Probability distribution of total life time of a machine with two parts in parallel system

Question

Two identical components having lifetimes $A$ and $B$ are connected in parallel in a system

.Suppose the distributions of $A$ and $B$ independently follow exponential with mean $\frac 1a, a>0$. But whenever one component fails the lifetime distribution of another changes to exponential with mean $\frac 1c ,c>0$ .Let $T$ denote the overall lifetime of the system

Find $P(T\ge t),\:t>0$

I am understanding that the total time will be $\min (A,B)+ \exp(c)$ but not getting how to write integral and also the limits of integration please help!!

Answer 1

Let $T_1$ be the time until the first component fails, and $T_2$ be the time between the first component failing and the second component failing. Since $A,B\stackrel{iid}{\sim}\text{Exp}(a)$, $T_1=\min(A,B)\sim \text{Exp}(2a)$, and $T_2\sim \text{Exp}(c)$. Since $T=T_1+T_2$, \begin{align*} \mathbb P(T\ge t)&= \mathbb P(T\ge t|T_1\ge t)\mathbb P(T_1\ge t)+\mathbb P(T\ge t|T_1<t)\mathbb P(T_1<t)\\ &= \mathbb P(T_1\ge t)+\mathbb P(T_1<t)\mathbb P(T_2\ge t-T_1|T_1<t)\\ \mathbb P(T_1\ge t)&= \exp(-2at)\\ \mathbb P(T_1<t)&= 1-\exp(-2at)\\ \mathbb P(T_2\ge t-T_1|T_1<t)&= \frac{\mathbb P(T_2\ge t-T_1\cap T_1<t)}{\mathbb P(T_1<t)}\\ &=\frac{\int _0^t\int _{t-t_1}^\infty (2a\exp(-2at_1)\cdot c\exp(-ct_2)\,dt_2\,dt_1}{1-\exp(-2at)}\\ &= \frac{2a\exp(-ct)(\exp(2at)-\exp(ct))}{(2a-c)(\exp(2at)-1)} \end{align*} Combining these results should give you the answer.

Answer 2

The time until the first failure is the minimum of two exponential random variables; one can show that this is itself an exponential random variable with mean $\frac{1}{2a}$.

The time between the first failure and the second failure is, by assumption, exponential with mean $1/c$.

Thus $T$ is the sum of two independent exponential random variables with different means. You can use the convolution formula to get the PDF of $T$, e.g. see here.

Answer 3

Let $T\sim X+Y$ with $X\sim \text{Exponential}(2a)$ and $Y\sim \text{Exponential}(c)$. Then the pdf is given by $$f(t)=\frac {2ac}{c-2a}\left(e^{-2at}-e^{-ct}\right), t>0$$

Then $$Pr(T>t)=\frac{2ac}{c-2a}\left(\frac{e^{-2at}}{2a}-\frac{e^{-ct}}{c}\right)$$