{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Consumer Theory for Cheap Information (Discrete case)**\n",
"\n",
"This simulation illustrates some of the main results of \"Consumer theory for cheap information.\" This code simulates the standard case where information is only consumed in discrete samples. See the other file for the case of infinitely divisible samples considered in the main text of the paper."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Code setup"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"First, we need to import a number of standard Python packages:"
]
},
{
"cell_type": "code",
"execution_count": 91,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np # Basic array stuff\n",
"import scipy.optimize as optim # For finding function mins\n",
"from scipy.stats import multinomial # Multinomial probability computation\n",
"import matplotlib.pyplot as plt # Plotting\n",
"from matplotlib import rc\n",
"from tabulate import tabulate # For nicer printing of some data\n",
"import math\n",
"import time\n",
"from tqdm import tqdm # Basic progress bar"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Model"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"A decision maker must take one of finitely many actions $a\\in A$ facing an uncertain state of the world, $\\theta$, that is one of finitely many possible states. She has a state-dependent payoff function, $u(a,\\theta)$ and chooses her action to maximize expected payoff. Her prior is given by $p\\in\\Delta\\Theta$.\n",
"\n",
"Prior to acting, she may purchase information about the state and update her prior based on that information. As standard, define an information as a *Blackwell experiment*, that is, a collection of state-dependent distributions, $F(x\\ |\\ \\theta)$ over some realization space, X:\n",
"\n",
"$$ \n",
"\\mathcal{E} \\equiv \n",
"\\{X, \\langle F(\\cdot\\ |\\ \\theta)\\rangle_{\\theta\\in\\Theta}\\} \n",
"$$\n",
"\n",
"(In a fully formal treatment, the definition would also include a σ-algebra. For the purposes of this paper, we can ignore such measure-theoretic complications). \n",
"\n",
"After observing a realization from an information source, the decision maker can update with Bayes rule:\n",
"\n",
"$$ \n",
"p'_\\theta(x) = \\frac{p_\\theta f(x\\ | \\ \\theta)}\n",
" {\\sum_{\\theta'\\in\\Theta}p_{\\theta'}f(x\\ |\\ \\theta')}\n",
"$$\n",
"\n",
"To avoid trivialities, assume that no realization perfectly rules in or out any subset of the states, that is, if realization has positive probability (density) under one state, it must have positive probability under all states. (In technical terms, assume the $F(\\cdot\\ |\\ \\theta)$ are all mutually absolutely continuous so the Radon-Nikodym derivatives, $dF(\\cdot\\ |\\ \\theta')/dF(\\cdot\\ |\\ \\theta)$ all exist.) For notational simplicity in this illustration, I'll assume each state-dependent distribution has finitely many possible realizations and thus pmf given by, $f$. \n",
"\n",
"We can the define an *amount* of information by a number of conditionally independent samples from such a source. \n",
"\n",
"For illustration, consider a two-state world, $\\theta\\in\\{H,L\\}$. An information source might be a coin that is fairly waited in the $L$ state, and biased 70% to heads in the $H$ state. Then samples from this source would simply be the number of coin flips. In an experimental setting, samples would be literal samples under some experimental design.\n",
"\n",
"The DM has a collection of information sources $\\mathcal{E}_i,\\ldots, \\mathcal{E}_I$ from each of which she can purchase an arbitrary number of samples, $\\mathbf{n}=[n_i]$, at some cost $[c_i]$ each.\n",
"\n",
"The goal is to characterize the substitutability of different information sources under the normal Bayesian (ex ante) information value---that is, the expected payoff gain from acting after observing a realization from information source $\\mathcal{E}$:\n",
"\n",
"$$\n",
"V(\\mathcal{E}) \\equiv \n",
"\\sum_{x\\in R} \\max_a \n",
"\\bigg\\{\\sum_\\theta p'_\\theta(x) u(a,\\theta)\\bigg\\} \n",
"f(x) \n",
"-\n",
"\\max_a \n",
"\\bigg\\{\\sum_\\theta p_\\theta u(a,\\theta)\\bigg\\}\n",
"$$\n",
"where $f(x)\\equiv \\sum_\\theta p_\\theta f(x\\ |\\ \\theta)$ is the unconditional realization probability for the given source. (NB: the paper works with expected *loss* rather than expected value. This doesn't affect the ordinal ranking of experiments, and thus has no effect on the budget-constrained model).\n",
"\n",
"Information value is typically a very poorly behaved function, so I approach the problem with an asymptotic approach using large deviations methods.\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Throughout this notebook, I'll be working with a 3 state decision problem."
