TITLE: Expectation of joint life span
QUESTION [7 upvotes]: The life span of a particular mechanical part is a random variable described by the following PDF:

If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.
I can get the PDF:
$$
f_L(l) = 0.4 (0 \leq l \leq 2) \\
f_L(l) = -0.4l + 1.2 (2 < l \leq 3)
$$
and the expectation:
$$
E(l) = \int_0^3 l f_L(l) dl \approx 1.27
$$
I think 'majority' means 2 or more, so we can focus on two parts of the three, and pay no attention to the third. The translation is $E(max(l1, l2))$, how will this be derived I currently have no idea.

Sorry about the misleading remark "$E(max(l_1, l_2))$", it's wrong to neglect the third part, because if that one fails early, then we only need one of the rest to fail.

REPLY [3 votes]: Let $X$ denote the life span of any given component and $T$ the first time when at least 2 out of 3 components fail. The event $[T\gt t]$ means either that none of the 3 components fails before time $t$ or that exactly 1 component out of 3 fails before that time, hence, for every $t\gt0$,
$$
\mathbb P(T\gt t)=\mathbb P(X\gt t)^3+3\mathbb P(X\gt t)^2\mathbb P(X\lt t),
$$
that is,
$$
\mathbb P(T\gt t)=1-3u(t)^2+2u(t)^3=v(t)^2(3-2v(t)),
$$
with
$$
u(t)=\mathbb P(X\lt t),\qquad v(t)=1-u(t)=\mathbb P(X\gt t).
$$
Furthermore,
$$
\mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\mathrm dt.
$$
In the present case, the density of $X$ is $f_X(t)=\frac25$ if $0\lt t\lt 2$ and $f_X(3-t)=\frac25t$ if $0\lt t\lt 1$. Hence
$u(t)=\frac25t$ if $0\lt t\lt 2$ and $v(3-t)=\frac15t^2$ if $0\lt t\lt 1$. This yields
$$
\mathbb E(T)=\int_0^2(1-3u(t)^2+2u(t)^3)\mathrm dt+\int_0^1v(3-t)^2(3-2v(3-t))\mathrm dt,
$$
that is,
$$
\mathbb E(T)=\int_0^2(1-\tfrac{12}{25}t^2+\tfrac{16}{125}t^3)\mathrm dt+\int_0^1\tfrac1{25}t^4(3-\tfrac25t^2)\mathrm dt,
$$
or,
$$
\mathbb E(T)=\left[t-\tfrac{4}{25}t^3+\tfrac{4}{125}t^4\right]_{t=0}^{t=2}+\left[\tfrac3{125}t^5-\tfrac2{125}\tfrac17t^7\right]_{t=0}^{t=1},
$$
that is,
$$
\mathbb E(T)=2-\tfrac{32}{25}+\tfrac{64}{125}+\tfrac3{125}-\tfrac2{125\cdot7}=\tfrac{1097}{875}=1.253\overline{714285}.
$$