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If I have a system which has an exponential failure rate distribution, i.e. \begin{align*} F(t) := P(T \leq t) = 1 - e^{-\lambda t} \end{align*}

where $T$ is the time of failure and $t$ is measured in seconds.

How can I compute the probability that the component fails between time $t$ and $t + \delta$ seconds?

My intuition is to use the derivative of $F(t)$, i.e. $f(t) = F'(t)$ and to integrate from $t$ to $t + \delta$, giving the probability of failure in this time period as ($T$ being the time of failure): \begin{align*} P(t \leq T \leq t + \delta) = \int_t^{t+\delta}{\lambda e^{-\lambda x}}\mathrm{d}x \end{align*}

Given that $F(t) = 1 - e^{-\lambda t}$, wouldn't this just be $F(t + \delta) - F(t)$?

Is this the correct way to approach the problem?

Átila Correia
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2 Answers2

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Yes, $F(u)$ is cumulative distribution function, and it is related to the probability density function by $F'(u)=f(u)$, so the probability between $a$ and $b$ is:

$$P(a\le X\le b)=F(b)-F(a)=\int_a^b f(u) du$$

MathFail
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Just nitpicking: if $F$ is not continuous, then you are interested in the expression: \begin{align*} \mathbb{P}(T\leq t + \delta) - \mathbb{P}(T < t) = \mathbb{P}(t \leq T \leq t + \delta) = F(t + \delta) - F(t^{-}) \end{align*}

If $F$ is continuous, then the expression proposed by @MathFail does the job.

Since $T\sim\text{Exp}(\lambda)$, the latter observation holds.

Átila Correia
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