]
},
{
"cell_type": "code",
"execution_count": 92,
"metadata": {},
"outputs": [],
"source": [
"numstates = 3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Composite information sources and information values"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Defining information sources"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In order to simulate information values, we need a way to define Blackwell experiments in a way amenable to computation: Define an information source as a $|\\Theta|\\times|X|$ matrix, so each row of the matrix lists the probability of each realization in that state. I will typically use Q to denote such a matrix.\n",
"\n",
"The following code specifies a pair of information sources in this manner. By default, it will use a pair of experiments that produces relatively nice plots (including Fig. 3 in the paper). If those lines are removed/commented, the code will generate a random pair of information sources."
]
},
{
"cell_type": "code",
"execution_count": 93,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Conditional probability of each realization:\n",
"\n",
"Q1 X1 = 0 X1 = 1 X1 = 2\n",
"------- -------- -------- --------\n",
"State 0 0.07 0.18 0.75\n",
"State 1 0.45 0.19 0.36\n",
"State 2 0.45 0.05 0.5\n",
"\n",
"Q2 X2 = 0 X2 = 1\n",
"------- -------- --------\n",
"State 0 0.42 0.58\n",
"State 1 0.63 0.37\n",
"State 2 0.03 0.97\n"
]
}
],
"source": [
"Qperfect = np.eye(numstates) # Perfect information source\n",
"\n",
"## Generate random experiments for testing\n",
"def rand_source(numstates, numrealizations):\n",
" # generate a random array with the appropriate dimension\n",
" Q = np.random.rand(numstates, numrealizations)\n",
" # normalize so each row sums to 1\n",
" for state in range(numstates):\n",
" Q[state, :] = Q[state, :] / np.sum(Q[state, :])\n",
" return Q\n",
"\n",
"# Q1 = rand_source(numstates, 3)\n",
"# Q2 = rand_source(numstates, 3)\n",
"\n",
"## The following are Blackwell ordered but have interior solutions\n",
"# Q1 = np.array([[0.50, 0.50],\n",
"# [0.60, 0.40],\n",
"# [0.65, 0.35]])\n",
"# Q2 = np.array([[0.10, 0.35, 0.55],\n",
"# [0.35, 0.30, 0.35],\n",
"# [0.10, 0.05, 0.85]])\n",
"\n",
"Q1 = np.array([[0.07, 0.18, 0.75],\n",
" [0.45, 0.19, 0.36],\n",
" [0.45, 0.05, 0.50]])\n",
"Q2 = np.array([[0.42, 0.58],\n",
" [0.63, 0.37],\n",
" [0.03, 0.97]])\n",
"\n",
"# state labels. Will use for nice output tables\n",
"states = []\n",
"for stateidx in range(Q1.shape[0]):\n",
" states.append([\"State {i}\".format(i=stateidx)])\n",
"\n",
"print(\"Conditional probability of each realization:\\n\")\n",
"\n",
"# Output table and print\n",
"table1 = np.append(states, Q1, axis=1)\n",
"realizations1 = [\"Q1\"]\n",
"for realization in range(Q1.shape[1]):\n",
" realizations1.append(\"X1 =\" + \" \" + str(realization))\n",
"print(tabulate(table1, headers=realizations1)+\"\\n\")\n",
"\n",
"table2 = np.append(states, Q2, axis=1)\n",
"realizations2 = [\"Q2\"]\n",
"for realization in range(Q2.shape[1]):\n",
" realizations2.append(\"X2 =\" + \" \" + str(realization))\n",
"print(tabulate(table2, headers=realizations2))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Testing Blackwell ordering (optional)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In the above code block, there is a (commented) example of a Blackwell-ordered pair of experiments that illustrates that interior solutions may be optimal even with Blackwell-ordered experiments. Thus we will need to test if a test Blackwell dominates another by determining whether a valid garbling matrix exists. We can do this with a simple linear algebra exercise: For each column (realization) of $Q_2$, premultiply by $(Q_1^T Q_1)^{-1}Q_1^T$. This operation will find the weight matrix, $G$ such that $Q_1 G$ is as close as possible to $Q_2$. We then only need check that $G$ is a valid garbling matrix (is row stochastic) and that $Q_1 G=Q_2$."
