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I'm stuck in a problem where I have distribution distribution $P(\boldsymbol{x})$, from which I know how to sample from (i.i.d.) and two functions of the random variable $\boldsymbol{x}$: $E(\boldsymbol{x})$ and $A(\boldsymbol{x})$ (both uninvertible). The aim is to compute the conditional mean of $A$ given $E$, $\mathbb{E}[A|E]$.

The question is how to do it.

Specifically, I'm writing that

$$P(A|E) = \int P(\boldsymbol{x},A|E)d\boldsymbol{x} = \int P(A|\boldsymbol{x},E)P(\boldsymbol{x}|E)d\boldsymbol{x} = \frac{1}{P(E)}\int P(A|\boldsymbol{x},E)P(E|\boldsymbol{x})P(\boldsymbol{x}) d\boldsymbol{x}$$

I then use $P(A|\boldsymbol{x}) = \delta(A-A(\boldsymbol{x}))$ ($\delta$ is the Dirac-delta) to write

$$\mathbb{E}[A|E] = \int A P(A|E) dA = \frac{1}{P(E)}\int \int A \delta(A-A(\boldsymbol{x}) P(E|\boldsymbol{x})P(\boldsymbol{x}) d\boldsymbol{x} dA $$ Integrating the $\delta(A-A(\boldsymbol{x}))$ out, and using $P(E|\boldsymbol{x}) = \delta(E-E(\boldsymbol{x}))$ leads to

$$\mathbb{E}[A|E] = \frac{1}{P(E)}\int A(\boldsymbol{x}) \delta(E - E(\boldsymbol{x})) P(\boldsymbol{x}) d\boldsymbol{x} \equiv \frac{I(E)}{P(E)}$$

Both $P(E)$ and $I(E)$ are consistently estimated (denoted as $\overline{P}$ and $\overline{I}$) by the calculation of a multidimensional integral using Monte Carlo. However:

  1. is $\overline{I}/\overline{P}$ a consistent estimator of $\mathbb{E}[A|E]$?

  2. How do I compute the error estimate of this estimator? $\overline{I}$ is not independent of $\overline{P}$...

  3. What is the error estimate of this estimator if $\boldsymbol{x}$ are now correlated?

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  • $\begingroup$ I don't see how $P(E|x)$ can possibly equal $\delta(E - E(x))$, as what happens if $\delta > 0$ and $E < E(x)$ or the reverse? $\endgroup$ – jbowman Jan 4 '16 at 21:51
  • $\begingroup$ I forgot to mention that $\delta$ is the Dirac delta, $\int P(E|x) dE = 1$. $E < E(x)$ implies $\delta(E - E(x)) = 0$. Does this answer you question @jbowman? $\endgroup$ – Jorge Leitao Jan 4 '16 at 21:55
  • $\begingroup$ Yes it does. In that case, is $E$ invertible, as in can you calculate $x | E(x)$? If so, then given $E$ you can calculate $x$ then $A(x)$. I assume not, but just checking... $\endgroup$ – jbowman Jan 4 '16 at 22:11
  • $\begingroup$ yap, $E$ is not invertible. I incorporated both comments in the question. Thanks! $\endgroup$ – Jorge Leitao Jan 4 '16 at 22:15
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    $\begingroup$ Why is your first formula $P(A|E)$ correct? When you marginalize out a variable, you either integrate joint only, or integrate a conditional (of that variable) times prior, but not the joint times prior. $\endgroup$ – Artem Sobolev Jan 4 '16 at 23:04
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I do not know what functions you are using (I am assuming that they are not stochastic here) but this is the way I would deal with the problem:

  1. Instead of defining $P(E|X)$ as a Dirac Delta, define it as a discrete probability distribution such that $$ P(E=E(x)|x)=\begin{cases} 1\;\;\mathrm{if}\;\;E=E(x) \\ 0\;\;\mathrm{otherwise} \end{cases} $$

This makes sense since because $E(x)$ is a function you know the value of $E$ with 100% certainty once you know $x$. As suggested in your comment (A), you can express the above distribution as the Kronecker delta; $$ \delta_{E,E(x)} = P(E=E(x)|x) $$ 2. Using Bayes Theorem... $$ P(X|E) = \frac{P(E=E(x)|x)P(X)}{P(E)} = \delta_{E,E(x)} \cdot \frac{P(X)}{P(E)} $$ The key here is that the support for $P(X|E)$ is a subset of the support for $P(X)$. i.e. Let $x \in \mathcal{X}$ and $x|E \in \mathcal{X_E}$ where $\mathcal{X_E} \subseteq \mathcal{X}$. Note that for this particular problem the Kronecker Delta $\delta_{x \in \mathcal{X_E}}$ is equivalent to $\delta_{E,E(x)}$.*

You can then define

$$ P(E) =\begin{cases} \sum_{x \in \mathcal{X_E}} P(x)\;\;\mathrm{if}\;\;\mathcal{X_E}\;\mathrm{is\;countable}\\ \;\\ \int_{\mathcal{X_E}} P(x)dx\;\;\mathrm{Otherwise} \end{cases} $$

With this notation you can also redefine your conditional expectation, $$ \mathbb{E}[A|E] = \mathbb{E}[A|x \in \mathcal{X_E}] $$

Once you know $E$ you should be able to define the set $\mathcal{X_E}$. In the trivial case that $E(x)$ was invertible, $\mathcal{X_E}$ would only contain a single number. Sampling from $P(X|E)$ can be accomplished with Monte-Carlo, Metropolis-Hastings, or direct calculation depending upon the nature of $\mathcal{X_E}$.

