(and that Abraham Wald solved by inventing sequential analysis)
This lecture describes a statistical decision problem presented to Milton Friedman and W. Allen Wallis during World War II when they were analysts at the U.S. Government’s Statistical Research Group at Columbia University.
This problem led Abraham Wald [Wald, 1947] to formulate sequential analysis, an approach to statistical decision problems intimately related to dynamic programming.
In this lecture, we apply dynamic programming algorithms to Friedman and Wallis and Wald’s problem.
Key ideas in play will be:
We’ll begin with some imports:
import numpy as np
import matplotlib.pyplot as plt
from numba import jit, prange, float64, int64
from numba.experimental import jitclass
from math import gamma
This lecture uses ideas studied in this lecture, this lecture. and this lecture.
On pages 137-139 of his 1998 book Two Lucky People with Rose Friedman [Friedman and Friedman, 1998], Milton Friedman described a problem presented to him and Allen Wallis during World War II, when they worked at the US Government’s Statistical Research Group at Columbia University.
Note
See pages 25 and 26 of Allen Wallis’s 1980 article [Wallis, 1980] about the Statistical Research Group at Columbia University during World War II for his account of the episode and for important contributions that Harold Hotelling made to formulating the problem. Also see chapter 5 of Jennifer Burns book about
Milton Friedman [Burns, 2023].
Let’s listen to Milton Friedman tell us what happened
In order to understand the story, it is necessary to have an idea of a
simple statistical problem, and of the standard procedure for dealing with it. The actual problem out of which sequential analysis grew will serve. The Navy has two alternative designs (say A and B) for a projectile. It wants to determine which is superior. To do so it undertakes a series of paired firings. On each round, it assigns the value 1 or 0 to A accordingly as its performance is superior or inferior to that of B and conversely 0 or 1 to B. The Navy asks the statistician how to conduct the test and how to analyze the results.
The standard statistical answer was to specify a number of firings (say
1,000) and a pair of percentages (e.g., 53% and 47%) and tell the client that if A receives a 1 in more than 53% of the firings, it can be regarded as superior; if it receives a 1 in fewer than 47%, B can be regarded as superior; if the percentage is between 47% and 53%, neither can be so regarded.
When Allen Wallis was discussing such a problem with (Navy) Captain
Garret L. Schyler, the captain objected that such a test, to quote from Allen’s account, may prove wasteful. If a wise and seasoned ordnance officer like Schyler were on the premises, he would see after the first few thousand or even few hundred [rounds] that the experiment need not be completed either because the new method is obviously inferior or because it is obviously superior beyond what was hoped for $ \ldots $.
Friedman and Wallis struggled with the problem but, after realizing that they were not able to solve it, described the problem to Abraham Wald.
That started Wald on the path that led him to Sequential Analysis [Wald, 1947].
We’ll formulate the problem using dynamic programming.
The following presentation of the problem closely follows Dmitri Berskekas’s treatment in Dynamic Programming and Stochastic Control [Bertsekas, 1975].
A decision-maker can observe a sequence of draws of a random variable $ z $.
He (or she) wants to know which of two probability distributions $ f_0 $ or $ f_1 $ governs $ z $.
Conditional on knowing that successive observations are drawn from distribution $ f_0 $, the sequence of random variables is independently and identically distributed (IID).
Conditional on knowing that successive observations are drawn from distribution $ f_1 $, the sequence of random variables is also independently and identically distributed (IID).
But the observer does not know which of the two distributions generated the sequence.
For reasons explained in Exchangeability and Bayesian Updating, this means that the sequence is not IID.
The observer has something to learn, namely, whether the observations are drawn from $ f_0 $ or from $ f_1 $.
The decision maker wants to decide which of the two distributions is generating outcomes.
We adopt a Bayesian formulation.
The decision maker begins with a prior probability
$$ \pi_{-1} = \mathbb P \{ f = f_0 \mid \textrm{ no observations} \} \in (0, 1) $$After observing $ k+1 $ observations $ z_k, z_{k-1}, \ldots, z_0 $, he updates his personal probability that the observations are described by distribution $ f_0 $ to
$$ \pi_k = \mathbb P \{ f = f_0 \mid z_k, z_{k-1}, \ldots, z_0 \} $$which is calculated recursively by applying Bayes’ law:
$$ \pi_{k+1} = \frac{ \pi_k f_0(z_{k+1})}{ \pi_k f_0(z_{k+1}) + (1-\pi_k) f_1 (z_{k+1}) }, \quad k = -1, 0, 1, \ldots $$After observing $ z_k, z_{k-1}, \ldots, z_0 $, the decision-maker believes that $ z_{k+1} $ has probability distribution
$$ f_{{\pi}_k} (v) = \pi_k f_0(v) + (1-\pi_k) f_1 (v) , $$which is a mixture of distributions $ f_0 $ and $ f_1 $, with the weight on $ f_0 $ being the posterior probability that $ f = f_0 $ [1].
