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Thread: To integrate static?

  1. #1
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    To integrate static?

    A curious question occurred to me, and I am not familiar enough with real analysis to answer it. Consider a real function $\displaystyle f(x): [a,b]\to[c,d]$ defined as follows: for each element $\displaystyle x\in[a,b]$, let $\displaystyle f(x)$ be some random element from $\displaystyle [c,d]$, a function obviously completely discontinuous. What would the value of $\displaystyle \int_a^b f(x)dx$ be? It would make intuitive sense that the area under the curve is equal to the length of the interval, $\displaystyle b-a$, times the average value of the function in that interval. Therefore the answer to the question would be $\displaystyle (b-a)\frac{c+d}2$. But how to prove this rigorously? Is the function even integrable and by what definition?
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  2. #2
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    A correct statement could be: Let $\displaystyle (f(x))_{x\in[a,b]}$ be a family of independent random variables uniformly distributed on $\displaystyle [c,d]$. What can be said about the (random variable) $\displaystyle \int_a^b f(x)dx$, if this is well-defined?

    As a matter of fact, I would bet that $\displaystyle f$ is almost-surely not measurable, hence the integral wouldn't make sense.

    You can try another way, for instance as an approximation: for $\displaystyle n\geq 0$, define $\displaystyle f_n:[0,1]\to [0,1]$ (or with a,b,c,d) to be a step function where the steps have width $\displaystyle \frac{1}{n}$ and independent uniformly distributed heights in $\displaystyle [0,1]$. This won't converge to a function when $\displaystyle n\to\infty$. However, $\displaystyle \int_0^1 f_n(t)dt\to_n \frac{1}{2}$ almost-surely by the law of large numbers, thus for large $\displaystyle n$ this function $\displaystyle f_n$ nearly satisfies what you said.

    By the way, there is a name for such a function $\displaystyle f$: it is called a white noise, you can look that up.
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  3. #3
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    Laurent: Thank you for the suggestion. I believe the following sketch is much closer to a formal proof.

    Define $\displaystyle f_n:[0,1]\to[0,1]$ as follows: Divide the interval $\displaystyle [0,1]$ into $\displaystyle n$ subintervals $\displaystyle \Delta_i$ of width $\displaystyle \frac1n$. For each of these subintervals, let $\displaystyle f_n(\Delta_i)$ be a random element of $\displaystyle [0,1]$. Now define $\displaystyle g$ by sorting these subintervals from shortest to tallest, $\displaystyle g_n(\Delta_{i-1})\leq g_n(\Delta_i)$ for all $\displaystyle i$. It should be plain that $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$.

    1. Define $\displaystyle g$ as the limiting function of $\displaystyle g_n$, ($\displaystyle g=\lim_{n\to\infty} g_n$), which is $\displaystyle g(x)=x$. Of course, $\displaystyle \int_0^1 g(x)dx=\frac12$.

    Define $\displaystyle f$ as the limiting function of $\displaystyle f_n$, ($\displaystyle f=\lim_{n\to\infty} f_n$)

    2. Since $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$, $\displaystyle \int_0^1 f(x)dx=\int_0^1 g(x)dx=\frac12$.

    Are steps 1 and 2 valid?
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  4. #4
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    Quote Originally Posted by Media_Man View Post
    Laurent: Thank you for the suggestion. I believe the following sketch is much closer to a formal proof.

    Define $\displaystyle f_n:[0,1]\to[0,1]$ as follows: Divide the interval $\displaystyle [0,1]$ into $\displaystyle n$ subintervals $\displaystyle \Delta_i$ of width $\displaystyle \frac1n$. For each of these subintervals, let $\displaystyle f_n(\Delta_i)$ be a random element of $\displaystyle [0,1]$. Now define $\displaystyle g$ by sorting these subintervals from shortest to tallest, $\displaystyle g_n(\Delta_{i-1})\leq g_n(\Delta_i)$ for all $\displaystyle i$. It should be plain that $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$.

    1. Define $\displaystyle g$ as the limiting function of $\displaystyle g_n$, ($\displaystyle g=\lim_{n\to\infty} g_n$), which is $\displaystyle g(x)=x$. Of course, $\displaystyle \int_0^1 g(x)dx=\frac12$.

    Define $\displaystyle f$ as the limiting function of $\displaystyle f_n$, ($\displaystyle f=\lim_{n\to\infty} f_n$)

    2. Since $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$, $\displaystyle \int_0^1 f(x)dx=\int_0^1 g(x)dx=\frac12$.

    Are steps 1 and 2 valid?
    I don't feel like this would be any more proof-like than what I did: you give no argument for the limits, as if they were obvious. Perhaps you should re-read my post. The fact that $\displaystyle g_n$ converges to $\displaystyle x\mapsto x$ almost surely (you need to specify this) is probably true but the proof is not immediate for sure. On the other hand, when I said that $\displaystyle \int_0^1 f_n(t) dt\to\frac{1}{2}$ a.s. by the law of large numbers, that was immediate since $\displaystyle \int_0^1 f_n(t)dt=\frac{1}{n}\sum_{i=1}^n f(\frac{i}{n})$. Then the fact that $\displaystyle f_n$ converges to a function $\displaystyle f$ should also be justified and as a matter of fact, like I wrote, you just can't: the sequence $\displaystyle (f_n)_n$ doesn't converge... I gave no justification since this should be clear: indeed, for any $\displaystyle x$, the sequence $\displaystyle (f_n(x))_{n\geq 0}$ is a sequence of independent random variables uniformly distributed on $\displaystyle [0,1]$...
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