]
},
{
"cell_type": "code",
"execution_count": 94,
"metadata": {},
"outputs": [],
"source": [
"def garbling_matrix(Q1, Q2):\n",
" numstates = Q1.shape[0]\n",
" numrealizations1 = Q1.shape[1]\n",
" numrealizations2 = Q2.shape[1]\n",
" projectionmatrix = np.linalg.inv(Q1.T @ Q1) @ Q1.T\n",
" garbling = np.empty([numrealizations1, numrealizations2])\n",
" for r2 in range(numrealizations2):\n",
" garbling[:, r2] = projectionmatrix @ Q2[:, r2]\n",
" return garbling\n",
"\n",
"def check_stochastic(garbling):\n",
" if ~np.all(garbling > -1E-6):\n",
" return False\n",
" if np.all(abs(sum(garbling.T) - 1) < 1E-6):\n",
" return True\n",
" return False\n",
"\n",
"def check_BD(Q1, Q2):\n",
" garbling = garbling_matrix(Q1, Q2)\n",
" # first check if the matrix is a valid garbling\n",
" if check_stochastic(garbling):\n",
" # then check if Q2 is the garbled Q1\n",
" if np.all(abs((Q1 @ garbling) - Q2) < 1E-6):\n",
" return True\n",
" # if either check fails, return false\n",
" return False"
]
},
{
"cell_type": "code",
"execution_count": 95,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 95,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"check_BD(Q1,Q2)"
]
},
{
"cell_type": "code",
"execution_count": 96,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 96,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"check_BD(Q2,Q1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Composite experiments"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now, we need to be able to quickly compute the matrix for composite experiments. First, we need to be able to compute the matrix for $n$ i.i.d. samples from 1 experiment. \n",
"\n",
"Since the total number of realizations of each type is a sufficient statistic for the entire vector of realizations, we can simplify things by first computing all of the partitions of $n$ with $|X|$ components (all possible realization sums), the use a multinomial distribution.\n",
"\n",
"The output matrix will be $|\\Theta|\\times$(number of ways to sum $|X|$ postive integers to add up to $n$) "
]
},
{
"cell_type": "code",
"execution_count": 97,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"conditional probabilities of each sample combination:\n",
"\n",
"3 samples of Q1 (0, 0, 3) (0, 1, 2) (0, 2, 1) (0, 3, 0) (1, 0, 2) (1, 1, 1) (1, 2, 0) (2, 0, 1) (2, 1, 0) (3, 0, 0)\n",
"----------------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- -----------\n",
"State 0 0.421875 0.30375 0.0729 0.005832 0.118125 0.0567 0.006804 0.011025 0.002646 0.000343\n",
"State 1 0.046656 0.073872 0.038988 0.006859 0.17496 0.18468 0.048735 0.2187 0.115425 0.091125\n",
"State 2 0.125 0.0375 0.00375 0.000125 0.3375 0.0675 0.003375 0.30375 0.030375 0.091125\n"
]
}
],
"source": [
"# Compute appropriate partitions (returns a generator)\n",
"def partitions(n, numrealizations):\n",
" if numrealizations == 1:\n",
" if n >= 0:\n",
" yield (n,)\n",
" return\n",
" for i in range(n+1):\n",
" for result in partitions(n-i, numrealizations-1):\n",
" yield (i,) + result\n",
"\n",
"\n",
"# Compute matrix for the n-sample source\n",
"def n_samples(Q, n):\n",
" numstates = Q.shape[0]\n",
" numrealizations = Q.shape[1]\n",
" if n == 0: # return trivial experiment if 0 samples\n",
" return np.ones((numstates, 1))\n",
" QnT = [] # transpose of Qn\n",
" for outcome in partitions(n, numrealizations):\n",
" outcomeprobs = [] # column of state-dep outcome probs\n",
" for state_idx in range(numstates):\n",
" # create a multinomial with the given outcome probs\n",
" multinom = multinomial(n, Q[state_idx, :])\n",
" outcomeprobs.append(multinom.pmf(outcome))\n",
" QnT.append(outcomeprobs)\n",
" Qn = np.array(QnT).T # convert to array and transpose\n",
" return Qn\n",
"\n",
"n = 3\n",
"realizations1 = [str(n) + \" samples of Q1\"]\n",
"for outcome in partitions(n, Q1.shape[1]):\n",
" realizations1.append(outcome)\n",
"table = np.append(states, n_samples(Q1, n), axis=1)\n",
"print(\"conditional probabilities of each sample combination:\\n\")\n",
"print(tabulate(table, headers=realizations1))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Second, we need to be able to composite two *distinct* information sources. If info source $\\mathcal{E}_1$ and $\\mathcal{E}_2$ have $|X_1|$ and $|X_2|$ possible realizations respectively, then the composite source consisting of 1 sample from each has $|X_1|\\times|X_2|$ outcomes. We can get a matrix of all possible combination probabilities by simply by listing out each element of the outer product of the rows of each matrix:"
