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Math Help - expected value of a normally distributed variable

  1. #1
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    expected value of a normally distributed variable

    Hey...I've been thinking about this problem all day, and can't seem to come up with an exact answer.

    consider a standard normally distributed random variable, say X. What is the expected value of X, given it is between two values (say 1 and -2)?
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  2. #2
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    Quote Originally Posted by minivan15 View Post
    Hey...I've been thinking about this problem all day, and can't seem to come up with an exact answer.

    consider a standard normally distributed random variable, say X. What is the expected value of X, given it is between two values (say 1 and -2)?
    Apply the ideas found in this link: Conditional Expectation
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  3. #3
    Super Member Random Variable's Avatar
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    In general, let X be any continuous random variable.

    Then  E[X|a<X<b] = \int^{b}_{a} \frac {x f(x)}{P(a<x<b)} \ dx
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    MHF Contributor matheagle's Avatar
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    The X is a rv, not a variable of integration...

    Quote Originally Posted by Random Variable View Post
    In general, let X be any continuous random variable.

    Then  E[X|a<X<b] = {\int^{b}_{a} x f_X(x)dx\over P(a<X<b)}
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  5. #5
    Super Member Random Variable's Avatar
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    I stated that X is a random variable. And of course I meant to capitalize the X in the denominator. P(a<x<b) has no meaning. And there is nothing wrong with the way I wrote it. The denominator is a number.

    I know you think I'm incredibly stupid, but give me some credit. Sheesh.
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  6. #6
    MHF Contributor matheagle's Avatar
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    The comment was not for you, but for the poster.
    The probability could be in or out of the integral, but with it inside and being lower case as with the x in the numerator it is misleading.
    It could easily be interpreted as part of the integrand.

    And I do not think you're stupid at all.
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    Super Member Random Variable's Avatar
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    Quote Originally Posted by matheagle View Post
    The comment was not for you, but for the poster.
    The probability could be in or out of the integral, but with it inside and being lower case as with the x in the numerator it is misleading.
    It could easily be interpreted as part of the integrand.

    And I do not think you're stupid at all.
    I apologize. I thought you were talking directly to me.

    But yes, it could cause confusion.
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    so, just to make sure I have this...

    the expected value of a std normal variable X, given X is between -2 and 1, is about -.229 (can someone verify this answer?)
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  9. #9
    Super Member Random Variable's Avatar
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    Quote Originally Posted by minivan15 View Post
    so, just to make sure I have this...

    the expected value of a std normal variable X, given X is between -2 and 1, is about -.229 (can someone verify this answer?)
     E[X|-2<X<1] = \frac {\frac {1}{\sqrt{2 \pi}} \int^{1}_{-2} x e^{- \frac {x^{2}}{2}} \ dx}{ \frac {1}{\sqrt{2 \pi}} \int^{1}_{-2}e^{- \frac {x^{2}}{2}}\ dx} \approx -0.2296

    yep
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