You can then write the conditional expectation as $$ \mathbb {E}[A|x \in \mathcal{X_E}] =\begin{cases} \sum_{x \in \mathcal{X_E}} A(x) \frac{P(X)}{P(E)}\;\; \mathrm{if} \;\; \mathcal{X_E}\;\mathrm{is\;countable} \\ \; \\ \int_{\mathcal{X_E}} A(x) \frac{P(X)}{P(E)}dx \;\; \mathrm{otherwise} \end{cases} $$ and under some regularity conditions this would be a consistent estimator.

To address your concerns in comment (B), it may very well be more computationally efficient to simulate a joint distribution $(A(x_i),E(x_i))$ if you want to estimate the conditional value of $A$ in respect to many $E^*$. In such a setting you could either average $A(x_i)$ over a weighted sample (with larger weights assigned to values closer to $E^*$), or average $A$ over a smaller sub-sample interval around $E^*$. Whether or not this technique is efficient in the statistical sense depends a lot on the functions you are using. I think it would be a consistent estimator....but I am not entirely sure.

I guess part of the problem is that I do not know what functions you are using. If, for example, $E(x) = x^2$ (totally trivial I know...) then it would be easy to implement the technique I discussed above. With very complicated functions I could see how it would be a total pain, especially if you had to implement an MCMC for each individual $E^*$.

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  • $\begingroup$ A) This derivation is in spirit equivalent to the above for discrete $E$. In step 1, shouldn't it be that $P(X=x|E) = \delta_{x \in \chi_E} P(X=x)/P(E)$ where $\delta$ is the Kronecker-delta? Otherwise, for a given $E$, when $x \notin \chi_E$, $P(X=x|E) = P(x)/P(E) \neq 0$, while it should be $P(X=x|E) = 0$ (another way to see is that marginalising $x$ (over $\chi$), does not give 1. The Kronecker-delta enforces that the marginal is only over $\chi_E$. When $E$ is continuous, the Kronecker-delta is replaced by the Dirac-delta because $P(E=E(x)|x)=\delta(E-E(x))$ fulfils $\int P(E|x) dE = 1$. $\endgroup$ – Jorge Leitao Jan 5 '16 at 6:45
  • $\begingroup$ B) that in practice I don't need to know $P(E)$ is true if I want to estimate the mean for a fixed $E^*$ only. In that case you are right: I only accept states inside $\chi_{E^*}$. However, I want to estimate the mean as a function of $E$, and it would be inefficient to run a MC simulation for every value of $E$: even sampling $x$ from $P(x)$ already gives pairs $(E(x_i),A(x_i))$ that can be used to estimate the mean of A for each value of $E$. I'm using MH with a weight (e.g. $e^{-\beta E}$) to favor some regions of $E$, but this will not change the fact that I have a ratio of two estimates. $\endgroup$ – Jorge Leitao Jan 5 '16 at 7:01
  • $\begingroup$ (A) I was lazy with my notation. The answer is yes but in many disciplines, the fact that $P(X|E) =0$ for $x \not\in \mathcal{X_E}$ is implied. I changed my notation in the answer and hopefully that helps. (B) See edits, I sympathize $\endgroup$ – Zachary Blumenfeld Jan 5 '16 at 10:03
  • $\begingroup$ Thank you for the comment and the edit. "...but I am not entirely sure." My exact feeling. :-) Yes, $E$ in my case is a complicated function: e.g. it has a zero-measure set of minima whose dimension is non-integer (a fractal set). If you are curious, these functions appear in chaotic systems, systems that I've been sampling, e.g. arxiv.org/abs/1302.4672 $\endgroup$ – Jorge Leitao Jan 5 '16 at 10:27
  • $\begingroup$ @J.C.Leitão I have no experience with chaotic systems. I would say that if you could easily find the set $\mathcal{X_E}$ then the method I proposed would be more efficient. Using plain MC on the joint will estimate the conditional mean with values that fall outside of $\mathcal{X_E}$ which could potentially lead to some inefficiencies. Informally, if you only sampled from an epsilon ball around $E^*$, as the ball became arbitrarily small, your sample would converge on $P(A|E^*)$, so you may be able to establish consistency with that sort of logic. $\endgroup$ – Zachary Blumenfeld Jan 5 '16 at 22:58

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