r‑statistics.co by Selva Prabhakaran
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Ancillary Statistics & Basu's Theorem in R: Advanced Statistical Theory

An ancillary statistic is a function of the sample whose distribution does not depend on the unknown parameter. Basu's theorem is the bridge that connects this idea to sufficiency: if a statistic is complete and sufficient, it is independent of every ancillary statistic.

What is an ancillary statistic?

The idea sounds paradoxical at first. A statistic built from data that carries no information about the parameter you care about? To see why that is useful, we will start with a location family. Shift the distribution up and down along the number line, and watch what stays put. That is the signature of ancillarity, and we can spot it in one short simulation.

Consider the uniform distribution $U(\mu, \mu + 1)$. Here $\mu$ is a location parameter: changing $\mu$ slides the whole distribution left or right without changing its shape. The range of a sample, $R = X_{(n)} - X_{(1)}$, is the difference between the largest and smallest observations. Because every observation shifts by the same amount when $\mu$ changes, their difference does not move at all.

RRange is ancillary for a location family
# Empirical distribution of the range for two different mu set.seed(2026) n <- 30 sim_range <- function(mu, n, reps = 3000) { replicate(reps, { x <- runif(n, min = mu, max = mu + 1) max(x) - min(x) }) } range_a <- sim_range(mu = 0, n = n) range_b <- sim_range(mu = 10, n = n) data.frame( mu = c(0, 10), mean = c(mean(range_a), mean(range_b)), sd = c(sd(range_a), sd(range_b)) ) #> mu mean sd #> 1 0 0.9361812 0.02953314 #> 2 10 0.9352974 0.02991722

  

  





The two rows are nearly identical, and the tiny gap is Monte Carlo noise rather than a real signal. Shifting the uniform distribution by ten units does not change the distribution of the range. In the language of theory, the range is ancillary for the location parameter $\mu$. It carries information about the spread of the sample, but none about where the sample lives on the number line.



Note
Ancillary means "invariant under the parameter". For a location family, any statistic that is unchanged by adding a constant to every observation is ancillary for the location parameter. The range, the sample variance, and the interquartile range are all members of that family.





Try it: Modify the simulation so the uniform span is $U(\mu, \mu + 2)$ instead of width one. Verify the range is still ancillary for $\mu$.



  

    
RYour turn: verify ancillarity for width-2 uniform

    
# Modify sim_range to sample from Uniform(mu, mu + 2)
ex_sim_range_w2 <- function(mu, n, reps = 3000) {
  # your code here
}

# Test at mu = 0 and mu = 7
ex_a <- ex_sim_range_w2(0, 30)
ex_b <- ex_sim_range_w2(7, 30)
c(mean_a = mean(ex_a), mean_b = mean(ex_b))
#> Expected: two very close values near 1.87

    

      
      
    

    

  

  





Click to reveal solution



  

    
RWidth-2 uniform solution

    
ex_sim_range_w2 <- function(mu, n, reps = 3000) {
  replicate(reps, {
    x <- runif(n, min = mu, max = mu + 2)
    max(x) - min(x)
  })
}

ex_a <- ex_sim_range_w2(0, 30)
ex_b <- ex_sim_range_w2(7, 30)
c(mean_a = mean(ex_a), mean_b = mean(ex_b))
#> mean_a mean_b
#> 1.8720 1.8714

    

      
      
    

    

  

  





Explanation: Doubling the width just rescales the range, but the distribution still does not depend on $\mu$ because $\mu$ only translates the endpoints.








How is ancillary different from sufficient?



Sufficient and ancillary statistics sit at opposite ends of an information spectrum. A sufficient statistic condenses all parameter information in the sample. An ancillary statistic carries none of it. Everything else lives somewhere in between.



Here is a compact contrast of the three concepts that matter in this post.







Concept
Depends on $\theta$?
Intuition






Sufficient
Yes, fully. Contains all information about $\theta$.
Conditional on $T$, the rest of the data tells you nothing more.




Ancillary
No. Its distribution is the same for every $\theta$.
Knowing $V$ tells you nothing about $\theta$.




Complete
Extra condition on sufficient.
No non-trivial function of $T$ has mean zero for every $\theta$.







To make this concrete, take a Poisson sample with rate $\lambda$. The total $\sum X_i$ is sufficient for $\lambda$, because once we know the total the individual $X_i$ values are a multinomial draw whose distribution no longer involves $\lambda$. Let us verify the sufficiency claim by looking at a conditional distribution.




