Three system components are connected in series (so the system fails if any one of them fails). The lifetimes of the components are independent random variables exponentially distributed with parameter λ. The system fails after 10 days

I'm trying to find the method of moments estimate of $\lambda$ and I need to find the PDF first. It seems to be similar to a binomial but I can't figure it out. I was given the PDF but I need an explanation.

$$f(t) = \binom{3}{1} \lambda e^{-\lambda t}(e^{-\lambda t})^2 = 3\lambda e^{-3\lambda t}$$

  • If the rate for one to fail on its own is $\lambda$ and you have three of them then the rate for any one of the three to fail is $ 3 \lambda$, which is what you have found – Henry Oct 12 '16 at 8:26

You are looking at the first order statistic for a sequence of three iid exponential values. That is the minimum of the three time-until-failures.

Any one of the three can be first, the density of that at time $t$ is $\lambda e^{-\lambda t}$ and the probability the other two both exceed that time is $e^{-\lambda t}$

$$\begin{align} f_{\min\{X_i\}}(t) ~=~& \binom 3 1~f_X(t)~(1-F_X(t))^2 \\[1ex] =~& \binom 3 1 \lambda~e^{-\lambda t}(e^{-\lambda t})^2 \\[1ex]=~& 3~\lambda~e^{-3\lambda t} \end{align}$$


This can also be obtained by considering $$\begin{align}f_{\min\{X_i\}}(t) ~ = ~ & \tfrac{\mathrm d~~}{\mathrm d ~t}\mathsf P(\min\{X_i\}\leq t) \\[1ex] =~& \tfrac{\mathrm d~~}{\mathrm d ~t}(1-\mathsf P(X_1> t, X_2> t, X_3> t)) \\[1ex] =~& -\tfrac{\mathrm d~~}{\mathrm d ~t}(1-F_X(t))^3 \\[1ex] =~& 3 \dot F_X(t)(1-F_X(t))^2 \\[1ex] =~& 3~\lambda~e^{-\lambda t}(e^{-\lambda t})^2 \\[1ex] =~& 3~\lambda~e^{-3\lambda t} \end{align}$$

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