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Math Help - [SOLVED] Question about justification (pdf)

  1. #1
    Moo
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    [SOLVED] Question about justification (pdf)

    Hi

    I'd like to know something... Let's assume I'm given a function ( f(x_1,x_2)=\dots). I'm then asked to prove that this function is a probability density function.

    So I have to check that \iint_{A \times B} f(x_1,x_2) ~dx_1dx_2=1
    Is it enough ?

    And do you guys have to prove first that the function is integrable ? Or do you even prove that it's integrable ?


    Thanks in advance


    These stats courses are driving me nuts
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    Quote Originally Posted by Moo View Post
    Hi

    I'd like to know something... Let's assume I'm given a function ( f(x_1,x_2)=\dots). I'm then asked to prove that this function is a probability density function.

    So I have to check that \iint_{A \times B} f(x_1,x_2) ~dx_1dx_2=1
    Is it enough ?

    And do you guys have to prove first that the function is integrable ? Or do you even prove that it's integrable ?


    Thanks in advance


    These stats courses are driving me nuts
    Hi,

    first of all f has to be non-negative.

    Then, integrating f must make sense: i.e., depending on the level of study, f has to be, say, piecewise continuous (for simple Riemann integration) or Lebesgue measurable. This part is usually skipped because it is readily checked.

    And finally, the integral of f on its domain must equal 1. Note that it is not mandatory to first prove that the integral is finite. Indeed f is non-negative, so that the integral always has a value, either finite or infinite, and the computation will show that it is 1.

    Good luck with your probability/stats course, it's worth it!
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    Moo
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    Quote Originally Posted by Laurent View Post
    Hi,

    first of all f has to be non-negative.

    Then, integrating f must make sense: i.e., depending on the level of study, f has to be, say, piecewise continuous (for simple Riemann integration) or Lebesgue measurable. This part is usually skipped because it is readily checked.

    And finally, the integral of f on its domain must equal 1. Note that it is not mandatory to first prove that the integral is finite. Indeed f is non-negative, so that the integral always has a value, either finite or infinite, and the computation will show that it is 1.
    Arf this is indeed simple.
    Hmmm how would you translate in French "integration makes sense" ? Is it just proving that it's Lebesgue measurable ?

    erm... I'll continue on measures... How can one prove that a function is measurable (with respect to any measure - Lebesgue, or counting, mostly) ?
    And is there an equivalence between a function being continuous and the function being measurable ?

    Good luck with your probability/stats course, it's worth it!
    Haha I must trust you on it
    I've got 3 courses of probability & stats


    Thanks =)
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    Quote Originally Posted by Moo View Post
    Arf this is indeed simple.
    Hmmm how would you translate in French "integration makes sense" ? Is it just proving that it's Lebesgue measurable ?
    I wasn't sure whether you knew measure theory; then I indeed meant "Lebesgue measurable".
    Otherwise, Riemann integration suffices for piecewise continuous density functions (and a few others), like the usual ones.

    erm... I'll continue on measures... How can one prove that a function is measurable (with respect to any measure - Lebesgue, or counting, mostly) ?
    And is there an equivalence between a function being continuous and the function being measurable ?
    Measurability is actually not defined with respect to a measure but to a \sigma-field ("une tribu"). I shouldn't have said "Lebesgue measurable", it was a slight mistake... (Or I could pretend I was talking about the Lebesgue \sigma-field)

    The Lebesgue measure is defined on the Borel \sigma-field ("la tribu des boréliens"). Continuous functions are measurable with respect to this \sigma-field, but there are plenty of other measurable functions. So many that it is somewhat difficult to find a non-measurable one! In fact it is not possible to build non-measurable functions (with respect to the borelian \sigma-field) without using the axiom of choice. In practice, any function is measurable, except if it was meant to be a counter-example to this property

    The counting measure is defined on the discrete \sigma-field, for which any function is measurable.
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    Moo
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    Quote Originally Posted by Laurent View Post
    I wasn't sure whether you knew measure theory; then I indeed meant "Lebesgue measurable".
    Otherwise, Riemann integration suffices for piecewise continuous density functions (and a few others), like the usual ones.

    Measurability is actually not defined with respect to a measure but to a \sigma-field ("une tribu"). I shouldn't have said "Lebesgue measurable", it was a slight mistake... (Or I could pretend I was talking about the Lebesgue \sigma-field)

    The Lebesgue measure is defined on the Borel \sigma-field ("la tribu des boréliens"). Continuous functions are measurable with respect to this \sigma-field, but there are plenty of other measurable functions. So many that it is somewhat difficult to find a non-measurable one! In fact it is not possible to build non-measurable functions (with respect to the borelian \sigma-field) without using the axiom of choice. In practice, any function is measurable, except if it was meant to be a counter-example to this property

    The counting measure is defined on the discrete \sigma-field, for which any function is measurable.
    Okay, it all looks clear

    So basically, we just have to check that :
    - the function is measurable
    - the unfction is positive
    - the integral is 1, there is no need to prove that the integral is finite.

    One last question : in order to prove that a function is the pdf, is it necessary to check the other axioms of a pdf (continuity from the right, limits at + and - infinity, etc...) ?
    Maybe I'm just getting confused lol... the lessons are not acquired yet ><
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    Quote Originally Posted by Moo View Post
    One last question : in order to prove that a function is the pdf, is it necessary to check the other axioms of a pdf (continuity from the right, limits at + and - infinity, etc...) ?
    Maybe I'm just getting confused lol... the lessons are not acquired yet ><
    To prove that a function f is a p.d.f., depending on how it is defined, either you can simply write it as the pdf of some r.v. X: f(x)=P(X\leq x) for every x\in\mathbb{R}, otherwise you have to check that f is increasing, right-continuous, and with respective limits 0 and 1 at -\infty and +\infty, that's all.
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