0
$\begingroup$

I am reading this paper Knowledge-Gradient Policy for Correlated Normal Beliefs for Rank and Selection Problem. The idea is as follows: We have $M$ distinct alternatives and samples from alternative $i$ are iid with $\mathcal{N}(\theta_i,\lambda_i)$, where $\theta_i$ is unknown and $\lambda_i$ is the known variance. At every time step, we can sample from only one of the arms and at the end of $N$ timesteps, we pick the alternative that we think that has the highest mean reward.

More concretely using the notation and text in the paper,

Let $\mathbf \theta = (\theta_1, \dots, \theta_M)'$ be column vector of unknown means. We initially assume our belief about $\bf \theta$ as:

\begin{align} \bf \theta \sim \mathcal{N}(\mu^0,\Sigma^0) \label{1} \tag{1} \end{align}

Consider a sequence of $N$ sampling decisions, $x^{0}, x^{1}, \ldots, x^{N-1} .$ The measurement decision $x^{n}$ selects an alternative to sample at time $n$ from the set $\{1, \ldots, M\}$. The measurement error $\varepsilon^{n+1} \sim \mathcal{N}\left(0, \lambda_{x^{n}}\right)$ is independent conditionally on $x^{n}$, and the resulting sample observation is $\hat{y}^{n+1}=\theta_{x^{n}}+\varepsilon^{n+1}$. Conditioned on $\theta$ and $x^{n}$, the sample has conditional distribution $\hat{y}^{n+1} \sim \mathcal{N}\left(\theta_{x^{n}}, \lambda_{x^{n}}\right)$. Note that our assumption that the errors $\varepsilon^{1}, \ldots, \varepsilon^{N}$ are independent differentiates our model from one that would be used for common random numbers. Instead, we introduce correlation by allowing a non-diagonal covariance matrix $\Sigma^{0}$.

We may think of $\theta$ as having been chosen randomly at the initial time 0 , unknown to the experimenter but according to the prior distribution (1), and then fixed for the duration of the sampling sequence. Through sampling, the experimenter is given the opportunity to better learn what value $\theta$ has taken.

We define a filtration $\left(\mathcal{F}^{n}\right)$ wherein $\mathcal{F}^{n}$ is the sigma-algebra generated by the samples observed by time $n$ and the identities of their originating alternatives. That is, $\mathcal{F}^{n}$ is the sigma-algebra generated by $x^{0}, \hat{y}^{1}, x^{1}, \hat{y}^{2}, \ldots, x^{n-1}, \hat{y}^{n} .$ We write $\mathbb{E}_{n}$ to indicate $\mathbb{E}\left[\cdot \mid \mathcal{F}^{n}\right]$, the conditional expectation taken with respect to $\mathcal{F}^{n}$, and then define $\mu^{n}:=\mathbb{E}_{n}[\theta]$ and $\Sigma^{n}:=\operatorname{Cov}\left[\theta \mid \mathcal{F}^{n}\right]$. Conditionally on $\mathcal{F}^{n}$, our posterior predictive belief for $\theta$ is multivariate normal with mean vector $\mu^{n}$ and covariance matrix $\Sigma^{n} .$

We can obtain the updates of $\mu^{n}$ and $\Sigma^{n}$ as functions of $\mu^{n-1}, \Sigma^{n-1}, \hat{y}^{n}$, and $x^{n-1}$ as follows:

\begin{align} \mu^{n+1} &=\mu^{n}+\frac{\hat{y}^{n+1}-\mu_{x}^{n}}{\lambda_{x}+\sum_{x x}^{n}} \Sigma^{n} e_{x} \tag{2} \label{2}\\\ \Sigma^{n+1} &=\Sigma^{n}-\frac{\Sigma^{n} e_{x} e_{x}^{\prime} \Sigma^{n}}{\lambda_{x}+\Sigma_{x x}^{n}} \tag{3} \label{3} \text { where } e_{x} \text { is a column } M \text { -vector of } 0 \text { s with a single } 1 \text { at index } x \end{align}

My question:

The authors claim that in equation \ref{2}, $\hat{y}^{n+1}-\mu_{x}^{n}$ when conditioned on $\mathcal{F}^n$ has zero mean ; this claim seems wrong to me. My understanding is that $\hat{y}^{n+1}$ still follows $\mathcal{N}(\theta^*_x,\lambda_x)$ where $\theta^*_x$ is some realisation sampled from $\mathcal{N}(\mu^0_x,\Sigma^0_{xx})$ and this true $\theta^{*}_x$ need not be the same as $\mu^n_{x}$.