  

    
RConditional distribution of X_1 given the sum is free of lambda

    
set.seed(7)
# For two different lambdas, collect pairs (sum, X_1) and keep only those where sum = 10
cond_x1_given_sum <- function(lambda, target_sum = 10, n = 4, reps = 20000) {
  x1_vals <- integer(0)
  for (i in seq_len(reps)) {
    x <- rpois(n, lambda = lambda)
    if (sum(x) == target_sum) x1_vals <- c(x1_vals, x[1])
  }
  x1_vals
}

cond_l2 <- cond_x1_given_sum(lambda = 2, target_sum = 10)
cond_l5 <- cond_x1_given_sum(lambda = 5, target_sum = 10)

rbind(
  lambda_2 = table(factor(cond_l2, levels = 0:10)) / length(cond_l2),
  lambda_5 = table(factor(cond_l5, levels = 0:10)) / length(cond_l5)
)
#>             0    1    2    3    4    5    6    7    8    9   10
#> lambda_2 0.055 0.187 0.286 0.257 0.141 0.058 0.014 0.002 0.000 0 0
#> lambda_5 0.057 0.188 0.281 0.255 0.143 0.059 0.015 0.002 0.000 0 0

    

      
      
    

    

  

  





The two rows are almost identical because they are both sampling from the same theoretical distribution: the conditional of $X_1$ given $\sum X_i = 10$ is Binomial$(10, 1/n)$, with no $\lambda$ inside. This is exactly what sufficiency means operationally.



Key Insight
Sufficient and ancillary statistics are complementary halves of the same information accounting. The sufficient statistic absorbs all $\theta$-dependence; the ancillary statistic is everything the model can still say about the data after the parameter has been accounted for. That is why Basu's theorem, which links the two, is so clean.





Try it: For a normal sample $X_i \sim N(\mu, 1)$, the sample mean is sufficient for $\mu$. Verify sufficiency empirically by fixing the observed mean at, say, $\bar{x} = 0$ and checking that the conditional distribution of $X_1 - \bar{x}$ does not depend on $\mu$.



  

    
RYour turn: sufficiency of the sample mean

    
# Generate many samples, keep only those with xbar near 0, record X_1 - xbar
ex_cond_resid <- function(mu, n = 5, reps = 20000, tol = 0.02) {
  # your code here: return a vector of (x[1] - mean(x)) values conditioned on |mean(x)| < tol
}

ex_mu0 <- ex_cond_resid(0)
ex_mu5 <- ex_cond_resid(5)
c(mean_mu0 = mean(ex_mu0), sd_mu0 = sd(ex_mu0),
  mean_mu5 = mean(ex_mu5), sd_mu5 = sd(ex_mu5))
#> Expected: means ~ 0 and sds ~ 0.9 for BOTH mu values

    

      
      
    

    

  

  





Click to reveal solution



  

    
RSufficiency of the sample mean solution

    
ex_cond_resid <- function(mu, n = 5, reps = 20000, tol = 0.02) {
  out <- numeric(0)
  for (i in seq_len(reps)) {
    x <- rnorm(n, mean = mu, sd = 1)
    if (abs(mean(x)) < tol) out <- c(out, x[1] - mean(x))
  }
  out
}

ex_mu0 <- ex_cond_resid(0)
ex_mu5 <- ex_cond_resid(5)
c(mean_mu0 = mean(ex_mu0), sd_mu0 = sd(ex_mu0),
  mean_mu5 = mean(ex_mu5), sd_mu5 = sd(ex_mu5))
#>   mean_mu0     sd_mu0   mean_mu5     sd_mu5
#> -0.0010...  0.895...  -0.0015...  0.898...

    

      
      
    

    

  

  





Explanation: Conditional on the sample mean, the residual $X_1 - \bar{X}$ has a distribution that does not depend on $\mu$. That is the empirical fingerprint of a sufficient statistic.








What does Basu's theorem state?



Basu's theorem is a one-sentence result with surprising reach. Here is the statement.



Basu's Theorem. If $T(X)$ is complete and sufficient for $\theta$, and $V(X)$ is ancillary, then $T$ and $V$ are independent for every value of $\theta$.



