To integrate static?

**Media_Man** · January 19th 2010, 11:15 AM

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

**Laurent** · January 19th 2010, 01:49 PM

A correct statement could be: Let $(f(x))_{x\in[a,b]}$ be a family of independent random variables uniformly distributed on $[c,d]$ . What can be said about the (random variable) $\int_a^b f(x)dx$ , if this is well-defined?

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

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

By the way, there is a name for such a function $f$ : it is called a white noise, you can look that up.

**Media_Man** · January 19th 2010, 06:45 PM

Laurent: Thank you for the suggestion. I believe the following sketch is much closer to a formal proof.

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

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

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

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

Are steps 1 and 2 valid?

**Laurent** · January 20th 2010, 12:30 AM

Originally Posted by Media_Man

Laurent: Thank you for the suggestion. I believe the following sketch is much closer to a formal proof.

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

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

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

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

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 $g_n$ converges to $x\mapsto x$ 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 $\int_0^1 f_n(t) dt\to\frac{1}{2}$ a.s. by the law of large numbers, that was immediate since $\int_0^1 f_n(t)dt=\frac{1}{n}\sum_{i=1}^n f(\frac{i}{n})$ . Then the fact that $f_n$ converges to a function $f$ should also be justified and as a matter of fact, like I wrote, you just can't: the sequence $(f_n)_n$ doesn't converge... I gave no justification since this should be clear: indeed, for any $x$ , the sequence $(f_n(x))_{n\geq 0}$ is a sequence of independent random variables uniformly distributed on $[0,1]$ ...

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