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Let Tbe the random variable of the lifetime(in years) of an LED lamp.The lifetime $Pr[T>t]=100/t^2$ for $t>10$

a. please find $Pr[T\le3]$ ,and the expected value of T

b. Find the conditional probability of failure in the next year for a working LED lamp that is now t years old for each $t>10$

for a.question,well, i cant solve it,i have been thinked that $Pr[T>t]=1-Pt[T\le t]$,then differential it to let it become pdf,and i can know the expected value. but the t still have to bigger than 10,$t>10$,so is question wrong?or does there someway to solve it?

for b.question,i think Pr[broken next year|have been worked for more than 10 yeasrs],that is,$[(100/t^2)\times[1-(100/(t+1)^2)]]/(100/t^2) $ for$t>10$, is my ideal wrong?

2 Answers 2

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for a. question,well, i cant solve it,i have been thinked that $\Pr[T>t]=1−\Pr[T≤t]$,then differential it to let it become pdf, and i can know the expected value. but the t still have to bigger than 10,$t>10$,so is question wrong? or does there someway to solve it?

The complete tail distribution function should be $\Pr(T>t) =\begin{cases} 100/t^2 &:& t>10 \\ 1 &:& t\leq 10\end{cases}$

Then $\Pr(T\leq 3) = 0$.

For the expectation, use that for strictly non-negative integer random variables, $$\begin{align} \mathsf E(X) &= \sum_{k=1}^\infty k\,\Pr(X=k) \\ & = \sum_{k=1}^\infty\sum_{x=0}^{k-1}\Pr(X=k) \\ & = \sum_{x=0}^\infty\sum_{k=x+1}^\infty \Pr(X=k) \\ & = \sum_{x=0}^\infty \Pr(X> x) \end{align}$$

And in particular $\mathsf E(T) = 11 +\sum\limits_{t=11}^\infty \Pr(T>t)$

for b.question,i think Pr[broken next year|have been worked for more than 10 yeasrs],that is,$[(100/t^2)×[1−(100/(t+1)^2)]]/(100/t^2)$ for $t>10$, is my ideal wrong?

Yes, it is wrong; but almost okay.

That is $\frac{\Pr(T>t)\times\big(1-\Pr(T>t+1)\big)}{\Pr(T>t)}$, which is simply $\Pr(T\leq t+1)$ and not at all for what you were asked.

You want to find the conditional probability that the device fails on or before $t+1$ given that it hasn't failed by $t$.

$\Pr(T\leq t+1\mid T>t) = \dfrac{\Pr(tt)} = \dfrac{\bbox[white]{\color{white}{\Pr(T>t)-\Pr(T> t+1)}}}{\Pr (T>t)}$

Hint: The events $\{T>t\}$ and $\{T>t+1\}$ are not independent, so the product rule is inapplicable, but $\{T>t+1\}\subseteq \{T>t\}$, so...

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For a, as you have $Pr(T \gt 10)=1$ you also have it for any lower $T$ because it has to be nonincreasing with $t$. The question could have been more clear, but maybe this is what you were supposed to realize. For b you are supposed to comput the chance the lamp fails in the next year, given that it has survived so far. So for $t=15$ you are given that it has not failed by year $15$. Without the condition, the chance that a lamp fails in year $15$ is $\frac {100}{15^2}-\frac {100}{16^2}$ where the first term is the chance that it survives the first $15$ years and the second is the chance it does not survive $16$

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    $Pr(T>t)=1−Pr(T\le t)$,so $P[T\le 3]=1−Pr[T>3]=1−100/9$ it is obviously wrong – 2017-02-01
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    @蕭仰恩 Indeed $\Pr(T\leq 3)=1-100/9$ is wrong because, in fact, $\Pr(T>3)=1$ . The complete expression for this tail distribution is $$\Pr(T>t) =\begin{cases} 100/t^2 &:& t>10 \\ 1 &:& t\leq 10\end{cases}$$ Thus... – 2017-02-01
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    thx for answering a.,however,is my ideal for b.wrong?i understand the way u think for b,but,i still dont know if my ideal for b were wrong,where did i think wrong?$[(100/t^2)×[1−(100/(t+1)^2)]]$ means he worked for more than 10 years and die,$1−(100/(t+1)^2)$, next year ;$(100/t^2)$ means worked for more than 10 years – 2017-02-01
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    @蕭仰恩 That is $\tfrac{\Pr(T>t)\times\big(1-\Pr(T>t+1)\big)}{\Pr(T>t)}$, which is simply $\Pr(T\leq t+1)$ and not at all for what you were asked. Hint: the events $\{T\leq t+1\}$ and $\{T>t\}$ are not independent, so multiplication is inapplicable. However, $\{T>t+1\}\subseteq \{T>t\}$, so.... – 2017-02-01