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Math Help - Notation problem inside integral

  1. #1
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    Notation problem inside integral

    Hi all

    I have a problem with formal mathematics notation. I guess this is related to probability as well as calculus. Probably to do with conditional expectation too.
    I was told that in mathematics, upper case means random variables and lower case means observed variables.

    I have two random variables, B and C. They are unknowns.
    Then I have a model:
    \int_x^y \int_t^u f(B-C) dC dB

    This is kind of limiting C to t to u and B to x to y.
    I was told that inside a integral I should write in small case, e.g. \int_x^y \int_t^u f(b-c) dc db

    But why?
    Last edited by avisccs; February 25th 2013 at 11:06 AM.
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  2. #2
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    Re: Notation problem inside integral

    Quote Originally Posted by avisccs View Post
    I was told that inside a integral I should write in small case, e.g. \int_x^y \int_t^u f(b-c) dc db

    But why?
    Maybe because random variables are in fact functions from the sample space to real numbers. On the other hand, the variable of integration ranges over real numbers, not functions.

    Another reason may be that random variables B and C are fixed, even though they are unknown, while the variable of integration is supposed to range over an interval.

    P.S. You may wrap LaTeX code in [TEX]...[/TEX] tags.
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    Re: Notation problem inside integral

    I guess I'd better simiplfy my question.
    In a integral, for example, \int_x^y t dt ,
    why t have to be a lower case even it can be any points between x and y?
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    Re: Notation problem inside integral

    Quote Originally Posted by avisccs View Post
    ...even it can be any points between x and y?
    I did not understand this phrase.

    The first reason I gave explains why variables of integration should not be given names that are reserved to random variables according to your convention. Does it make sense? I am not sure if this is the same reason why you were corrected, but I would also think that using random variable names inside an integral is wrong.

    By the way, what are observed variables?
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    Re: Notation problem inside integral

    Your first answer probably is the reason why it should be lower case.

    [[Maybe because random variables are in fact functions from the sample space to real numbers. On the other hand, the variable of integration ranges over real numbers, not functions.]]

    Using your terms, Observed variable = real numbers.
    But it should be real numbers inside the integral?
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  6. #6
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    Re: Notation problem inside integral

    Quote Originally Posted by avisccs View Post
    But it should be real numbers inside the integral?
    Every expression must make type sense. The idea of types is standard in programming languages, but in fact it is rooted in mathematics. One of the properties of a well-typed expression is that if it contains f(x) for some function f and argument x, the type of x must coincide with the domain of f, or, simply speaking, x must belong to the domain of f.

    Let \Omega denote the sample space and \mathbb{R} denote the set of real numbers. You probably consider integrals that take functions of type \mathbb{R}\to\mathbb{R}. The function being integrated is applied to the variable of integration, so the latter must have type \mathbb{R}. Also, since the variable of integration ranges over the segment between the limits of integration, the type of the variable of integration must be the same as that of the limits, i.e., \mathbb{R}.

    In your expression

    \int_x^y \int_t^u f(b-c) dc db

    is f:\mathbb{R}\to\mathbb{R} or f: (\Omega\to\mathbb{R})\to\mathbb{R}? If the former, then applying f to B - C is a type error because the type of B - C is \Omega\to\mathbb{R}. (Note that subtraction here has to be redefined to act on functions, probably, as pointwise subtraction.) If f: (\Omega\to\mathbb{R})\to\mathbb{R} and x, y, t, u\in\Omega\to\mathbb{R}, it is not even clear what it means to integrate between two functions. And if one make a sense of it, the result of the integral would probably also be a random variable, i.e., a function.
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    Re: Notation problem inside integral

    Thank you very much taking time to answer my questions.

    In this case, it is a type error. b and c themselves are not a function and therefore they are not random variables. Am I right thinking this?

    However, b and c can be anywhere in  [-\infty,\infty] . I guess this means they are not from a sample space, right?

    This  f(b-c) is actually a pdf of S, which S=b-c.

    Does this make S a random variable?

    I still have confusion about the terms 'random variable', because a real number sometimes can turn into a function. For example, define x a real number, say my height, and I assume it to be constant when I use it in a model. Then, in another model, it can also become a function that x(y), where y is the tireness level. I'm slightly shorter when I'm tired.
    What is the type of x?
    Last edited by avisccs; February 26th 2013 at 03:34 AM.
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    Re: Notation problem inside integral

    Quote Originally Posted by avisccs View Post
    In this case, it is a type error. b and c themselves are not a function and therefore they are not random variables. Am I right thinking this?
    I am not sure. Where exactly is a type error? There is a type mismatch if you write f(B - C) because f:\mathbb{R}\to\mathbb{R}, but (B - C):\Omega\to\mathbb{R} instead of just \mathbb{R}. In contrast, there is no error in f(b - c) because (b - c)\in\mathbb{R}.

    Quote Originally Posted by avisccs View Post
    However, b and c can be anywhere in  [-\infty,\infty] . I guess this means they are not from a sample space, right?
    I need more context to answer this.

    Quote Originally Posted by avisccs View Post
    This  f(b-c) is actually a pdf of S, which S=b-c.
    Here you need to write S = B - C because when you say "pdf of S", it means that S is a random variable, i.e., S:\Omega\to\mathbb{R}, and therefore B and C must also be random variables are not just real numbers.

    Quote Originally Posted by avisccs View Post
    I still have confusion about the terms 'random variable', because a real number sometimes can turn into a function. For example, define x a real number, say my height, and I assume it to be constant when I use it in a model. Then, in another model, it can also become a function that x(y), where y is the tireness level. I'm slightly shorter when I'm tired.
    What is the type of x?
    It depends on the context. If X depends on some outcome, then it is not a constant. In this case, if there is a measure on the set of outcomes that for each subset returns a number from 0 to 1, then X can be considered a random variable. Even when X returns the same value for each outcome, it is still a random variable. On the other hand, if x\in\mathbb{R}, then x is not a random variable but just a number.
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    Re: Notation problem inside integral

    Quote Originally Posted by emakarov View Post
    Here you need to write S = B - C because when you say "pdf of S", it means that S is a random variable, i.e., S:\Omega\to\mathbb{R}, and therefore B and C must also be random variables are not just real numbers.
    Ok, now you have confirmed that B and C are random variables (not just my imagination) and therefore there is no type error, so we are now back to square one. Am I correct writing \int_x^y \int_t^u f(B-C) dC dB?
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  10. #10
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    Re: Notation problem inside integral

    Quote Originally Posted by avisccs View Post
    Ok, now you have confirmed that B and C are random variables (not just my imagination) and therefore there is no type error, so we are now back to square one. Am I correct writing \int_x^y \int_t^u f(B-C) dC dB?
    No, I would not write this, as I repeatedly said. On the other hand, you should use the names reserved for random variables when you are defining another random variable S in S = B - C. If you don't understand any of my arguments, please ask specifically about that argument. Basically, in post #2 I gave the reason why the integral expression should use variable names for real numbers instead of those for random variables.

    Quote Originally Posted by emakarov View Post
    Maybe because random variables are in fact functions from the sample space to real numbers. On the other hand, the variable of integration ranges over real numbers, not functions.
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