Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

2 Answers

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

share|improve this answer
+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
show 3 more comments

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*}

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.