To illustrate such a distribution, let’s inspect some mixtures of beta distributions.
The density of a beta probability distribution with parameters $ a $ and $ b $ is
$$ f(z; a, b) = \frac{\Gamma(a+b) z^{a-1} (1-z)^{b-1}}{\Gamma(a) \Gamma(b)} \quad \text{where} \quad \Gamma(t) := \int_{0}^{\infty} x^{t-1} e^{-x} dx $$The next figure shows two beta distributions in the top panel.
The bottom panel presents mixtures of these distributions, with various mixing probabilities $ \pi_k $
@jit
def p(x, a, b):
r = gamma(a + b) / (gamma(a) * gamma(b))
return r * x**(a-1) * (1 - x)**(b-1)
f0 = lambda x: p(x, 1, 1)
f1 = lambda x: p(x, 9, 9)
grid = np.linspace(0, 1, 50)
fig, axes = plt.subplots(2, figsize=(10, 8))
axes[0].set_title("Original Distributions")
axes[0].plot(grid, f0(grid), lw=2, label="$f_0$")
axes[0].plot(grid, f1(grid), lw=2, label="$f_1$")
axes[1].set_title("Mixtures")
for π in 0.25, 0.5, 0.75:
y = π * f0(grid) + (1 - π) * f1(grid)
axes[1].plot(y, lw=2, label=f"$\pi_k$ = {π}")
for ax in axes:
ax.legend()
ax.set(xlabel="$z$ values", ylabel="probability of $z_k$")
plt.tight_layout()
plt.show()
After observing $ z_k, z_{k-1}, \ldots, z_0 $, the decision-maker chooses among three distinct actions:
Associated with these three actions, the decision-maker can suffer three kinds of losses:
If we regard $ f=f_0 $ as a null hypothesis and $ f=f_1 $ as an alternative hypothesis, then $ L_1 $ and $ L_0 $ are losses associated with two types of statistical errors
So when we treat $ f=f_0 $ as the null hypothesis
Before proceeding, let’s try to guess what an optimal decision rule might look like.
Suppose at some given point in time that $ \pi $ is close to 1.
Then our prior beliefs and the evidence so far point strongly to $ f = f_0 $.
If, on the other hand, $ \pi $ is close to 0, then $ f = f_1 $ is strongly favored.
Finally, if $ \pi $ is in the middle of the interval $ [0, 1] $, then we are confronted with more uncertainty.
This reasoning suggests a decision rule such as the one shown in the figure
As we’ll see, this is indeed the correct form of the decision rule.
Our problem is to determine threshold values $ \alpha, \beta $ that somehow depend on the parameters described above.
You might like to pause at this point and try to predict the impact of a parameter such as $ c $ or $ L_0 $ on $ \alpha $ or $ \beta $.
Let $ J(\pi) $ be the total loss for a decision-maker with current belief $ \pi $ who chooses optimally.
With some thought, you will agree that $ J $ should satisfy the Bellman equation
$$ J(\pi) = \min \left\{ (1-\pi) L_0, \; \pi L_1, \; c + \mathbb E [ J (\pi') ] \right\} \tag{51.1} $$
where $ \pi' $ is the random variable defined by Bayes’ Law
$$ \pi' = \kappa(z', \pi) = \frac{ \pi f_0(z')}{ \pi f_0(z') + (1-\pi) f_1 (z') } $$when $ \pi $ is fixed and $ z' $ is drawn from the current best guess, which is the distribution $ f $ defined by
$$ f_{\pi}(v) = \pi f_0(v) + (1-\pi) f_1 (v) $$In the Bellman equation, minimization is over three actions:
We can represent the Bellman equation as
$$ J(\pi) = \min \left\{ (1-\pi) L_0, \; \pi L_1, \; h(\pi) \right\} \tag{51.2} $$
where $ \pi \in [0,1] $ and
The optimal decision rule is characterized by two numbers $ \alpha, \beta \in (0,1) \times (0,1) $ that satisfy
$$ (1- \pi) L_0 < \min \{ \pi L_1, c + \mathbb E [J(\pi')] \} \textrm { if } \pi \geq \alpha $$and
$$ \pi L_1 < \min \{ (1-\pi) L_0, c + \mathbb E [J(\pi')] \} \textrm { if } \pi \leq \beta $$The optimal decision rule is then
$$ \begin{aligned} \textrm { accept } f=f_0 \textrm{ if } \pi \geq \alpha \\ \textrm { accept } f=f_1 \textrm{ if } \pi \leq \beta \\ \textrm { draw another } z \textrm{ if } \beta \leq \pi \leq \alpha \end{aligned} $$Our aim is to compute the cost function $ J $, and from it the associated cutoffs $ \alpha $ and $ \beta $.
