User thierry de la rue - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:40:51Z http://mathoverflow.net/feeds/user/19603 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121830/if-mathcalf-t-is-separable-why-is-mathcalf-infty-generated-by-a-random/121834#121834 Answer by Thierry de la Rue for If $\mathcal{F}_t$ is separable why is $\mathcal{F}_\infty$ generated by a random variable? Thierry de la Rue 2013-02-14T20:19:27Z 2013-02-14T20:19:27Z <p>Yes, you can define separable by saying that it means "generated by a countable collection of (real) random variables", and you can always assume that all these variables take values in $[0,1]$. Then you can find a single random variable generating the same $\sigma$-algebra by intertwining the digits of all your initial random variables (the countable collection of all the digits of all random variables is itself countable, so that you can encode them in a single sequence of digits).</p> http://mathoverflow.net/questions/93700/de-finettis-theorem-the-pointwise-ergodic-theorem-and-reverse-martingales/93738#93738 Answer by Thierry de la Rue for De Finetti's theorem, the pointwise ergodic theorem, and reverse martingales Thierry de la Rue 2012-04-11T06:02:37Z 2012-04-11T06:02:37Z <p>It was recently pointed out by Bill Johnson in a comment to a <a href="http://mathoverflow.net/questions/93635/fully-exchangeable-random-sequences" rel="nofollow">question concerning "Fully exchangeable random sequences</a>" that if an infinite sequence of random variables is exchangeable, then it is "fully exchangeable", that is to say its distribution is in fact invariant by any permutation (even if the permutation has no fixed point). The same argument shows that the distribution of an infinite exchangeable sequence of random variables is invariant under the shift map. Hence you can view (A) as the pointwise ergodic theorem for the shift map.</p> http://mathoverflow.net/questions/92571/conditional-probabilities-the-mad-kings-draft/92686#92686 Answer by Thierry de la Rue for Conditional Probabilities - The Mad Kings' Draft Thierry de la Rue 2012-03-30T14:08:37Z 2012-03-30T14:08:37Z <p>Let us formalize the story in the following way: Let $U_1,U_2,U_3,U_4$ be 4 independent random variables, uniformly distributed on $[0,1]$ (the possibly drafted citizens), and let $M$ be an independent Bernoulli random variable, with $\mathbb{P}(M=0) = \mathbb{P}(M=1) = 1/2$. ($M=1$ means that the king is mad.) Then we construct the random point process $\eta$ on the unit interval by $$\eta := \delta_{U_1}+\delta_{U_2} + (1-M)(\delta_{U_3}+\delta_{U_4}).$$ ($\delta_U$ is the Dirac measure on the point $U$.) The question is, given some fixed $j\in[0,1]$, to compute the conditional probability that $M=1$ knowing that $\eta(j)=1$. </p> <p>If the set modelizing the kingdom were finite, this would pose no problem since the event with respect to which we condition would be of positive probability. But here we condition with respect to a negligible event and this is not a priori clear what this conditional probability means. Fortunately a tool has been developped to deal with this kind of problems: Palm probabilities. In this context, it is a family $(P_j)_{j\in [0,1]}$ of probability measures on the set $I^*$ of point measures on the unit interval $I$, such that for any Borel set $A\subset I$ and any measurable subset $B\subset I^*$: $$\mathbb{E} [\mathbb{1}_B(\eta) \ \eta(A)] = \int_A P_j(B)\ d\mu(j),$$ where $\mu$ is the measure on $I$ given by $$\mu(A):= \mathbb{E} [ \eta(A) ].$$ If $I$ were finite, then we could easily check that this family of Palm probabilities would be given by $$P_j(B)= \mathbb{P}(B | \eta(j)=1).$$ In the case of a continuous space $I$, $P_j$ can still be interpreted as the conditional distribution of our point process knowing that it charges $j$. </p> <p>In the setting of the question, we observe that $\mu(A)=3 |A|$ where $|A|$ denotes the Lebesgue measure of $A$, and then we can easily compute $P_j(B)$, which is given by $$P_j(B)= \frac{1}{3}\int_I \mathbb{1}_B (\delta_j+\delta_{u_2}) du_2 + \frac{2}{3}\int_{I\times I\times I} \mathbb{1}_B (\delta_j+\delta_{u_2}+\delta_{u_3}+\delta_{u_4}) du_2 du_3 du_4.