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A simple derivation for the Lebesgue measure of the euclidean unit ball in $\mathbb{R}^n$ follows from computing $$ \int_{\mathbb{R}^n}e^{-\|x\|^2}\,dx $$ in two different ways. See, for instance, Keith Ball, An Elementary Introduction to Modern Convex Geometry, page $5$. Now I was wondering about the following slightly unusual variation:

Let $X_1,\ldots,X_n$ independent random variables with the Cantor distribution.

What is the probability that $X_1^2+\ldots+X_n^2\leq 1$?

I bet this can be tackled by exploiting the fact that the cumulants of the Cantor distribution are given by:

$$ \kappa_{2n}=\frac{2^{2n-1}(2^{2n}-1)\,B_{2n}}{n (3^{2n}-1)},\tag{CM}$$ but how to prove $(\mathrm{CM})$? - This has been answered, but the main question is still open.

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HINT:

For $\mu(x)$ the Cantor measure supported on the Cantor set $\subset [0,1]$ we have the change of variable formula: $$\int f(x)\, d\mu(x) = \frac{1}{2} \int f(1/3 x)\, d\mu(x) + \frac{1}{2} \int f(1/3 x + 2/3)\, d \mu(x)$$ analogous to $\int_0^1 f(x)\, dx =\frac{1}{2} \int_0^1 f(1/2 x)\, d x + \frac{1}{2} \int_0^1 f(1/2 x + 1/2)\, d x $

$\bf{Added:}$ It's easy to see that the first moment $E(X) = \int x \, d \mu(x)= \frac{1}{2}$, and this can be obtained readily from the above formula for $f(x) = x$.

Consider now the central moment generating function $$F(t)\colon =E[e^{t(X-\frac{1}{2})}] = \int e^ {t(x-\frac{1}{2})} \, d\mu(x) $$ From the above equality for $f_t(x) = e^{t(x-\frac{1}{2})}$ we get $$\int e^ {t(x-\frac{1}{2})} \, d\mu(x) = \frac{1}{2}\left( \int e^ {t(\frac{x}{3}-\frac{1}{2})} \, d\mu(x) + \int e^ {t(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})} \, d\mu(x) \right) $$ Now we notice that \begin{eqnarray} t\,(\frac{x}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) - \frac{t}{3}\\ t\,(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) + \frac{t}{3} \end{eqnarray} Therefore we get the equality $$F(t) = \frac{e^{\frac{t}{3}} + e^{-\frac{t}{3}}}{2} \cdot F(\frac{t}{3}) $$

$\bf{Added:}$ Rewrite the above equality as $$F(3t) = \frac{e^t + e^{-t}}{2} F(t)$$ Let $G(t) = \log F(t)$. From the above we get $$G(3 t) - G(t) = \log ( \frac{e^t + e^{-t}}{2})$$

$\bf{Added:}$ Some (moment) calculations: $$\int x d \mu(x) = \frac{1}{2} \left( \ \int (\frac{1}{3} x + \frac{1}{3} x + \frac{2}{3}) d\mu(x) \right )$$ implies $\int x d \mu(x) = \frac{1}{2}$ as expected.

Let's apply the same formula for $f(x) = (x-\frac{1}{2})^n$. We have \begin{eqnarray} \int (x-\frac{1}{2})^n d \mu(x) = \frac{1}{2}\left( \int ( \frac{x}{3} - \frac{1}{2})^n + ( \frac{x}{3} + \frac{2}{3}- \frac{1}{2})^n d\mu(x) \right ) = \\ =\frac{1}{2\cdot 3^n}\left( \int ( x - \frac{1}{2}-1)^n + ( x-\frac{1}{2} + 1)^n d\mu(x) \right ) \end{eqnarray} that is $$M_n = \frac{1}{3^n}\sum_{k \ge 0} \binom{n}{2k} M_{n-2k}$$ which is basically a formula from above $F(3t) = \frac{e^t + e^{-t}}{2} F(t)$.

We get from here $m_2 = \frac{1}{8}$, $m_4 = \frac{7}{320}$, etc. Note that the formula provided in Wikipedia is for the cumulants, not the central moments, as $\kappa_4 = \frac{1}{40}$.

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  • $\begingroup$ And so? $\phantom{}$ $\endgroup$ – Jack D'Aurizio Jun 20 '15 at 23:19
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    $\begingroup$ @Jack D'Aurizio: Using this, you can calculate the central moments, and perhaps more. $\endgroup$ – Orest Bucicovschi Jun 20 '15 at 23:47
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    $\begingroup$ Ok, so we have a derivation for the moment generating function, and the Bernoulli numbers just come from expanding $\log\cosh$ as a Taylor series. That is good, but it is not the main point of my question, i.e. to compute $$\mathbb{P}[X_1^2+\ldots+X_n^2]\leq 1.$$ Anyway, I am upvoting your answer since I'm quite curious about it leading to an effective answer. $\endgroup$ – Jack D'Aurizio Jun 23 '15 at 17:09
  • $\begingroup$ When a question contains two or more questions, how do you know what the "main question" is? $\endgroup$ – GEdgar Jun 26 '15 at 15:20
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    $\begingroup$ @GEdgar: The answer only deals with the cumulants ( central moments). The main question about vol$ ( C^n \cap B_n)$ ( $C^n$ Cantor set with the product measure, $B_n$ the unit ball ) is still unanswered. $\endgroup$ – Orest Bucicovschi Jun 26 '15 at 16:42

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