To make our computations manageable, using (51.2), we can write the continuation cost $ h(\pi) $ as
$$ \begin{aligned} h(\pi) &= c + \mathbb E [J(\pi')] \\ &= c + \mathbb E_{\pi'} \min \{ (1 - \pi') L_0, \pi' L_1, h(\pi') \} \\ &= c + \int \min \{ (1 - \kappa(z', \pi) ) L_0, \kappa(z', \pi) L_1, h(\kappa(z', \pi) ) \} f_\pi (z') dz' \end{aligned} \tag{51.3} $$
The equality
$$ h(\pi) = c + \int \min \{ (1 - \kappa(z', \pi) ) L_0, \kappa(z', \pi) L_1, h(\kappa(z', \pi) ) \} f_\pi (z') dz' \tag{51.4} $$
is a functional equation in an unknown function $ h $.
Using the functional equation, (51.4), for the continuation cost, we can back out optimal choices using the right side of (51.2).
This functional equation can be solved by taking an initial guess and iterating to find a fixed point.
Thus, we iterate with an operator $ Q $, where
$$ Q h(\pi) = c + \int \min \{ (1 - \kappa(z', \pi) ) L_0, \kappa(z', \pi) L_1, h(\kappa(z', \pi) ) \} f_\pi (z') dz' $$First, we will construct a jitclass
to store the parameters of the model
wf_data = [('a0', float64), # Parameters of beta distributions
('b0', float64),
('a1', float64),
('b1', float64),
('c', float64), # Cost of another draw
('π_grid_size', int64),
('L0', float64), # Cost of selecting f0 when f1 is true
('L1', float64), # Cost of selecting f1 when f0 is true
('π_grid', float64[:]),
('mc_size', int64),
('z0', float64[:]),
('z1', float64[:])]
@jitclass(wf_data)
class WaldFriedman:
def __init__(self,
c=1.25,
a0=1,
b0=1,
a1=3,
b1=1.2,
L0=25,
L1=25,
π_grid_size=200,
mc_size=1000):
self.a0, self.b0 = a0, b0
self.a1, self.b1 = a1, b1
self.c, self.π_grid_size = c, π_grid_size
self.L0, self.L1 = L0, L1
self.π_grid = np.linspace(0, 1, π_grid_size)
self.mc_size = mc_size
self.z0 = np.random.beta(a0, b0, mc_size)
self.z1 = np.random.beta(a1, b1, mc_size)
def f0(self, x):
return p(x, self.a0, self.b0)
def f1(self, x):
return p(x, self.a1, self.b1)
def f0_rvs(self):
return np.random.beta(self.a0, self.b0)
def f1_rvs(self):
return np.random.beta(self.a1, self.b1)
def κ(self, z, π):
"""
Updates π using Bayes' rule and the current observation z
"""
f0, f1 = self.f0, self.f1
π_f0, π_f1 = π * f0(z), (1 - π) * f1(z)
π_new = π_f0 / (π_f0 + π_f1)
return π_new
As in the optimal growth lecture, to approximate a continuous value function
We define the operator function Q
below.
