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For an exponential distribution, the hazard rate has a clear interpretation as the inverse of the expected time until the next failure (let's say we are modelling machine failures here). For other distributions however, this does not seem to hold. Take for example, the Lomax distribution with probability density function -

$$f(x) = \frac{\lambda \kappa}{(1+\lambda x)^{\kappa+1}}$$

The Hazard rate becomes -

$$H(x) = \frac{\lambda \kappa}{1+\lambda x}$$

And the expected value of the random variable being modelled given it is greater than $x$ is -

$$E[X|X>x] = \frac{1}{\lambda (\kappa-1)}+ \frac{\kappa x}{\kappa-1}$$

Unlike the exponential distribution,

$$H(x) \neq \frac{1}{E[X|X>x] - x}$$ (see comment from @Math1000 below)

Although they are both linear.

So, what is the difference between these two quantities and why are they the same only for the exponential distribution?

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    Recall that the exponential distribution is the unique continuous probability distribution with the memoryless property. So if $T\sim\mathrm{Exp}(\mu)$ then for $t>0$ ,$$\mathbb E[T\mid T>t] = \mathbb E[T] + t = \frac1\mu + t.$$ – 2017-02-20
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    What I'm wondering is, what do I make of the hazard rate (in general) if not the inverse of the expected time until the next failure? – 2017-02-20
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    Updated my post with the correction you pointed out. – 2017-02-20
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    The hazard function $h(t)$ is the instantaneous rate of occurrence of failure at time $t$. The exponential distribution is has a constant hazard rate (which is the reciprocal of its mean) *precisely* because of its unique memoryless property. – 2017-02-20
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    What I'm struggling with is - if h(t) is the instantaneous rate of occurrence of failure, then shouldn't $1/h(t)$ be instantaneous expected time till the next failure? Or is that completely off? – 2017-02-20
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    @RohitPandey Expected time depends on future behavior, and not just what is going on right now. So "instantaneous expected time" is meaningless. – 2017-03-02
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    @btilly - we can still talk of - at this point in time, given all the information I have, the future is modelled by this distribution. And as soon as we do that, we can talk about the expected value of that distribution. No? – 2017-03-02
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    @RohitPandey I can talk about the time that I expect to take in the future at this moment. But that time depends on not just the current hazard rate, but the future hazard rates that I haven't encountered yet. So it is not uniquely determined by the current hazard rate. – 2017-03-02
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    But you can estimate the future hazard rates based on your distribution. So what's stopping you from taking those estimates into account for the expected value? – 2017-03-03
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    Remember, we are talking not about real world messy data but a very well defined distribution here. – 2017-03-03

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The instantaneous hazard rate and additional expected time are the same only for cases of deterministic processes or the exponential distribution.

Otherwise, the hazard rate is like instantaneous velocity and the reciprocal of additional expected time is like average velocity.