Reliability of a system with non-independent parallel components

Question

Consider a system $Z$ that consists of two components $X$ and $Y$, whose failure rates per time unit are $\lambda_x$ and $\lambda_y$ respectively.

Assume that if $X$ fails, then the failure rate of $Y$ doubles. What is the probability density function of the time to failure of the overall system $Z$?

What I could gather so far is the following:

The failure probability of the single components can be modelled as $$ F_X(t) = 1-e^{-\lambda_X t} \text{ and } F_Y(t) = 1-e^{-\lambda_Y t},$$ where $$ F_X(t) = \int_0^t f_X(t) dt, f_X(t) = \lambda_X e^{-\lambda_X t} $$ (and the same for Y) is the cumulative density function of an exponential distribution.

Since the two components that make up the system are not independent, the reliability of the system, defined as $R(t) = 1-F(t)$, should begiven by $$ R(t) = 1-F(t) = 1 - F_X(t) \cdot F_{Y|X}(t).$$

How can one determined the conditioned cdf $F_{Y|X}(t)$?

Answer 1

Comment:

Clues from simulation of a million such systems: Let $\lambda_x = 2; \lambda_y = 3.$

Then $P(X < Y) = 2/5 = 0.4.$

set.seed(2021)
x = rexp(m,2); y = rexp(m,3);  yy = rexp(m,6)
v = pmin(x,y); w = pmax(x,y)
mean(x < y);  mean(x == v)
[1] 0.400608
[1] 0.400608

Then, by the no-memory property, the total lifetime of the system is $T$ as shown below.

t = 0.4*(v + yy) + 0.6*w
summary(t)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.003752 0.280261 0.450555 0.526562 0.689709 4.661313 

hist(t, prob=T, col="skyblue2", main="System Lifetime")

Notes: The CDFs of the minimum (exponential) and maximum (not exponential) are easily obtained, and their PDFs from them. The density of the sum of two exponentials with different rates is nontrivial. But then the PDF of the mixture distribution is easy.