@jit(nopython=True, parallel=True)
def Q(h, wf):
c, π_grid = wf.c, wf.π_grid
L0, L1 = wf.L0, wf.L1
z0, z1 = wf.z0, wf.z1
mc_size = wf.mc_size
κ = wf.κ
h_new = np.empty_like(π_grid)
h_func = lambda p: np.interp(p, π_grid, h)
for i in prange(len(π_grid)):
π = π_grid[i]
# Find the expected value of J by integrating over z
integral_f0, integral_f1 = 0, 0
for m in range(mc_size):
π_0 = κ(z0[m], π) # Draw z from f0 and update π
integral_f0 += min((1 - π_0) * L0, π_0 * L1, h_func(π_0))
π_1 = κ(z1[m], π) # Draw z from f1 and update π
integral_f1 += min((1 - π_1) * L0, π_1 * L1, h_func(π_1))
integral = (π * integral_f0 + (1 - π) * integral_f1) / mc_size
h_new[i] = c + integral
return h_new
To solve the key functional equation, we will iterate using Q
to find the fixed point
@jit
def solve_model(wf, tol=1e-4, max_iter=1000):
"""
Compute the continuation cost function
* wf is an instance of WaldFriedman
"""
# Set up loop
h = np.zeros(len(wf.π_grid))
i = 0
error = tol + 1
while i < max_iter and error > tol:
h_new = Q(h, wf)
error = np.max(np.abs(h - h_new))
i += 1
h = h_new
if error > tol:
print("Failed to converge!")
return h_new
Let’s inspect outcomes.
We will be using the default parameterization with distributions like so
wf = WaldFriedman()
fig, ax = plt.subplots(figsize=(10, 6))
ax.plot(wf.f0(wf.π_grid), label="$f_0$")
ax.plot(wf.f1(wf.π_grid), label="$f_1$")
ax.set(ylabel="probability of $z_k$", xlabel="$z_k$", title="Distributions")
ax.legend()
plt.show()
To solve the model, we will call our solve_model
function
h_star = solve_model(wf) # Solve the model
We will also set up a function to compute the cutoffs $ \alpha $ and $ \beta $ and plot these on our cost function plot
@jit
def find_cutoff_rule(wf, h):
"""
This function takes a continuation cost function and returns the
corresponding cutoffs of where you transition between continuing and
choosing a specific model
"""
π_grid = wf.π_grid
L0, L1 = wf.L0, wf.L1
# Evaluate cost at all points on grid for choosing a model
payoff_f0 = (1 - π_grid) * L0
payoff_f1 = π_grid * L1
# The cutoff points can be found by differencing these costs with
# The Bellman equation (J is always less than or equal to p_c_i)
β = π_grid[np.searchsorted(
payoff_f1 - np.minimum(h, payoff_f0),
1e-10)
- 1]
α = π_grid[np.searchsorted(
np.minimum(h, payoff_f1) - payoff_f0,
1e-10)
- 1]
return (β, α)
β, α = find_cutoff_rule(wf, h_star)
cost_L0 = (1 - wf.π_grid) * wf.L0
cost_L1 = wf.π_grid * wf.L1
fig, ax = plt.subplots(figsize=(10, 6))
ax.plot(wf.π_grid, h_star, label='sample again')
ax.plot(wf.π_grid, cost_L1, label='choose f1')
ax.plot(wf.π_grid, cost_L0, label='choose f0')
ax.plot(wf.π_grid,
np.amin(np.column_stack([h_star, cost_L0, cost_L1]),axis=1),
lw=15, alpha=0.1, color='b', label='$J(\pi)$')
ax.annotate(r"$\beta$", xy=(β + 0.01, 0.5), fontsize=14)
ax.annotate(r"$\alpha$", xy=(α + 0.01, 0.5), fontsize=14)
plt.vlines(β, 0, β * wf.L0, linestyle="--")
plt.vlines(α, 0, (1 - α) * wf.L1, linestyle="--")
ax.set(xlim=(0, 1), ylim=(0, 0.5 * max(wf.L0, wf.L1)), ylabel="cost",
xlabel="$\pi$", title="Cost function $J(\pi)$")
plt.legend(borderpad=1.1)
plt.show()
The cost function $ J $ equals $ \pi L_1 $ for $ \pi \leq \beta $, and $ (1-\pi )L_0 $ for $ \pi \geq \alpha $.
The slopes of the two linear pieces of the cost function $ J(\pi) $ are determined by $ L_1 $ and $ - L_0 $.
The cost function $ J $ is smooth in the interior region, where the posterior probability assigned to $ f_0 $ is in the indecisive region $ \pi \in (\beta, \alpha) $.
The decision-maker continues to sample until the probability that he attaches to model $ f_0 $ falls below $ \beta $ or above $ \alpha $.
The next figure shows the outcomes of 500 simulations of the decision process.
On the left is a histogram of stopping times, i.e., the number of draws of $ z_k $ required to make a decision.
The average number of draws is around 6.6.
On the right is the fraction of correct decisions at the stopping time.