$$ When $B$ is the set $(M=1)$, $\mathbb{1}_B(\eta)=1$ if $\eta$ charges 2 points, 0 if $\eta$ charges 4 points, hence we get $$P_j(B)=1/3.$$</p> http://mathoverflow.net/questions/89064/what-does-it-mean-to-say-almost-always/89089#89089 Answer by Thierry de la Rue for What does it mean to say "almost always" ? Thierry de la Rue 2012-02-21T07:38:46Z 2012-02-21T07:38:46Z <p>I see some contradiction in your hypotheses: since the $A_j$'s are disjoint and have non empty interiors, the union should be at most countable (there can't exist more $A_j$'s than the cardinal of the set of rational points in $\mathbb{R}^{m\times n}$).</p> http://mathoverflow.net/questions/88621/what-is-the-adic-realization-of-a-bernoulli-shift/88637#88637 Answer by Thierry de la Rue for What is the adic realization of a Bernoulli shift ? Thierry de la Rue 2012-02-16T14:51:45Z 2012-02-16T14:51:45Z <p>I think you can realize the Bernoulli shift on $k$ symbols as an adic transformation on the following Bratelli-Vershik diagram: put $k$ nodes on the first level. Suppose levels 1 to $n$ have been defined, and call $L_n$ the set of nodes in the $n$-th level. Then nodes on the $(n+1)$-th level are pairs $(i,j)\in L_n\times L_n$, where $(i,j)$ is connected to $i$ and to $j$ (in this order) to the $n$-th level. Does not this work?</p> http://mathoverflow.net/questions/88486/probability-of-iid-sum-being-positive/88497#88497 Answer by Thierry de la Rue for probability of IID sum being positive Thierry de la Rue 2012-02-15T07:53:56Z 2012-02-15T07:53:56Z <p>I think this is false without the second moment assumption. Construct $X_n$ of the form $$X_n = -1+\sum_{k\ge 1} Y_n^k,$$ where $Y_n^k\ge 0$, $E[Y_n^k]=2^{-k}$, $P(Y_n^k>0)=\epsilon_k$ and $(Y_n^{k+1}>0)\subset(Y_n^k>0)$. Choose the $\epsilon_k$ inductively as follows: Assume $\epsilon_1,\ldots,\epsilon_k$ have already been chosen, so that $$X_n^k := -1+\sum_{j=1}^k Y_n^j$$ is already defined. Observe that $X_n^k$ has negative expectation, so that by the law of large numbers you can find an arbitrarily large integer $n_k$ such that $$P\left(\frac{1}{n_k} \sum_{n=1}^{n_k} X_n^k &lt; 0 \right) > 99/100.$$ Then choose $\epsilon_{k+1}$ so small that $$P\left( Y_1^{k+1}=0,\ldots,Y_{n_k}^{k+1}=0 \right) > 99/100,$$ then $$P\left(\frac{1}{n_k} \sum_{n=1}^{n_k} X_n &lt; 0 \right) > 98/100.$$</p> http://mathoverflow.net/questions/82661/book-recommendation-for-ergodic-theory-and-or-topological-dynamics/82677#82677 Answer by Thierry de la Rue for Book recommendation for ergodic theory and/or topological dynamics? Thierry de la Rue 2011-12-05T06:12:56Z 2011-12-05T06:12:56Z <p>Let me suggest you a recent book by Steve Kalikow and Randall McCutcheon: "An Outline of Ergodic Theory". This is a nice book to get a solid background in isomorphism theory of measurable dynamical systems. I like the way proofs of theorems are presented through guided exercises.</p> http://mathoverflow.net/questions/92571/conditional-probabilities-the-mad-kings-draft/92686#92686 Comment by Thierry de la Rue Thierry de la Rue 2012-04-02T08:18:52Z 2012-04-02T08:18:52Z I am not sure you can avoid the concept of Palm probabilities, since the conditional expectation you need for the mad king problem is precisely what Palm probabilities are designed for. If I understand correctly your generalized definition of a conditional expectation, the formula you give ($\int_\Omega\ldots$) is the exact analog of the formula I gave in my answer for the definition of the Palm probabilities. Let me suggest you some references: Stoyan, Kendall and Mecke: &quot;Stochastic geometry and its applications&quot; Scheider and Weil: &quot;Stochastic and Integal Geometry&quot;. http://mathoverflow.net/questions/88621/what-is-the-adic-realization-of-a-bernoulli-shift/88637#88637 Comment by Thierry de la Rue Thierry de la Rue 2012-02-16T20:03:53Z 2012-02-16T20:03:53Z Indeed, the graph I proposed is homogeneous: Each node of level n has exactly k^n paths to the root of the diagram. But then I must admit that this poses some problem, since this introduces eigenvalues for the associated adic transformation. So this adic transformation is rather an extension of the Bernoulli shift (probably a direct product with an odometer).