most recent 30 from http://mathoverflow.net 2013-06-20T10:52:02Z http://mathoverflow.net/feeds/user/1954 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3912/question-on-sigma-fields/6330#6330 Jeff Hussmann 2009-11-20T22:56:39Z 2010-07-24T07:37:22Z

<p>It is trivially true but maybe worth noting that the converse is also true - if there exists such an f, then Y is &sigma;(X)-measurable.</p> <p>This and the question asked are theorem 20.1(ii) in Billingsley's <em>Probability and Measure</em>, 3rd edition.</p>

http://mathoverflow.net/questions/9608/can-you-explain-a-step-in-an-expectation-maximization-algorithm-in-a-nature-artic/20556#20556 Jeff Hussmann 2010-04-06T22:18:44Z 2010-04-06T22:33:12Z

<p>The numbers in question are the conditional probabilities that each data set was generated using coin A or coin B given the number of heads that occured in the data set and the current estimate of each coin's bias. To compute these numbers, we calculate likelihoods of head counts given coin assignments and biases, then use Bayes' theorem to relate these likelihoods to the desired conditional probabilities. </p> <p>As in the paper, let $\mathbf{z}$ be the current vector of (random variables representing) assignments of coin A or coin B to each data set, let $\mathbf{X}$ be the vector of head counts in each data set, and let $\theta = (\theta_A, \theta_B)$ be the current estimate of the coins' biases. </p> <p>Then (for example) the first red number in Figure 1(b), step 2 is $\mathbb{P}(z_1 = A | X_1 = x_1, \theta).$ </p> <p>Apply Bayes' theroem to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) \mathbb{P}(z_1 = A | \theta)}{\mathbb{P}(X_1 = x_1 | \theta)}$.</p> <p>Marginalize the denominator over $z_1$ and use the fact that $\mathbf{z}$ is assumed a priori to be uniformly distributed and independent of $\theta$ to cancel terms of the form $\mathbb{P}(z_1 | \theta)$ to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta)}{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) + \mathbb{P}(X_1 = x_1 | z_1 = B, \theta)} = \frac{\theta_A^{x_1} (1 - \theta_A)^{n - x_1}}{\theta_A^{x_1} (1 - \theta_A)^{10 - x_1} + \theta_B^{x_1} (1 - \theta_B)^{10 - x_1}}$. </p>

http://mathoverflow.net/questions/6233/order-statistics-for-components-of-a-random-unit-vector Jeff Hussmann 2009-11-20T07:31:20Z 2009-12-18T19:22:22Z

<p>Suppose you sample uniformly from the unit vectors in R^n. What are the distributions of the order statistics of the magnitudes of the components of the sampled vectors? That is, for 1 &lt;= i &lt;= n and x in [0,1], what is the probability that the i'th largest component of the vector (in absolute value) is less than or equal to x? </p>

http://mathoverflow.net/questions/7584/what-are-the-most-misleading-alternate-definitions-in-taught-mathematics/7867#7867 Jeff Hussmann 2010-05-27T20:02:12Z 2010-05-27T20:02:12Z

Resolve it how, short of a serious digression into conditional expectation?

http://mathoverflow.net/questions/21065/how-do-you-measure-the-relative-randomness-of-a-function/21066#21066 Jeff Hussmann 2010-04-12T04:22:15Z 2010-04-12T04:22:15Z

I don't know about non-trivial examples, but X = 0 (or 1) a.s. works.

http://mathoverflow.net/questions/13171/how-many-trial-picks-expectedly-sufficient-to-cover-a-sample-space/13186#13186 Jeff Hussmann 2010-01-28T17:42:16Z 2010-01-28T17:42:16Z

The r=1 case is just the coupon collector's problem - <a href="http://en.wikipedia.org/wiki/Coupon_collector%27s_problem" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a>. Look at the 'calculating the expectation' section of the linked page for a much simpler and more elegant way of getting your result.

http://mathoverflow.net/questions/3912/question-on-sigma-fields/3925#3925 Jeff Hussmann 2010-01-11T22:22:26Z 2010-01-11T22:22:26Z

Don't we need the sequence Y_n to converge to Y pointwise, not just almost surely, since we want Y = f(X) pointwise, not just almost surely? Of course, this isn't a problem for the proof, since such a sequence exists. Also, a way to avoid the distinction between when lim_n f_n exists and when it doesn't is to just take f = limsup_n f_n.