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At any time, a dog has the probability of p to bark. What's the probability that this dog did not bark in the past T seconds?

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    Sounds like HW...2011-06-03
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    Does "at any time" mean "in any second"? (If it means "in any millisecond", for instance, the answer will be rather different.)2011-06-03
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    The answer is $1$ if $p = 0$, else it is $0$.2011-06-03
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    Dan, I guess you are right. ShreevatsaR's comment explained it.2011-06-03
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    Dan, what would be the "right" hypothesis to consider for this problem (instead of "at any time, a dog has the probability of p to bark")?2011-06-03
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    Elliott, this isn't physics.SE! :)2011-06-03
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    Fine, I'll drop the word "hypothesis" from my vocabulary.2011-06-03
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    I was just suggesting that in mathematics we shouldn't assume time is quantized in any particular way. But actually I was wrong for the same reason! :)2011-06-03
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    Dan, that's not how stochastic processes in continuous time work.2011-06-03
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    Is my humor misunderstood. or is it not funny? After I rigorously inspect it I'll know for sure.2011-06-03
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    I was referring to your first comment. The humor in the second one is appreciated.2011-06-03
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    Related: http://math.stackexchange.com/questions/28088/salt-concentration-as-a-function-of-time/28090#28090 :)2011-06-03

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This is phrased as a continuous time question: the probability $p$ isn't referring to barking within a minute, or second, or microsecond, but at any time. That indicates continuous time.

The distribution that describes the probability of an event that occurs at a constant rate is the exponential distribution.

Having a probability $p$ of barking at a "moment" -- an infinitesimal unit of time, means that:

$p = \lim_{t \rightarrow 0} P($Bark at time $

For the exponential. So the rate parameter is $p$.

The question asks for the probability of not barking in an interval of time $T$.

That is given by $1 - F(T) = 1-(1-e^{-pT}) = e^{-pT}$.

That makes the answer $e^{-pT}$.

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    How did you decide the parameter? You missed a p if it is exponential distribution.2011-06-03
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    Nope, this is 1-cdf so I didn't miss a p (you must be thinking of the pdf). The rate is p because that's the probability of barking at any given moment. The time is T because... well that's given.2011-06-03
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    Here: answer explained better.2011-06-03
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    If an event happens with nonzero probability in infintessimal time, then it is always happening.2011-06-03
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    I guess you're right. I would just write it off as a badly worded problem then. But I bet that what was intended is that $p$ is a barking rate, which I think is a safer assumption than a discrete time assumption, without any further context.2011-06-03
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    The question says "at any time, a dog has the probability of p to bark" so the unit of p is not a unit of time.2011-06-03
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    Neither is $\lambda$. It is a _probability rate_ measured in units of _probability per unit time_.2011-06-03
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It should be (1-P)^T.Because 1-P is the probability that the dog has not barked in the last 1 sec (assuming that p is the probability that it does not bark in a given second).

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Hint: What are the odds that an event happens twice in a row? three times? in relation to it happening once?

Also, what are the odds of an event not happening, as opposed to it happening?

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    I guess a continuity makes it different.2011-06-03
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    What do you mean?2011-06-03
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This should have Poisson distribution. http://en.wikipedia.org/wiki/Poisson_distribution