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exkor5000
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Joined: 29 Sep 2006
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 Posted: Sun, 17 Dec 2006 13:21:32 UTC Post subject: A quick question about continuous uniform random variable |
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If  then is it safe to say that for this specific variable
due to the fact that 
I have this confusion for some reason, because the probability of any given value is 0. can anybody give me some intuition please.
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mathisfun
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| Posted: Mon, 18 Dec 2006 02:43:27 UTC Post subject: Re: A quick question about continuous unfirom random varialb |
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| exkor5000 wrote: | If then is it safe to say that for this specific variable
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Not quite. Without knowing the distribution of  (i.e.  ) you don't know  . |
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Zone Ranger
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| Posted: Mon, 18 Dec 2006 03:12:43 UTC Post subject: |
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I believe that exkor5000 is referring to a uniform(0,1) random variable.
"Post subject: A quick question about continuous uniform random variable"
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royhaas
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| Posted: Mon, 18 Dec 2006 17:31:48 UTC Post subject: |
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It is true that for any continuous random variable that the probability of assuming a specific value is zero, so that it doesn't matter whether you include it as part of an inequality or not. On the other hand, any continuous and symmetric distribution on the unit interval will have a median of 1/2.
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exkor5000
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Joined: 29 Sep 2006
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| Posted: Tue, 19 Dec 2006 01:56:40 UTC Post subject: |
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| Quote: |
It is true that for any continuous random variable that the probability of assuming a specific value is zero, so that it doesn't matter whether you include it as part of an inequality or not
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ok, so how would we go about say calculating the value for
 ?
Where  is some constant, representing a probability of a specific point.
When talking about the [0,1] I frequently heard the term "Lebesgue Integral". However I don't know anything about that term or how to apply it in any situation.
Any suggestions?
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Bilbo
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Posts: 690
Location: Yountville, California
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| Posted: Tue, 19 Dec 2006 03:33:56 UTC Post subject: |
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| exkor5000 wrote: | | Quote: |
It is true that for any continuous random variable that the probability of assuming a specific value is zero, so that it doesn't matter whether you include it as part of an inequality or not
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ok, so how would we go about say calculating the value for
?
Where is some constant, representing a probability of a specific point.
When talking about the [0,1] I frequently heard the term "Lebesgue Integral". However I don't know anything about that term or how to apply it in any situation.
Any suggestions?
Thanks |
I am puzzled by your interpretation of  We know for a continuous random variable the "probability of a specific point" is zero. Usually would be chosen so the total probability is one. For a random variable that is uniform on   and
Lebesgue integration is a different way of defining integration that replaces Riemann integration. It has better theoretical properties, e.g., it is easier to prove limit theorems using Lebesgue integration. But IIRC the Riemann and Lebesgue integrals for a function will have the same value when the Riemann integral is properly defined, say for continuous functions. And in practice when an integral has to be evaluated, the techniques learned from Riemann integration will be used. |
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exkor5000
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Joined: 29 Sep 2006
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| Posted: Thu, 21 Dec 2006 21:42:00 UTC Post subject: |
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Thanks for the intuition guys!
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ehremo
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Joined: 24 Dec 2006
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| Posted: Sun, 24 Dec 2006 23:34:21 UTC Post subject: |
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No, not at all -- an analogous argument would be that because the area of the line  is zero means that  .
Reasoning about the probability of a continuous random variable being equal to a specific value  doesn't work, since as you point out it's zero. However, the density  , since in some intuitive sense it's the probability of the set  (the mass of the set) divided by the measure of  (the area/volume of the set - hence the term probability density), which in our case, is the indeterminate form  which of course doesn't have to be zero. |
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