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The life span of a particular mechanical part is a random variable described by the following PDF: enter image description here

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.

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What are you thoughts? – Henning Makholm Dec 16 '12 at 15:47
Your question asks for the first time when 2 out of 3 components fail. You accepted an answer which studies the first time when 2 out of 2 components fail. This is quite different. – Did Dec 20 '12 at 0:14
I agree with did. The accepted answer computes the first time when 2 out of 2 components fail. – Mike Spivey Dec 21 '12 at 4:49
CravingSpirit: Sorry but I am afraid that you lost me: you accepted an answer THEN offered a bounty, without unaccepting the answer nor commenting on the other answer? Could you keep me posted on the status of this question and the answers you received? – Did Dec 21 '12 at 21:38
@did, sorry, I've been busy with other things. I will check both answers and give the bounty to the preferred one. Thanks! – qed Dec 24 '12 at 17:39

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, $$ \mathbb E(T)=2-\tfrac{32}{25}+\tfrac{64}{125}+\tfrac3{125}-\tfrac2{125\cdot7}=\tfrac{1097}{875}=1.253\overline{714285}. $$

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+1, although I think you should have $\frac{16}{125}t^3$ rather than $\frac{16}{25}t^3$ in the first integral in the third-to-last displayed line, as well as $2$ instead of $1$ right after the = sign in the last line. That changes the final value of $E(T)$ as well. – Mike Spivey Dec 21 '12 at 4:48
@MikeSpivey Thanks a lot for the checking. I hope the numerics are correct now. – Did Dec 21 '12 at 6:22
Your new final number for $E(T)$ agrees with what I have. – Mike Spivey Dec 21 '12 at 16:10
@did, could you please explain this a bit more: $$ \mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\mathrm dt. $$ – qed Dec 24 '12 at 18:01
Yes: write $P(T\gt t)$ as an integral on $(t,+\infty)$ and use Tonelli to exchange the order of the two integration signs. – Did Dec 25 '12 at 10:49

The value for $E \left[ \max \left( L_1, L_2 \right) \right]$ is computed in the following way. First, the distribution of the maximum of two identically independently distribued random variable $L_1$ and $L_2$ is given by $2 f \left( \ell \right) F \left( \ell \right)$ where $f \left( \ell \right)$ is the density and $F \left( \ell \right)$ is the cumulative distribution function. This is well known, you could find the formula here. It is not difficult to derive: \begin{eqnarray*} \Pr \left[ \max \left( L_1, L_2 \right) \leqslant \ell \right] & = & \Pr \left[ \left\{ L_1 \leqslant \ell \right\} \cap \left\{ L_2 \leqslant \ell \right\} \right]\\ & = & \Pr \left[ L_1 \leqslant \ell \right] \Pr \left[ L_2 \leqslant \ell \right]\\ & = & F \left( \ell \right)^2 \end{eqnarray*} Taking derivative gives the density $2 f \left( \ell \right) F \left( \ell \right)$.

The probability density function $f(\ell)$ is given by (as you indicated) $$ f \left( \ell \right) = \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) $$ where the notation $1_{\ell}A$ with interval $A$ is that of an indicator variable. This means $$ 1_{\ell} \left( A \right) = \left\{ \begin{array}{lll} 1 & & \text{if } \ell \in A\\ 0 & & \text{otherwise} \end{array} \right. $$ Therefore the cumlative distribution function is given by $$ F( \ell)= \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) $$ Multiplying both we get the density \begin{eqnarray*} 2 f \left( \ell \right) F \left( \ell \right) & = & 2 \left\{ \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) \right\}\\ & \times & \left\{ \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) \right\}\\ & = & \frac{8 \ell}{25} 1_{\ell} \left[ 0, 2 \right) + \frac{4}{25} \left( - \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) \end{eqnarray*} and therefore \begin{eqnarray*} E \left[ \max \left( L_1, L_2 \right) \right] & = & \frac{8}{25} \int_0^2 \ell^2 \mathrm{d} \ell + \frac{4}{25} \int_2^3 \ell \left( - \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) \mathrm{d} \ell\\ & = & \frac{637}{375}\\ & \approx & 1.69867 \end{eqnarray*}

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Thanks! I am not sure though the CDF $F( \ell)= \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) + 1_{\ell} \left[ 3, \infty \right)$ is correct though, could you please check it? – qed Dec 19 '12 at 13:13
@CravingSpirit It looks fine to me because the CDF is increasing between 0 and 1 on the support and $2fF$ is a density since it integrates to one. Where do you see an error? – Learner Dec 19 '12 at 13:37
sorry, I made a mistake. Thanks for explaining! – qed Dec 19 '12 at 14:27
The question asks for the first time when 2 out of 3 components fail. Your answer studies the first time when 2 out of 2 components fail. This is quite different. – Did Dec 19 '12 at 23:13

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