## going to war [a riddle]

Posted in Books, Kids, Statistics with tags , , , , , on December 16, 2016 by xi'an

On the Riddler this week, a seemingly obvious riddle:

A game consists of Alice and Bob, each with a $1 bill, receiving a U(0,1) strength each, unknown to the other, and deciding or not to bet on this strength being larger than the opponent’s. If no player bets, they both keep their$1 bill. Else, the winner leaves with both bills. Find the optimal strategy.

As often when “optimality” is mentioned, the riddle is unclear because, when looking at the problem from a decision-theoretic perspective, the loss function of each player is not defined in the question. But the St. Petersburg paradox shows the type of loss clearly matters and the utility of money is anything but linear for large values, as explained by Daniel Bernoulli in 1738 (and later analysed by Laplace in his Essai Philosophique).  Let us assume therefore that both players live in circumstances when losing or winning \$1 makes little difference, hence when the utility is linear. A loss function attached to the experiment for Alice [and a corresponding utility function for Bob] could then be a function of (a,b), the result of both Uniform draws, and of the decisions δ¹ and δ² of both players as being zero if δ¹=δ²=0 and

$L(a,b,\delta^1,\delta^2)=\begin{cases}0&\text{if }\delta^1=\delta^2=0\\\mathbb{I}(ab)&\text{else}\\\end{cases}$

Considering this loss function, Alice aims at minimising the expected loss by her choice of δ¹, equal to zero or one, expected loss that hence depends  on the unknown and simultaneous decision of Bob. If for instance Alice assumes Bob takes the decision to compete when observing an outcome b larger than a certain bound α, her decision is based on the comparison of (when B is Uniform (0,1))

$\mathbb{P}(a\alpha)-\mathbb{P}(a>B,B>\alpha)=2(1-a\vee\alpha)-(1-\alpha)$

(if δ¹=0) and of 1-2a (if δ¹=1). Comparing both expected losses leads to Alice competing (δ¹=1) when a>α/2.

However, there is no reason Alice should know the value of α when playing the (single) game and so she may think that Bob will follow the same reasoning, leading him to choosing a new bound of α/4, and, by iterating the thought process, down all the way to α=0!  So this modelling leads to always play the game, with each player having a ½ probability to win… Alternatively, Alice may set a prior on α, which leads to another bound on a for playing or not the game. Which in itself is not satisfactory either. (The published solution is following the above argument. Except for posting the maths expressions.)

## Bayesian model averaging in astrophysics [guest post]

Posted in Books, pictures, Statistics, University life with tags , , , , , , , , , , , , , on August 12, 2015 by xi'an

.[Following my posting of a misfiled 2013 blog, Ewan Cameron told me of the impact of this paper in starting his own blog and I asked him for a guest post, resulting in this analysis, much deeper than mine. No warning necessary this time!]

Back in February 2013 when “Bayesian Model Averaging in Astrophysics: A Review” by Parkinson & Liddle (hereafter PL13) first appeared on the arXiv I was a keen, young(ish) postdoc eager to get stuck into debates about anything and everything ‘astro-statistical’. And with its seemingly glaring flaws, PL13 was more grist to the mill. However, despite my best efforts on various forums I couldn’t get a decent fight started over the right way to do model averaging (BMA) in astronomy, so out of sheer frustration two months later I made my own soapbox to shout from at Another Astrostatistics Blog. Having seen PL13 reviewed recently here on Xian’s Og it feels like the right time to revisit the subject and reflect on where BMA in astronomy is today.

As pointed out to me back in 2013 by Tom Loredo, the act of Bayesian model averaging has been around much longer than its name; indeed an early astronomical example appears in Gregory & Loredo (1992) in which the posterior mean representation of an unknown signal is constructed for an astronomical “light-curve”, averaging over a set of constant and periodic candidate models. Nevertheless the wider popularisation of model averaging in astronomy has only recently taken place through a variety of applications in cosmology: e.g. Liddle, Mukherjee, Parkinson & Wang (2006) and Vardanyan, Trotta & Silk (2011).

In contrast to earlier studies like Gregory & Loredo (1992)—or the classic review on BMA by Hoeting et al. (1999)—in which the target of model averaging is typically either a utility function, a set of future observations, or a latent parameter of the observational process (e.g. the unknown “light-curve” shape) shared naturally by all competing models, the proposal of cosmological BMA studies is to produce a model-averaged version of the posterior for a given ‘shared’ parameter: a so-called “model-averaged PDF”. This proposal didn’t sit well with me back in 2013, and it still doesn’t sit well with me today. Philosophically: without a model a parameter has no meaning, so why should we want to seek meaning in the marginalised distribution of a parameter over an entire set of models? And, practically: to put it another way, without knowing the model ‘label’ to which a given mass of model-averaged parameter probability belongs there’s nothing much useful we can do with this ‘PDF’: nothing much we can say about the data we’ve just analysed and nothing much we can say about future experiments. Whereas the space of the observed data is shared automatically by all competing models, it seems to me to be somehow “un-Bayesian” to place the further restriction that the parameters of separate models share the same scale and topology. I say “un-Bayesian” since this mode of model averaging suggests a formulation of the parameter space + prior pairing stronger than the statement of one’s prior beliefs for the distribution of observable data given the model. But I would be happy to hear arguments from the other side in the comments box below … ! Continue reading