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Properties of density function question

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  • Properties of density function question

    One of the properties is f(x) is greater than or equal to 0.
    f(x) being the probability density function.

    How come it's not emphasized that f(x) has to be equal to or be less than 1?

  • #2
    One of the conditions of a pdf is that it's integral from -inf to inf is equal to 1.

    Which is pretty much what you're saying I think.

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    • #3
      You can think of f(x) as the likelihood function if that helps.

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      • #4
        Oh ok I have another question. I know that the integral of e^-x is -e^-x.
        How come in probability we assume that the integral of e^-1 is (1-e^-1)?
        I know adding a constant doesn't matter because the derivative of a constant is zero.

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        • #5
          Originally posted by ronaldy27 View Post
          Oh ok I have another question. I know that the integral of e^-x is -e^-x.
          How come in probability we assume that the integral of e^-1 is (1-e^-1)?
          I know adding a constant doesn't matter because the derivative of a constant is zero.
          Huh? The integral of e^{-1} is not 1 - e^{-1} it's x*e^{-1} assuming you are integrating with respect to x. This has nothing to do with probability, it's Calculus rules.

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          • #6
            No! haha I meant the integral of e^-x is (1-e^-x).
            I got it now. This has to do with exponential distribution.
            I am kind of new to P and I didn't get to the section where the manual talks about all the different continuous distributions
            so I was confused but I figured it out.

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            • #7
              Originally posted by ronaldy27 View Post
              No! haha I meant the integral of e^-x is (1-e^-x).
              I got it now. This has to do with exponential distribution.
              I am kind of new to P and I didn't get to the section where the manual talks about all the different continuous distributions
              so I was confused but I figured it out.
              You figure out the constant due to the fact that the integral of the pdf over its support must be 1. The indefinite integral of e^{-x} dx is def. not 1 - e^{-x}

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