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In this question I sought to find the probability of a component failing in a given interval $[t, t + \delta]$ assuming an exponential distribution ($T$ being the time of failure).

\begin{align*} F(t)= P(T \leq t) = 1 - e^{-\lambda t} \end{align*}

So, initially I sought $P(t \leq T \leq t + \delta)$. However, after computing some values empirically, I found that the probability of the component failing was decreasing over time which seemed counter-intuitive to me, as with an exponential distribution one would expect that the longer the component lives, the higher probability that it will experience a failure.

My intuition is that the probability I was initially computing was actually the probability that the component would fail in that interval as computed from $t=0$, and the fact that the component has survived until some time $t>0$ must be taken into account using a conditional probability along the lines of: $$P(t \leq T \leq t + \delta ~|~ t \leq T)$$

How can I compute this conditional probability?

MathFail
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ose
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2 Answers2

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I found that the probability of the component failing was decreasing over time ...

My intuition is that the probability I was initially computing was actually the probability that the component would fail in that interval as computed from $t=0$

Yes, exactly. The computation of $\int_t^{t+\delta} p(t)\, dt = F(t+\delta)-F(t)$ is the probability that a brand new component is going to fail during the interval $[t,t+\delta]$. And this decreases with larger $t$ precisely because the component is less likely to survive up to time $t$ in the first place.

Bayes' Theorem says $$P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$$

If $A$ is the event $t \leq T \leq t+\delta$ and $B$ is the event $t \leq T$, then $P(B\mid A) = 1$: if $A$ is true then $B$ is necessarily true. We know $P(A) = F(t+\delta)-F(t)$ as explained in your previous question. And $P(B) = 1 - F(t)$ since $F(t)$ is the probability $T \leq t$, $P(B)$ is the probability $T \geq t$, and the probability the real number $T$ is exactly any particular $t$ is zero. So

$$ P(t \leq T \leq t + \delta \mid t \leq T) = \frac{F(t+\delta)-F(t)}{1-F(t)} = \frac{e^{-\lambda t} - e^{-\lambda (t+\delta)}}{e^{-\lambda t}} = 1 - e^{-\lambda \delta} $$

That's right, this probability doesn't depend on $t$ at all. The exponential distribution models a case where an old component is just as likely to fail in any given time interval as a brand new component. We see this a lot in particle physics: particles of the same type act just the same, and how long that particle has existed isn't relevant. A carbon-14 nucleus in a dinosaur fossil and a carbon-14 nucleus formed today in the upper atmosphere have the same chance of decaying in the next year.

aschepler
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    Thanks for the detailed explanation. Would it be correct to say that the numerator is essentially an expression of the derivative of the CDF (as delta approaches zero), i.e. the PDF, and the denominator is the survival function, therefore the whole fraction would express the hazard rate at the limit of delta? So your answer also proves why the hazard rate of an exponential distribution is a constant? – ose Jul 28 '22 at 22:21
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$$\begin{align} P(t \leq T \leq t + \delta | t \leq T)&=\frac{P(t \leq T \leq t + \delta \cap t \leq T)}{P(t\le T)}\\ \\ &=\frac{P(t \leq T \leq t + \delta )}{P(t\le T)}\\ \\ &=\frac{F(t+\delta)-F(t)}{1-F(t)} \end{align}$$

MathFail
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