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Standard Normal distribution probability

  1. Dec 1, 2011 #1
    Suppose X is the standard normal variable.

    is it possible to find P(X>x)?

    i.e does the function have a defined integral? How would you show that it is or is not defined?
    What information can we obtain about P(X>x).
     
  2. jcsd
  3. Dec 1, 2011 #2
    The standard normal distribution has a mean of x=0 and a standard deviation of 1, meaning that the value of x is in standard deviation (SD) units from the mean. So a value of x = -2 is exactly 2 SD from the mean. Unless I'm missing something in your question, [itex]P(X \geq x) [/itex] will be 1 minus the integral of the PDF from negative infinity to x. Remember that the integral of the PDF from negative to positive infinity must be 1. Although the mean has a value of x=0, the probability density at x=0 is 0.5.
     
    Last edited: Dec 1, 2011
  4. Dec 1, 2011 #3

    Simon Bridge

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    You mean that X~N(0,1) ... eg, X is Gaussian distributed about mean 0 with unit standard deviation.

    Thus x will only be on a std deviation if x is an integer.

    The probability is determined:
    [tex]p(X>x)= \frac{1}{\sqrt{2\pi}}\int_x^\infty \exp{(-\frac{x'^2}{2})}dx'[/tex]

    Tables normally give you p(0<X<x), so you want p(X>x) = 0.5 - p(0<X<x)

    For discussions on how to integrate this - look at error function:
    http://en.wikipedia.org/wiki/Error_function
    ... which I suspect you are asking about.
    There is no elementary indefinite integral for the Gaussian.

    Also see:
    http://mathworld.wolfram.com/GaussianIntegral.html
     
    Last edited: Dec 1, 2011
  5. Dec 1, 2011 #4
    Yes. So? Are you trying to say that SD measure can't be continuous? If not, it looks you're saying x will only be an integer if it's an integer. Also the Standard Normal Distribution isn't approximately normal. It's defined to be normal.
     
    Last edited: Dec 1, 2011
  6. Dec 1, 2011 #5

    Simon Bridge

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    Standard deviation cannot be continuous because it is a single value for a particular normal (or normalized gaussian) distribution. However, I would not like anyone to think that I was suggesting that the SD could only take integer values, in general.

    In this particular case, the distribution is standard normal ... so the SD tick marks are all on integer values because the standard deviation is 1. That's what I was trying to say - sorry if that was not clear.

    And that's what I said. Oh I think I see the confusion...

    Please note:
    "4. In statistics, the tilde is frequently used to mean "has the distribution (of)," for instance, X ~ N(0,1) means "the stochastic (random) variable X has the distribution N(0,1) (the standard normal distribution). If X and Y are stochastic variables then X~Y means "X has the same distribution as Y."
    http://mathworld.wolfram.com/Tilde.html
     
  7. Dec 1, 2011 #6
    You're not making any sense. I'm saying the SD is a continuous measure of the distance of x from the mean in the SND. Are you disputing that?
     
    Last edited: Dec 1, 2011
  8. Dec 1, 2011 #7

    Simon Bridge

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    What did I just say? Lets see:
    ... I have no idea why you think I'm disagreeing with you.

    I'll try another tack:

    In the std normal, x can be measured in units of the SD.
    (No that can't be the dispute - this is what "unit standard deviation" means - no, sorry, I have no idea what you are arguing about.)
     
  9. Dec 1, 2011 #8
    This:

    "Standard deviation cannot be continuous because it is a single value for a particular normal (or normalized gaussian) distribution."

    This does not make sense to me. Either it's a tautology or it's wrong. Of course a particular distribution is defined by the particular values of its parameters. However what you clearly say is that the SD cannot be continuous, This is very strange thing to say. Earlier you seem to indicate that the SD could only take integer values. Since you now seem to agree the SD measure is a continuous variable under the Normal Distribution, I can only understand that you meant the SD is an integer when it's an integer.
     
    Last edited: Dec 1, 2011
  10. Dec 2, 2011 #9

    So how would you go about proving for example that

    P[X>x] > f(x)

    if you're given some f(x)
     
  11. Dec 3, 2011 #10

    Simon Bridge

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    I would use a numerical method - fastest.
    The exact method used will be determined by the purpose of the exercise.

    eg.
    attachment.php?attachmentid=41466&stc=1&d=1322903809.png
    ... see, easy. Took about a minute.

    A student would be expected to be clever and use their knowledge of the properties of f(x) and the gaussian, their relationship etc. For example, if f(x)=-x2 then it will always be negative while p(X>x) is always positive. In the image, if f(x) is the unit normal, then the proposition is easily disproved.

    I am guessing that it is the absence of an indefinite integral that bothers you?

    As you'd expect, this is a very well studied class of functions. Approaches for dealing with them are very well documented - eg. in the links I gave you. In general one uses numerical methods or approximate-analytic solutions or combination - these can be made as accurate as needed.
     

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  12. Dec 11, 2011 #11

    Simon Bridge

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    From private communication: (sorry dude, but it benifits most people to do this in public)
    [tex]\frac{1}{\sqrt{2\pi}}\frac{1}{x}\exp{(-\frac{x^2}{2})} > p(X>x) > \frac{1}{\sqrt{2\pi}}\frac{x}{x^2 + 1}\exp{(-\frac{x^2}{2})}[/tex]
    ... realizing that x > 0.

    Observe that: [tex]p(X>x)=1-p(X<x)=1-\text{erf}(x)[/tex]

    I'd approach this by exploiting the properties of the curves.

    Notice the std unit normals in there?
    Denote the std normalized unit gaussian as n(x), and rewrite:

    [tex]f(x) > 1-\text{erf}(x) > g(x): \qquad f(x)=\frac{1}{x}n(x), \qquad g(x)=\frac{x}{x^2+1}n(x)[/tex]

    attachment.php?attachmentid=41849&stc=1&d=1323641744.png
    ... the curves are: blue: f(x), cyan: g(x), red: 1-erf(x), green: n(x)


    A strategy suggests itself:

    f(x) approaching x=0, is assymptotic, thus >> 1-erf(x)
    as x gets big, f(x) approaches 1-erf(x) from above.
    since they are both well behaved, f(x)>1-erf(x) in the whole domain.

    g(x) has a global maximum by x=0.3 which is less than 1-erf(x) there, and g(x) only gets smaller from there. As x gets big, g(x) approaches 1-erf(x) from below.

    So it looks like you want to know the shapes of the functions in the limit as x gets big.

    Probably someone else can do better.
     

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