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October 30th, 2009, 03:17 PM
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| |  Squareing a Normal[0,1] random variable
A certain random variable X has a a normal distribution with \[Mu] = 0 and \[Sigma] = 1. Come up with a formulas for the cumulative distribution function and probability density function of X^2 and plot each separately.
Would this be right ? Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]]. |
October 30th, 2009, 03:25 PM
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Quote:
Originally Posted by aeubz  A certain random variable X has a a normal distribution with \[Mu] = 0 and \[Sigma] = 1. Come up with a formulas for the cumulative distribution function and probability density function of X^2 and plot each separately.
Would this be right ? Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]]. | Read this thread: Distribution Theory
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October 30th, 2009, 04:01 PM
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I'm not understanding it very much.. I need to know if this is the probability: Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]]. |
October 30th, 2009, 04:11 PM
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That's correct, but there is no closed form for the cdf of N(0,1)
You have to use the transformation of the pdf.
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October 30th, 2009, 05:05 PM
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Quote:
Originally Posted by Moo That's correct, but there is no closed form for the cdf of N(0,1)
You have to use the transformation of the pdf. | He can do what he is doing and just take the derivative of the expression he already has to get the pdf. Transformation method is fine too, but some intro texts don't cover transformations that are only monotone on intervals (mine didn't, at least).
If  , you've already derived  . Just take the derivative with respect to y to get the pdf of Y in terms of the pdf of X, which you have a closed form expression for. You should recognize the form of the pdf straight away and know that there isn't a closed form formula for the cdf.
Last edited by theodds; October 30th, 2009 at 05:17 PM.
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October 30th, 2009, 09:19 PM
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Quote:
Originally Posted by aeubz I'm not understanding it very much.. I need to know if this is the probability: Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]]. | What part of the thread whose link I gave you don't you understand (answer in that thread, not here).
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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