13e−μα+23(1−e−λα) 13e−μα+23(1−e−λα) - correct23e−μα+13(1−e−λα)13(1−e−μα)+23e−λα23(1−e−μα)+13e−λα Assume that λ=2 and μ=3. 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 two decimal places.) LMS estimate of T2=?

diogenes · Accepted Answer

In 2019: 1. ln(mu/(3*lambda))/(mu-lambda) 2. (a): (1/4)*e^(-mu*alpha) + (3/4)(1-e^(-lambda*alpha)) 3. 0.3286

Jacob · Answer

2) Part a

RVE · Answer

2.) 1/3*e^(-μ*α) + 2/3*(1-(e^-λ*α)) please, share what you have got for the whole problem set!!

ubn · Answer

alpha = ? LMS estimate of T2 = ? Please

mary · Answer

can anyone answer the rest of the problem set please?

alec · Answer

part 3: LMS estimate of T2 = 0.48 part 1: alpha = ?

RVE · Answer

1. alpha = ln(mu/(2*lambda))/(mu-lambda)

Anonymous · Answer

===𝖤(𝑇2∣A)𝖯(A∣𝑇1=2)+𝖤(𝑇2∣B)𝖯(B∣𝑇1=2)(Using the Means of Exponential Distributions)
==>1/2*𝖯(A∣𝑇1=2)+1/3*𝖯(B∣𝑇1=2)
𝖯(A∣𝑇1=2)=4𝖾^2/(4𝖾^2+3)
𝖯(B∣𝑇1=2)=3/(4𝖾^2+3)

diogenes · Answer

In 2019 the answer to part 2 is (a). (1/4)*e^(-mu*alpha) + (3/4)(1-e^(-lambda*alpha))

Zorry · Answer

Answer of Diogenes Works in 2022 1. ln(mu/(3*lambda))/(mu-lambda) 2. (a): (1/4)*e^(-mu*alpha) + (3/4)(1-e^(-lambda*alpha)) 3. 0.3286