In this case, the decision-maker is correct 80% of the time
def simulate(wf, true_dist, h_star, π_0=0.5):
"""
This function takes an initial condition and simulates until it
stops (when a decision is made)
"""
f0, f1 = wf.f0, wf.f1
f0_rvs, f1_rvs = wf.f0_rvs, wf.f1_rvs
π_grid = wf.π_grid
κ = wf.κ
if true_dist == "f0":
f, f_rvs = wf.f0, wf.f0_rvs
elif true_dist == "f1":
f, f_rvs = wf.f1, wf.f1_rvs
# Find cutoffs
β, α = find_cutoff_rule(wf, h_star)
# Initialize a couple of useful variables
decision_made = False
π = π_0
t = 0
while decision_made is False:
# Maybe should specify which distribution is correct one so that
# the draws come from the "right" distribution
z = f_rvs()
t = t + 1
π = κ(z, π)
if π < β:
decision_made = True
decision = 1
elif π > α:
decision_made = True
decision = 0
if true_dist == "f0":
if decision == 0:
correct = True
else:
correct = False
elif true_dist == "f1":
if decision == 1:
correct = True
else:
correct = False
return correct, π, t
def stopping_dist(wf, h_star, ndraws=250, true_dist="f0"):
"""
Simulates repeatedly to get distributions of time needed to make a
decision and how often they are correct
"""
tdist = np.empty(ndraws, int)
cdist = np.empty(ndraws, bool)
for i in range(ndraws):
correct, π, t = simulate(wf, true_dist, h_star)
tdist[i] = t
cdist[i] = correct
return cdist, tdist
def simulation_plot(wf):
h_star = solve_model(wf)
ndraws = 500
cdist, tdist = stopping_dist(wf, h_star, ndraws)
fig, ax = plt.subplots(1, 2, figsize=(16, 5))
ax[0].hist(tdist, bins=np.max(tdist))
ax[0].set_title(f"Stopping times over {ndraws} replications")
ax[0].set(xlabel="time", ylabel="number of stops")
ax[0].annotate(f"mean = {np.mean(tdist)}", xy=(max(tdist) / 2,
max(np.histogram(tdist, bins=max(tdist))[0]) / 2))
ax[1].hist(cdist.astype(int), bins=2)
ax[1].set_title(f"Correct decisions over {ndraws} replications")
ax[1].annotate(f"% correct = {np.mean(cdist)}",
xy=(0.05, ndraws / 2))
plt.show()
simulation_plot(wf)
Now let’s consider the following exercise.
We double the cost of drawing an additional observation.
Before you look, think about what will happen:
wf = WaldFriedman(c=2.5)
simulation_plot(wf)
Increased cost per draw has induced the decision-maker to take fewer draws before deciding.
Because he decides with fewer draws, the percentage of time he is correct drops.
This leads to him having a higher expected loss when he puts equal weight on both models.
To facilitate comparative statics, we provide a Jupyter notebook that generates the same plots, but with sliders.
With these sliders, you can adjust parameters and immediately observe
For several reasons, it is useful to describe the theory underlying the test that Navy Captain G. S. Schuyler had been told to use and that led him to approach Milton Friedman and Allan Wallis to convey his conjecture that superior practical procedures existed.
Evidently, the Navy had told Captail Schuyler to use what it knew to be a state-of-the-art Neyman-Pearson test.
We’ll rely on Abraham Wald’s [Wald, 1947] elegant summary of Neyman-Pearson theory.
For our purposes, watch for there features of the setup:
Recall that in the sequential analytic formulation above, that
In chapter 1 of Sequential Analysis [Wald, 1947] Abraham Wald summarizes the Neyman-Pearson approach to hypothesis testing.
Wald frames the problem as making a decision about a probability distribution that is partially known.
(You have to assume that something is already known in order to state a well-posed problem – usually, something means a lot)
By limiting what is unknown, Wald uses the following simple structure to illustrate the main ideas:
To quote Abraham Wald,
A test procedure leading to the acceptance or rejection of the [null]
hypothesis in question is simply a rule specifying, for each possible sample of size $ n $, whether the [null] hypothesis should be accepted or rejected on the basis of the sample. This may also be expressed as follows: A test procedure is simply a subdivision of the totality of all possible samples of size $ n $ into two mutually exclusive parts, say part 1 and part 2, together with the application of the rule that the [null] hypothesis be accepted if the observed sample is contained in part 2. Part 1 is also called the critical region. Since part 2 is the totality of all samples of size $ n $ which are not included in part 1, part 2 is uniquely determined by part 1. Thus, choosing a test procedure is equivalent to determining a critical region.