On the basis of this claim, the authors design an algorithm and prove some theoretical results. This is a widely cited paper and so, I think I am missing something here with respect to bayesian setting and posterior distributions.

$\endgroup$
2
+50
$\begingroup$

The setup seems to imply that $(\theta_1,\ldots\theta_M)$ are only sampled once. $\theta_x$ for given $x$ will always be the same regardless of $n$. There is no $\theta^*_x$ different from the original $\theta_x$. Note that there is a certain confusion of notation between random variables and their realisations (which is often found in Bayesian notation). I denote by $\tilde\theta_x$ the random variable (over which the prior is defined) that has taken the value $\theta_x$.

In fact, $E(\hat y^{n+1}|\tilde\theta_x=\theta_x)=\theta_x$. Now we don't know $\theta_x$ at this point, but we know what we expect it to be: $E(\hat y^{n+1}|\mathcal{F}^n)=E_n(\tilde\theta_x)=\mu_x^n$ (it looks like you use $x=x^n$ for ease of notation), therefore $E(\hat y^{n+1}-\mu_x|\mathcal{F}^n)=0$.

$\endgroup$
12
  • $\begingroup$ "..but we know what we expect it to be: $E(\hat y^{n+1}|\mathcal{F}^n)=E_n(\tilde\theta_x)=\mu_x^n$ " - actually, this statement is causing my confusion and the one I am claiming as incorrect. Because, the truth is that $E(\hat y^{n+1}|\mathcal{F}^n) = \theta_x$. We don't know $\theta_x$ yet but it does not mean that it is equal to $\mu^n_x$. With the help of the information until $n$, the best we can do is to refine our prior distribution to a posterior distribution with mean $\mu^n_x$. $\endgroup$ Jul 21 '21 at 19:05
  • 1
    $\begingroup$ $\theta_x$ is not a (frequentist) constant (parameter), but a realisation of a random variable. In Bayesian reasoning, everything unknown is modelled as random. As long as $\theta_x$ is unknown, it has a distribution (or rather, to be precise, the random variable $\tilde\theta_x$ modelling its uncertainty) and an expected value, which at this point, given the information in $\mathcal{F}^n$, is $\mu_x^n$. $\endgroup$ Jul 21 '21 at 21:15
  • 1
    $\begingroup$ $E(\hat y^{n+1}|\mathcal{F}^n)$ is not $\theta_x$, because it is our expectation given $\mathcal{F}^n$, which cannot be $\theta_x$ as we don't know what that is. Only conditionally on $\tilde\theta_x=\theta_x$ it is $\theta_x$. Without this condition it is $\mu_x^n$. $\endgroup$ Jul 21 '21 at 21:20
  • 1
    $\begingroup$ The standard probability concept in Bayesian statistics is epistemic, meaning that probability models our uncertainty. Many Bayesians including de Finetti and Phil Dawid state that they don't believe there is any "true" objective probability, so there is no such thing as a "true" expectation as to be distinguished from our epistemic one. There are different opinions though and I agree that this is often presented by Bayesians in a confusing way. $\endgroup$ Jul 22 '21 at 11:07
  • 1
    $\begingroup$ Note by the way that the setup and even at times my own discussion of it seems to imply that there is a true but unknown value of $\theta_x$, but there is no postulation that this is indeed the case in reality. Rather it is a modelling device, meaning that part of the formalisation of our uncertainty is that the situation can be thought of as if there were such a value. $\endgroup$ Jul 22 '21 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.