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 lambda.
One of the engines will fail first, followed by the second, and followed by the last.
Let T1 be the time of the first failure, T2 be the time of the second failure, and T3 be the time of the third failure.
1. Determine the PDF of T1.
for t>0,
f_T1(t) = ?
3*lambda*exp(- 3*t*lambda)
To find the probability density function (PDF) of T1, we need to determine the probability that the first engine fails at time t (T1 = t) given that all three engines have started operating at time 0.
Since the lifetime of each engine is exponentially distributed with parameter lambda, the probability that an engine fails within a small time interval dt is given by lambda * e^(-lambda * t) * dt.
Now, let's consider the event that the first engine fails at time t (T1 = t). This event occurs if the first engine lasts at least t seconds before failing, and the second and third engines last longer than t seconds. Therefore, we have the following condition:
T1 = t if T2 > t and T3 > t.
To find the probability that T1 = t, we can express it in terms of conditional probabilities:
P(T1 = t) = P(T2 > t and T3 > t | T1 = t) * P(T1 = t)
Since we assume all engines fail independently, we can express P(T2 > t and T3 > t | T1 = t) in terms of independent probabilities:
P(T2 > t and T3 > t | T1 = t) = P(T2 > t | T1 = t) * P(T3 > t | T1 = t)
Since T2 and T3 represent the time until the second and third engine failures, we can consider them independently from T1. Thus, the conditional probabilities simplify to:
P(T2 > t | T1 = t) = P(T2 > t)
Similarly,
P(T3 > t | T1 = t) = P(T3 > t)
Since these probabilities only depend on the exponential distribution of each engine's lifetime, we can directly use the formula for the exponential CDF (cumulative distribution function). The CDF of an exponentially distributed random variable X with parameter lambda is given by:
F_X(t) = 1 - e^(-lambda * t)
Therefore, the probability that an engine lasts longer than t seconds is given by:
P(T2 > t) = 1 - F_2(t) = 1 - (1 - e^(-lambda * t)) = e^(-lambda * t)
Similarly,
P(T3 > t) = 1 - F_3(t) = 1 - (1 - e^(-lambda * t)) = e^(-lambda * t)
Combining all the results, we have:
P(T1 = t) = P(T2 > t) * P(T3 > t) = e^(-lambda * t) * e^(-lambda * t) = e^(-2 * lambda * t)
Thus, the PDF of T1 is given as:
f_T1(t) = d/dt (e^(-2 * lambda * t)) = -2 * lambda * e^(-2 * lambda * t)
Therefore, the PDF of T1 for t > 0 is f_T1(t) = -2 * lambda * e^(-2 * lambda * t).