# Interesting problem

1. Sep 15, 2009

### csopi

Hi everybody,

On a probability theory lecture I met an interesting problem:

Let f : [0;1] -> R a bounded, 3 times continously differentiable function. Calculate the following limit

$$n * \int_0^1 \int_0^1 ... \int_0^1 f(\frac{x_1+...+x_n}n)-f(0.5) dx_1...dx_n$$ while n -> infinity

Any ideas?

Last edited: Sep 15, 2009
2. Sep 15, 2009

### statdad

As a suggestion: treat the first term as a function of $$x_1, x_2, ..., x_n$$, and use the fact that $$f$$ is three times differentiable to expand the integrand in a multi-variable Taylor polynomial centered at 0.5. Simplify, use the fact that $$f$$ is bounded to help with the error term, and try to get an expression for the multiple integral at $$n$$ fixed. IF you can do that, multiply by $$n$$ and take the limit.

Disclaimer: I haven't tried this, so I cannot give any insight on its correctness or lack thereof, nor of its level of difficulty. It seems like the reasonable approach (thinking back to my probability/math stat courses), but (as my father used to say) it may be "as useless as a hog on ice". Let us know what you find.

3. Sep 16, 2009

### mXSCNT

Looks like the change of variables y = (x1 + x2 + ... + xn)/n might be a good first step. You integrate like:
$$\int_0^1 M(y)(f(y) - f(0.5)) dy$$
The tricky part is figuring out the multiplier M(y). For n=2 the multiplier is 2 sqrt(2) min(y, 1-y). For n=3 it is a little more complicated. M(y) is the "area" of the hyperplane portion (x_1 + x_2 + ... + x_n)/n = y, where 0 <= x_i <= 1.

Assuming that the limit as n -> infinity is 0, perhaps all you need to prove about M(y) is that it bulges in the middle more and more as n increases, so eventually nearly all of the integral ends up being near y=0.5.

Last edited: Sep 16, 2009
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