]
},
{
"cell_type": "code",
"execution_count": 98,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Conditional probabilties of each combination of realizations from Q1 and Q2:\n",
"\n",
"1 from each X1=0, X2=0 X1=0, X2=1 X1=1, X2=0 X1=1, X2=1 X1=2, X2=0 X1=2, X2=1\n",
"------------- ------------ ------------ ------------ ------------ ------------ ------------\n",
"State 0 0.0294 0.0406 0.0756 0.1044 0.315 0.435\n",
"State 1 0.2835 0.1665 0.1197 0.0703 0.2268 0.1332\n",
"State 2 0.0135 0.4365 0.0015 0.0485 0.015 0.485\n"
]
}
],
"source": [
"def composite_source(Q1, Q2):\n",
" numstates = Q1.shape[0]\n",
" numrealizations = Q1.shape[1] * Q2.shape[1]\n",
" Qcomp = np.empty((numstates, numrealizations)) # initialize output\n",
" for state in range(numstates):\n",
" # compute all possible combination probs with an outer product\n",
" Qcomp[state, :] = \\\n",
" np.reshape(np.outer(Q1[state, :], Q2[state, :]),\n",
" (numrealizations)) # reshape to vect.\n",
" return Qcomp\n",
"\n",
"Q12comp = composite_source(Q1, Q2)\n",
"realizations12 = [\"1 from each\"]\n",
"for x1idx in range(Q1.shape[1]):\n",
" for x2idx in range(Q2.shape[1]):\n",
" realizations12.append(\"X1={i1}, X2={i2}\".format(i1=x1idx, i2=x2idx))\n",
"table = np.append(states, Q12comp, axis=1)\n",
"print(\"Conditional probabilties of each combination of realizations from Q1 and Q2:\\n\")\n",
"print(tabulate(table, headers=realizations12))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Value of information"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In order to compute information value, we must now define a state-dependent utility function and a prior belief. I'll code the utility function as a $|A|\\times|\\Theta|$ matrix of payoffs where $U_{a\\theta}=u(a,\\theta)$. The prior can simply be coded as a vector of belief probabilities."
]
},
{
"cell_type": "code",
"execution_count": 99,
"metadata": {},
"outputs": [],
"source": [
"# example payoff matrix (payoff 1 only if choose the correct state\n",
"# plus an insurance action that always gives a low payoff)\n",
"#U = np.eye(numstates)\n",
"U = np.array([[1, 0, 0],\n",
" [0, 1, 0],\n",
" [0, 0, 1]])\n",
"# example prior vector (diffuse prior)\n",
"P = np.ones(numstates) / numstates\n",
"P = np.array([0.1, 0.1, 0.8])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Information value is typically a fairly tricky thing to compute. In order to maximize computational efficiency, I vectorize the problem where possible. For a given information matrix, $Q$, and payoff matrix $U$, we can write the value *with* information as\n",
"\n",
"$$ W(Q) = \\max_D\\{\\text{tr}(QDU\\pi)\\}$$\n",
"\n",
"where $\\pi$ is a matrix who's diagonal elements are the prior probabilities and $D$ is a $|X|\\times|A|$ matrix specifying the probability of taking each action after each realization (this is a linear program: $D$ generically is all zeros and ones since each realization generically has a unique optimal response.\n",
"\n",
"([Leshno, 1992](https://cpb-us-w2.wpmucdn.com/campuspress.yale.edu/dist/3/352/files/2013/01/LeshnoSpector92.pdf) uses this formulation to provide an elementary proof of Blackwell's theorem for the finite-action/finite-state case.)\n",
"\n",
"The value *of* information would then be $W(Q)$ minus the payoff from acting with no information. Such a subtraction is a monotone transformation, so it won't affect the ordinal properties I'm interested in.\n",
"\n",
"In order to evaluate how close a bundle is to perfect information, I will sometimes use the ratio of the full-information gap (the difference between the value of a perfect signal and $W(Q)$, relative to the full-info value. This will be a percentage that approaches zero as the amount of samples increases.\n",
"\n",
"None of these approximation would be particularly useful if they require so many samples as to be indistinguishable from a perfect source anyways. I will use thus use this relative info-gap as an ad hoc measure how useful the approximation is. That is, the relavent approximations are useful if they are accurate, even when the relative info-gap is large."
