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How do I show that a weibull is NOT memoryless? Q3A below

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asked Jun 8 in STA 3045F - Adv. Stochastic Processes by anonymous

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   Is showing that the exponential dbn is the only memoryless dbn equivalent to showing that the weibull is NOT memoryless?

This seemed like a lot of work for 2 marks (the proof we did on day 1 I mean).   

1 Answer

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answered Jun 9 by Njabulo.Dube (1,180 points)
selected Jun 28 by Rowan
Best answer

So there are two ways to show this for any continuous distribution.

  1. Showing that the exponential distribution is the only continuous distribution that has the memoryless property is equivalent to showing that any other continuous distribution does not have the memoryless property.
  2. Let \(X\) be any continuous random variable. In order to show that \(X\) does not have the memoryless property, you need to show that:
  3. $$ Pr(X>s+t|X>t) \neq Pr(X>s)$$
commented Jun 9 by ChanceTheRapper (650 points)


We were leaning towards 1 but seeing as the Q was only 2 marks - we thought this was too much work?

Also - 2 isn't really possible becuase we are only given the intensity funtion of the Weibull dbn and not its actual pdf or is there another way?

commented Jun 12 by Njabulo.Dube (1,180 points)
Firstly, from my understanding the intensity function given fully parameterises the Weibull distribution so you can obtain the pdf and use the second method to show that the Weibull is not memoryless.

That being said, both methods do take a bit of time so I suggest you use the method that is most comfortable for you.