Suppose that we have three engines, which we turn on at time 0. Each engine will eventually fail, and we model each engine"s lifetime as exponentially distributed with parameter λ. The lifetimes of different engines are independent. One of the engines will fail first, followed by the second, and followed by the last. Let Ti be the time of the first failure·T, be the time of the second failure, and 13 be the time of the third failure. For answers involving algebraic expressions, enter "lambda" for λ and use "exp "for exponentials. Follow standard notation.
1. Determine the PDF of T1. For t>0,
fT1(t)=
2. Let X=T2−T1. Determine the conditional PDF fX|T1(x|t). For x,t>0,
fX|T1(x∣t)=
To determine the PDF of T1, we need to consider the probability distribution function of an exponential random variable. The PDF of an exponential random variable with parameter λ is given by f(t) = λ * exp(-λ * t), for t > 0.
1. The time at which the first engine fails, T1, is an exponential random variable with parameter λ. Therefore, the PDF of T1 is given by fT1(t) = λ * exp(-λ * t), for t > 0.
2. Let X = T2 - T1. We want to determine the conditional PDF of X given T1, denoted as fX|T1(x|t), where x and t are both greater than 0.
To find the conditional PDF, we can use the properties of exponential random variables. The time until the second engine fails, T2, is also an exponential random variable with the same parameter λ. However, since the second engine cannot fail before the first engine, we have a lower bound on T2, which is T1. Therefore, the PDF of T2 given T1 (T2|T1) is given by fT2|T1(t|s) = λ * exp(-λ * (t-s)), for t >= s.
Given X = T2 - T1, we can substitute T2 = X + T1 in the conditional PDF and solve for fX|T1(x|t). Substituting, we have:
fX|T1(x|t) = fT2|T1(x+t|t) = λ * exp(-λ * (x+t - t)) = λ * exp(-λ * x).
Therefore, the conditional PDF of X given T1 is fX|T1(x|t) = λ * exp(-λ * x), for x,t > 0.
Note: This assumes that the lifetimes of the engines are independent.