]
},
{
"cell_type": "code",
"execution_count": 100,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"-------------------------------------------- ---\n",
"Expected value of acting after observing Q1: 0.8\n",
"-------------------------------------------- ---\n"
]
}
],
"source": [
"def info_value(Q, U, P):\n",
" numrealizations = Q.shape[1]\n",
"\n",
" Upi = U @ np.diag(P)\n",
" # compute (actions x realizations) matrix of payoff of each action\n",
" # times unconditional prob of each realization\n",
" Ua = Upi @ Q\n",
" # choose best action for each message then sum across messages\n",
" valuewithinfo = sum(np.amax(Ua, axis=0))\n",
" return valuewithinfo\n",
"\n",
"# value of info for the examples above\n",
"print(tabulate([[\"Expected value of acting after observing Q1:\",\n",
" info_value(Q1, U, P)]]))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Generalized precision"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In the paper, I show that two information sources are exchangeable with ratios of respective precision-like indices of each experiment. In order to define this precision, we must first take a brief detour into large deviations theory.\n",
"\n",
"I approach the problem of approximating information values by approximating the probability of a \"mistake\" (taking a suboptimal action in a given state). The normal form of Bayes's rule is a bit messy, so instead of working with probabilities, I work with log-likelihood ratios, where Bayes rule becomes a sum:\n",
"\n",
"$$\n",
"\\log\\bigg(\\frac{p'_{\\theta}(x)}{p'_{\\theta'}(x)}\\bigg) = \n",
"\\log\\bigg(\\frac{p_\\theta}{p_{\\theta'}}\\bigg) +\n",
"\\log\\bigg(\\frac{f(x\\ |\\ \\theta)}{f(x\\ |\\ \\theta)}\\bigg)\n",
"$$\n",
"\n",
"And, of course, we have no shortage of asymptotic results for approximating sums of many independent distributions.\n",
"\n",
"For a given pair of states, define the (Chernoff, 1952, Moscarini and Smith, 2002) efficiency index of an experiment, $\\mathcal{E}_1$, as the minimized value of the moment generating function (MGF) of the distribitution of log-likelihood ratios (LLR):\n",
"\n",
"$$\n",
"\\rho_1 \\equiv \\min_t \\sum_r f(x\\ |\\ \\theta)^t f(x\\ |\\ \\theta')^{1-t}\n",
"$$\n",
"(Define $\\tau_1$ as the minimizer)\n",
"\n",
"Note that the expected value of the distribution of the above mgf is the negative Kullback-Leibler divergence, $-D(F(\\cdot\\ |\\ \\theta')\\ ||\\ F(\\cdot\\ |\\ \\theta))$.\n",
"\n",
"Note that because MGF of an indpendent sum is the product of MGFs, we have that $n$ samples from $\\mathcal{E}_1$ will have efficiency index $\\rho_1^n$.\n",
"\n",
"Furthermore, because the minimum of a sum will be bigger than the sum of minima, we have that the efficiency index of a composite is more than the sum of its parts:\n",
"\n",
"$$\n",
"\\rho_{12} \\geq \\rho_1\\rho_2\n",
"$$\n",
"\n",
"Now, we can define the *precision* for a given state pair of a test by $\\beta \\equiv -\\log(\\rho)$. I call this a precision because, for Gaussian tests, it is, up to a multiplicative constant, the same as classical precision ($1/\\sigma^2$).\n",
"\n",
"This measure has a number of properties that you might expect for something called a precision\n",
"\n",
"1. For any non-trivial experiment, $\\beta>0$\n",
"2. $n$ samples from the same experiment has precision $n\\beta$\n",
"3. Blackwell dominant experiments have higher precision\n",
"\n",
"Intuitively, you can think of the precision as measuring how well an information source can distinguish between a given pair of states. \n",
"\n",
"Because the efficiency index of a composite is weakly higher than the product of the individual efficiency index, a composite experiment is weakly less precise, for a given state, than the sum of it's parts."