Let’s listen to Wald longer:
As a basis for choosing among critical regions the following
considerations have been advanced by Neyman and Pearson: In accepting or rejecting $ H_0 $ we may commit errors of two kinds. We commit an error of the first kind if we reject $ H_0 $ when it is true; we commit an error of the second kind if we accept $ H_0 $ when $ H_1 $ is true. After a particular critical region $ W $ has been chosen, the probability of committing an error of the first kind, as well as the probability of committing an error of the second kind is uniquely determined. The probability of committing an error of the first kind is equal to the probability, determined by the assumption that $ H_0 $ is true, that the observed sample will be included in the critical region $ W $. The probability of committing an error of the second kind is equal to the probability, determined on the assumption that $ H_1 $ is true, that the probability will fall outside the critical region $ W $. For any given critical region $ W $ we shall denote the probability of an error of the first kind by $ \alpha $ and the probability of an error of the second kind by $ \beta $.
Let’s listen carefully to how Wald applies law of large numbers to interpret $ \alpha $ and $ \beta $:
The probabilities $ \alpha $ and $ \beta $ have the
following important practical interpretation: Suppose that we draw a large number of samples of size $ n $. Let $ M $ be the number of such samples drawn. Suppose that for each of these $ M $ samples we reject $ H_0 $ if the sample is included in $ W $ and accept $ H_0 $ if the sample lies outside $ W $. In this way we make $ M $ statements of rejection or acceptance. Some of these statements will in general be wrong. If $ H_0 $ is true and if $ M $ is large, the probability is nearly $ 1 $ (i.e., it is practically certain) that the proportion of wrong statements (i.e., the number of wrong statements divided by $ M $) will be approximately $ \alpha $. If $ H_1 $ is true, the probability is nearly $ 1 $ that the proportion of wrong statements will be approximately $ \beta $. Thus, we can say that in the long run [ here Wald applies law of large numbers by driving $ M \rightarrow \infty $ (our comment, not Wald’s) ] the proportion of wrong statements will be $ \alpha $ if $ H_0 $is true and $ \beta $ if $ H_1 $ is true.
The quantity $ \alpha $ is called the size of the critical region, and the quantity $ 1-\beta $ is called the power of the critical region.
Wald notes that
one critical region $ W $ is more desirable than another if it
has smaller values of $ \alpha $ and $ \beta $. Although either $ \alpha $ or $ \beta $ can be made arbitrarily small by a proper choice of the critical region $ W $, it is possible to make both $ \alpha $ and $ \beta $ arbitrarily small for a fixed value of $ n $, i.e., a fixed sample size.
Wald summarizes Neyman and Pearson’s setup as follows:
Neyman and Pearson show that a region consisting of all samples
$ (z_1, z_2, \ldots, z_n) $ which satisfy the inequality
$$ \frac{ f_1(z_1) \cdots f_1(z_n)}{f_0(z_1) \cdots f_0(z_n)} \geq k $$is a most powerful critical region for testing the hypothesis $ H_0 $ against the alternative hypothesis $ H_1 $. The term $ k $ on the right side is a constant chosen so that the region will have the required size $ \alpha $.
Wald goes on to discuss Neyman and Pearson’s concept of uniformly most powerful test.
Here is how Wald introduces the notion of a sequential test
A rule is given for making one of the following three decisions at any stage of
the experiment (at the m th trial for each integral value of m ): (1) to accept the hypothesis H , (2) to reject the hypothesis H , (3) to continue the experiment by making an additional observation. Thus, such a test procedure is carried out sequentially. On the basis of the first observation, one of the aforementioned decision is made. If the first or second decision is made, the process is terminated. If the third decision is made, a second trial is performed. Again, on the basis of the first two observations, one of the three decision is made. If the third decision is made, a third trial is performed, and so on. The process is continued until either the first or the second decisions is made. The number n of observations required by such a test procedure is a random variable, since the value of n depends on the outcome of the observations.
We’ll dig deeper into some of the ideas used here in the following lectures:
[1] The decision maker acts as if he believes that the sequence of random variables $ [z_{0}, z_{1}, \ldots] $ is *exchangeable*. See [Exchangeability and Bayesian Updating](https://python.quantecon.org/exchangeable.html) and [[Kreps, 1988](https://python.quantecon.org/zreferences.html#id110)] chapter 11, for discussions of exchangeability.