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I googled and it seems not. Only exponential distribution is memoryless. Does anyone have an intuitive explanation why it is not? Thanks

hehe
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    Take a look at : https://en.wikipedia.org/wiki/Weibull_distribution The section on "Standard Parameterization" answers the question almost directly. It provides a nice interpretation of what a Weibull random variable represents, in terms of physical processes which makes it almost obvious why the Weibull distribution is not memoryless. – user303375 Dec 19 '20 at 00:59
  • In general, Weibull family is not memoryless. But, of course, its exponential subfamily (distinguished by using appropriate parameters) is memoryless; see @user303375's link. – BruceET Dec 19 '20 at 02:51
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    In discrete case: Geometric Distribution and its continuous cousin: Exponential Distribution are the ONLY distributions that are memory less.

    Continuous Weibull is Exponential when shape parameter becomes 1 and in Discrete Weibull is Geometric if shape parameter becomes 1, then it will be memory less otherwise Weibull is not memory less.

     – forecaster Dec 19 '20 at 03:24
  • I see. Thanks. Does it mean if the failure rate is not constant over time, then the distribution is not memoryless? Is failure rate mentioned in wikipedia the same as hazard rate? – hehe Dec 21 '20 at 19:02

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Does it mean if the failure rate is not constant over time, then the distribution is not memoryless?

From Wikipedia on Memorylessness:

[Memorylessness] usually refers to the cases when the distribution of a "waiting time" until a certain event does not depend on how much time has elapsed already. To model memoryless situations accurately, we must constantly 'forget' which state the system is in: the probabilities would not be influenced by the history of the process.

If the failure rate changes over time, then the system necessarily has a memory of how much time has already elapsed. From Wikipedia on the Weibull shape parameter $k$, as noted in a comment from @user303375:

A value of $k < 1$ indicates that the failure rate decreases over time... A value of $k = 1$ indicates that the failure rate is constant over time... A value of $k > 1$ indicates that the failure rate increases with time.

So the only Weibull distribution without memory of elapsed time is for $k=1$, the specific case of the exponential distribution.

Is failure rate mentioned in wikipedia the same as hazard rate?

The portion of the Wikipedia Weibull entry on its cumulative distribution function says specifically that the failure rate is equivalent to the hazard function (each as a function of time and the Weibull parameter values). Wikipedia notes "hazard rate" as a synonym for "hazard function", although some find the terminology "hazard rate" to be confusing.

EdM
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  • I thought that Markov process is considered memoryless if the transition probability is independent of the previous states, not independent of time. So, $P(X_t = x_i | X_k : 0 \le k \le t-1) = P(X_t = x_i | X_{t-1} = x_j) = P_{i,j}(t) $. – Confounded Jun 24 '21 at 12:35
  • @Confounded that might be the definition for a Markov process, but this question was about the memorylessness of a distribution of waiting times. In that sense, the Weibull is not "memoryless" except for the specific case of the exponential. – EdM Jun 24 '21 at 12:40
  • When a discrete-time Markov process is converted into a continuous-time one, I believe exponential distribution is used to represent time-homogeneous process. Wouldn't then a time-inhomogeneous process be represented by, say, Weibull distribution? – Confounded Jun 24 '21 at 12:56
  • @Confounded that depends on the type of inhomogeneity. Weibull has a monotone association of hazard with time, with hazard continually decreasing, constant, or increasing with time. It wouldn't fit the type of "bathtub"-shaped hazards that are seen, say, in human mortality over a full range of ages. – EdM Jun 24 '21 at 13:19
  • Right, I had a monotonically decreasing hazard rate in mind. What is not clear to me is how to get the rate matrix $Q$ for the continues-time Markov process with one of the transitions being governed be Weibull distribution. – Confounded Jun 24 '21 at 13:23