]
},
{
"cell_type": "code",
"execution_count": 101,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" Q1 Precisions Q2 Precisions\n",
"--------------- ---------------\n",
" 0.126584 0.0226118\n",
" 0.123481 0.152613\n",
" 0.0288732 0.314685\n"
]
}
],
"source": [
"# the following returns the full list of state-pair precisions\n",
"# and their respective MGF minimizers\n",
"def precisions(Q):\n",
" numstates = Q.shape[0]\n",
" # compute the efficiency index for each state\n",
" betalist = []\n",
" taulist = []\n",
" for state1 in range(numstates):\n",
" for state2 in range(state1+1, numstates):\n",
" # Define the llr mgf for the given dichtomy\n",
" def llrmgf(t):\n",
" Qstate1t = Q[state1, :]**t\n",
" Qstate2t = Q[state2, :]**(1-t)\n",
" return np.sum(Qstate1t * Qstate2t)\n",
" # Compute index for the dichotomy\n",
" optimizer = optim.minimize(llrmgf, 0.5)\n",
" rho = optimizer.fun\n",
" tau = optimizer.x[0]\n",
" betalist.append(-np.log(rho))\n",
" taulist.append(tau)\n",
" return betalist, taulist\n",
"\n",
"\n",
"beta1list, tau1list = precisions(Q1)\n",
"beta2list, tau2list = precisions(Q2)\n",
"table = np.array([beta1list, beta2list]).T\n",
"print(tabulate(table,\n",
" headers=[\"Q1 Precisions\", \"Q2 Precisions\"]))"
]
},
{
"cell_type": "code",
"execution_count": 102,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" composite precisions sum of parts\n",
"---------------------- --------------\n",
" 0.149026 0.149196\n",
" 0.271436 0.276094\n",
" 0.343391 0.343559\n"
]
}
],
"source": [
"beta12list, tau12list = precisions(composite_source(Q1, Q2))\n",
"table = np.array([beta12list, [beta1list[i]+beta2list[i]\n",
" for i in range(len(beta1list))]]).T\n",
"print(tabulate(table,\n",
" headers=[\"composite precisions\", \"sum of parts\"]))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now, it might seem that composites are always worse than the sum of their parts since, for any state pair, the composite is always less precise than the sum of its parts. But the value of information depends generically on a source's ability to distinguish any state from any other. Moscarini and Smith, 2002, showed that, for large sample sizes, the only state pair that matters is the pair hardest to tell apart---i.e. the pair with the least precision (highest efficiency index).\n",
"\n",
"Thus complementarity often arises when experiments differ in the pair of states they most struggle to distinguish:"
]
},
{
"cell_type": "code",
"execution_count": 103,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" min precision of composite sum of min precisions\n",
"---------------------------- -----------------------\n",
" 0.149026 0.051485\n"
]
}
],
"source": [
"beta1 = min(beta1list)\n",
"beta2 = min(beta2list)\n",
"beta12 = min(beta12list)\n",
"print(tabulate([[beta12, beta1+beta2]],\n",
" headers=[\"min precision of composite\", \"sum of min precisions\"]))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In this particular example, we can see that it is in fact the composite has a *higher* least precision than the sum of least precisions of its parts."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Plotting iso-precision curves"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In particular, at large samples, two bundles of samples will perform equally if they have equal least precision. Information value at large samples is ordinally equivalent to\n",
"\n",
"$$\n",
"v(n_1, n_2) \\simeq (n_1+n_2)\\beta_r\n",
"$$\n",
"\n",
"where $\\beta_r$ is the least precision dichotomy for a bundle that is composed of $r$ fraction of samples from $n_1$. Furthermore, we can breakdown $\\beta_r$ into *marginal* precisions:\n",
"\n",
"$$\n",
"\\beta_\\omega=\\omega\\beta_{r 1}+(1-r)\\beta_{r 2}\n",
"$$\n",
"\n",
"where $\\beta_{r i}$ is $-\\log M_i(\\tau_r)$, is the negative log of the LLR MGF for the composite's worst-case state pair, evaluated at the composite's minimizer. We can then write\n",
"\n",
"$$\n",
"v(n_1, n_2) \\simeq n_1\\beta_{r 1} + n_2\\beta_{r 2}\n",
"$$\n",
"\n",
"Heuristically, then it seems like the MRS between two samples at any bundle with $r$ fraction from $\\mathcal{E}_1$ must then be the ratio of the component precisions. (For small subsititutions, relative to total sample size, the fraction of samples from each source doesn't change much, so the component precisions don't change much.) \n",
"\n",
"Additionally, since the value is a min of a sums of precisions, there will be kinks when the least-precision state pair changes.\n",
"\n",
"Of course, in a discrete sample setting there is no true MRS. In the appendix of the paper, I discuss a the formal notion of subsitutability in this setting, but for the purpose of interpretting things here, it works well enough to just pretend samples are divisible."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"First, note that the component precisions only depend on the fraction of the bundle from each source (info value is homothetic). First, I compute the component precision for a given composite factor $r$."
