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    When using the definite integration method to convert a pdf to cdf and you have 3 functions in the pdf how do you work out the constant for the middle one?

    I know you put F(lower value of range)=0 and F(upper value of range)=1 but how would you know what to put for the middle function?

    Thanks
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    Zacken
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    (Original post by mp_x)
    When using the definite integration method to convert a pdf to cdf and you have 3 functions in the pdf how do you work out the constant for the middle one?

    I know you put F(lower value of range)=0 and F(upper value of range)=1 but how would you know what to put for the middle function?

    Thanks
    If you have the pdf just find integral of pdf from (-infinity) to upper value of the middle part then that gives you a value for F(upper value of middle part) which you can then use to find the constant.
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    (Original post by Zacken)
    If you have the pdf just find integral of pdf from (-infinity) to upper value of the middle part then that gives you a value for F(upper value of middle part) which you can then use to find the constant.
    I don't get what you mean
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    (Original post by mp_x)
    I don't get what you mean
    Say you have a PDF

    f(x) = \begin{cases}0.5x \quad 0< x < 0.5 \\ 0.6x \quad 0.5 < x < 1 \\ 0.7x \quad 1 < x < 1.5\end{cases}}

    Then to find F(1) = \int_{0}^{1} f(x) \, \mathrm{d}x and you can then use F(1) to find your constant of integration in your CDF.
 
 
 
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