# Game theory bayesian updating

By my reasoning, the posterior distribution is not straightforwardly uniform since it is formed from a non-conjugate prior. Does it make sense to say $\mathscr|\pi^ \ \sim Uniform[-1,\pi^*]$ or should some other distribution be specified?

Alternatively is it possible that the answer to the problem contains an error? Bayes formula: $$ p(t|\pi^*) \propto p(\pi^*|t)p(t). $$ Since $p(\pi^*|t)$ is

By my reasoning, the posterior distribution is not straightforwardly uniform since it is formed from a non-conjugate prior. Does it make sense to say $\mathscr|\pi^ \ \sim Uniform[-1,\pi^*]$ or should some other distribution be specified?Alternatively is it possible that the answer to the problem contains an error? Bayes formula: $$ p(t|\pi^*) \propto p(\pi^*|t)p(t). $$ Since $p(\pi^*|t)$ is $1$ only when $t\in [-1, \pi^*]$ and $0$ otherwise, you get the resulting posterior.Particle filtering methodology uses a genetic type mutation-selection sampling approach, with a set of particles (also called individuals, or samples) to represent the posterior distribution of some stochastic process given some noisy and/or partial observations.The state-space model can be nonlinear and the initial state and noise distributions can take any form required.In game theory, a Bayesian game is a game in which the players have incomplete information on the other players (e.g.on their available strategies or payoffs), but, they have beliefs with known probability distribution.

||By my reasoning, the posterior distribution is not straightforwardly uniform since it is formed from a non-conjugate prior. Does it make sense to say $\mathscr|\pi^ \ \sim Uniform[-1,\pi^*]$ or should some other distribution be specified?

Alternatively is it possible that the answer to the problem contains an error? Bayes formula: $$ p(t|\pi^*) \propto p(\pi^*|t)p(t). $$ Since $p(\pi^*|t)$ is $1$ only when $t\in [-1, \pi^*]$ and $0$ otherwise, you get the resulting posterior.

Particle filtering methodology uses a genetic type mutation-selection sampling approach, with a set of particles (also called individuals, or samples) to represent the posterior distribution of some stochastic process given some noisy and/or partial observations.

The state-space model can be nonlinear and the initial state and noise distributions can take any form required.

$ only when $t\in [-1, \pi^*]$ andIn addition to the actual players in the game, there is a special player called Nature.Does it mean that, if you test positive, you have a 95% chance of having the disease.While this sounds sensible, the answer is usually "no." The actual probability depends not only on the reliability of the test, but also the number of infections in the population to begin with. Go to the applet It is often difficult to observe effort on the part of employees, so companies are forced to reward employees based on success or failure (a measure of performance) which is only partially controlled by effort.I was working through a signaling game problem recently and the proof suggested the following: Actor A has a type: $\ \mathscr \sim Uniform[-1,1]$ Actor A gives signal $\pi^*$ that perfectly seperates types at $\pi^$.In other words, $pr(\pi^*|\mathscr\in [-1,\pi^*])=1\ \&\ pr(\pi^*|\mathscr\in [\pi^*,1])=0$ (this is the likelihood) Actor B observes $\pi^*$, yielding posterior beliefs about actor A: $\mathscr \sim Uniform[-1,\pi^*]$. It appears that this process, as i read it, has the same prior and posterior distributions (uniform), yet the likelihood distribution is unspecified and the uniform is not a conjugate prior for any common distribution.

$ otherwise, you get the resulting posterior.Particle filtering methodology uses a genetic type mutation-selection sampling approach, with a set of particles (also called individuals, or samples) to represent the posterior distribution of some stochastic process given some noisy and/or partial observations.

The state-space model can be nonlinear and the initial state and noise distributions can take any form required.

[[In addition to the actual players in the game, there is a special player called Nature.

Does it mean that, if you test positive, you have a 95% chance of having the disease.

While this sounds sensible, the answer is usually "no." The actual probability depends not only on the reliability of the test, but also the number of infections in the population to begin with. Go to the applet It is often difficult to observe effort on the part of employees, so companies are forced to reward employees based on success or failure (a measure of performance) which is only partially controlled by effort.

I was working through a signaling game problem recently and the proof suggested the following: Actor A has a type: $\ \mathscr \sim Uniform[-1,1]$ Actor A gives signal $\pi^*$ that perfectly seperates types at $\pi^$.

In other words, $pr(\pi^*|\mathscr\in [-1,\pi^*])=1\ \&\ pr(\pi^*|\mathscr\in [\pi^*,1])=0$ (this is the likelihood) Actor B observes $\pi^*$, yielding posterior beliefs about actor A: $\mathscr \sim Uniform[-1,\pi^*]$. It appears that this process, as i read it, has the same prior and posterior distributions (uniform), yet the likelihood distribution is unspecified and the uniform is not a conjugate prior for any common distribution.

||In addition to the actual players in the game, there is a special player called Nature.Does it mean that, if you test positive, you have a 95% chance of having the disease.While this sounds sensible, the answer is usually "no." The actual probability depends not only on the reliability of the test, but also the number of infections in the population to begin with. Go to the applet It is often difficult to observe effort on the part of employees, so companies are forced to reward employees based on success or failure (a measure of performance) which is only partially controlled by effort.I was working through a signaling game problem recently and the proof suggested the following: Actor A has a type: $\ \mathscr \sim Uniform[-1,1]$ Actor A gives signal $\pi^*$ that perfectly seperates types at $\pi^$.In other words, $pr(\pi^*|\mathscr\in [-1,\pi^*])=1\ \&\ pr(\pi^*|\mathscr\in [\pi^*,1])=0$ (this is the likelihood) Actor B observes $\pi^*$, yielding posterior beliefs about actor A: $\mathscr \sim Uniform[-1,\pi^*]$. It appears that this process, as i read it, has the same prior and posterior distributions (uniform), yet the likelihood distribution is unspecified and the uniform is not a conjugate prior for any common distribution.

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