]
},
{
"cell_type": "code",
"execution_count": 104,
"metadata": {},
"outputs": [],
"source": [
"# return the component precision for each test at the r composite factor\n",
"def comp_precision(Q1, Q2, r):\n",
" numstates = Q1.shape[0]\n",
" # loop over pairs of states\n",
" rho = 0\n",
" for state1 in range(numstates):\n",
" for state2 in range(state1+1, numstates):\n",
" # Define the Hellinger transform for the given dichotomy\n",
" def llrmgf1(t):\n",
" Q1state1_t = Q1[state1, :]**t\n",
" Q1state2_t = Q1[state2, :]**(1-t)\n",
" return np.sum(Q1state1_t*Q1state2_t)\n",
"\n",
" def llrmgf2(t):\n",
" Q2state1_t = Q2[state1, :]**t\n",
" Q2state2_t = Q2[state2, :]**(1-t)\n",
" return np.sum(Q2state1_t*Q2state2_t)\n",
"\n",
" def compllrmgf(t):\n",
" return llrmgf1(t)**r * llrmgf2(t)**(1-r)\n",
"\n",
" # Compute efficiency index for the dichotomy\n",
" optimizer = optim.minimize(compllrmgf, 0.5)\n",
" rhopair = optimizer.fun\n",
" taupair = optimizer.x\n",
" # if new rho is worse (higher), store it\n",
" if rho < rhopair:\n",
" rho = rhopair\n",
" # Store component rhos\n",
" rho1w = llrmgf1(taupair)\n",
" rho2w = llrmgf2(taupair)\n",
" return -np.log(rho1w), -np.log(rho2w)\n",
"\n",
"# return total precision for a composite factor w\n",
"def total_precision(Q1, Q2, w):\n",
" beta1w, beta2w = comp_precision(Q1, Q2, w)\n",
" return w*beta1w + (1-w)*beta2w"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can then compute points on an indifference curve by using the differential equation defined by the MRS ($dn_2/dn_1$) of iso-precision (note that precision can be computed at \"fractional\" samples even for discretely sampled experiments):\n",
"\n",
"$$\n",
"\\text{MRS}(r) = \\frac{\\beta_{r 1}}{\\beta_{r 2}}\n",
"$$\n",
"\n",
"In all the plots that follow, the reference point is the lower right corner bundle consisting entirely of samples from $\\mathcal{E}_1$."
]
},
{
"cell_type": "code",
"execution_count": 105,
"metadata": {},
"outputs": [],
"source": [
"def iso_least_precision(n1, n2, Q1, Q2):\n",
" # step size for sample proportions.\n",
" dr = 0.0005\n",
" rstart = n1 / (n1+n2)\n",
" startprecision = (n1+n2) * total_precision(Q1, Q2, rstart) # precision of initial sample bundle\n",
" # find iso-precision point on each ray from the origin\n",
" n1points, n2points = [], []\n",
" for r in np.arange(0,1+dr,dr):\n",
" N = startprecision / total_precision(Q1, Q2, r)\n",
" n1points.append(r*N)\n",
" n2points.append((1-r)*N)\n",
" return n1points, n2points\n",
"\n",
"n1starts = [40, 100, 160]\n",
"#n1starts = [15]\n",
"n1pointsapprox, n2pointsapprox = [], []\n",
"for n1 in n1starts:\n",
" #n1points, n2points = mrs_approx(n1start)\n",
" n1points, n2points = iso_least_precision(n1, 0, Q1, Q2)\n",
" n1pointsapprox.append(n1points)\n",
" n2pointsapprox.append(n2points)"
]
},
{
"cell_type": "code",
"execution_count": 106,
"metadata": {},
"outputs": [
{
"data": {
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