The intuition runs in three steps. First, $T$ has absorbed all of the parameter information, so the conditional distribution of $V$ given $T = t$ cannot be shifted around by $\theta$. Second, $V$ is ancillary, so its marginal distribution also does not depend on $\theta$. Third, completeness forces these two distributions to agree exactly, not just on average. The formal statement of the agreement is short and sharp.



$$P(V \in A \mid T = t) = P(V \in A) \quad \text{for every } t, \text{ every } A.$$



Where:





When the conditional and the marginal are identical, independence follows by definition. That is the whole theorem in a line.



A clean way to see independence in R is with an empirical independence helper that checks two things at once: the linear correlation between $T$ and $V$, and a contingency-table $\chi^2$ test on their quantile bins. If both signals are weak, we have no evidence against independence.




  

    
RHelper: empirical independence check

    
is_independent <- function(t, v, n_bins = 5) {
  # 1. Pearson correlation
  rho <- cor(t, v)

  # 2. Bin into quantiles and run a chi-square test of independence
  tb <- cut(t, breaks = quantile(t, probs = seq(0, 1, length.out = n_bins + 1)),
            include.lowest = TRUE)
  vb <- cut(v, breaks = quantile(v, probs = seq(0, 1, length.out = n_bins + 1)),
            include.lowest = TRUE)
  p_val <- suppressWarnings(chisq.test(table(tb, vb))$p.value)

  list(correlation = round(rho, 4), chisq_pvalue = round(p_val, 4))
}

# Sanity check: two unrelated N(0,1) columns
set.seed(42)
is_independent(rnorm(4000), rnorm(4000))
#> $correlation
#> [1] -0.0083
#>
#> $chisq_pvalue
#> [1] 0.5126

    

      
      
    

    

  

  





A correlation near zero and a $\chi^2$ p-value that is not small tell us the helper is calibrated correctly on a known-independent pair. We will reuse this helper in the next three sections to test Basu's theorem on three different models.



Warning
Completeness is essential. Without it, Basu's theorem can fail. There are classic examples where a sufficient statistic that is not complete turns out to be dependent on an ancillary statistic. Completeness is the ingredient that promotes "same distribution" into "same random variable almost surely".





Try it: Feed the helper a clearly dependent pair (e.g., $T = Z$ and $V = Z^2$ for $Z \sim N(0,1)$). Does the helper flag the dependence?



  

    
RYour turn: helper on a dependent pair

    
set.seed(11)
ex_z <- rnorm(4000)
ex_t <- ex_z
ex_v <- ex_z^2
# your code: call is_independent on ex_t and ex_v
#> Expected: chisq_pvalue very close to 0; correlation ~ 0 (nonlinear dependence!)

    

      
      
    

    

  

  





Click to reveal solution



  

    
RHelper on a dependent pair solution

    
set.seed(11)
ex_z <- rnorm(4000)
ex_t <- ex_z
ex_v <- ex_z^2
is_independent(ex_t, ex_v)
#> $correlation
#> [1] -0.0015
#>
#> $chisq_pvalue
#> [1] 0

    

      
      
    

    

  

  





Explanation: Correlation alone would have missed the dependence because $Z$ and $Z^2$ are uncorrelated yet perfectly dependent. The $\chi^2$ check on binned quantiles catches this non-linear structure and fires a near-zero p-value.








How do we verify sample mean and variance are independent under a normal model?



The textbook Basu example is also the most useful in practice. For $X_1, \ldots, X_n$ iid $N(\mu, \sigma^2)$ with $\sigma$ known, the sample mean $\bar{X}$ is complete and sufficient for $\mu$. The sample variance $S^2$ is ancillary for $\mu$, because shifting every $X_i$ by a constant leaves $S^2$ unchanged. Basu's theorem then gives us the independence of $\bar{X}$ and $S^2$ with no integrals.



Let us verify it by Monte Carlo.




  

    
RNormal model: sample mean and variance are independent

    
set.seed(101)
reps <- 5000
n    <- 30

mc_normal <- function(mu, sigma = 1, n, reps) {
  out <- replicate(reps, {
    x <- rnorm(n, mean = mu, sd = sigma)
    c(xbar = mean(x), s2 = var(x))
  })
  data.frame(xbar = out["xbar", ], s2 = out["s2", ])
}

norm_res <- mc_normal(mu = 0, n = n, reps = reps)
is_independent(norm_res$xbar, norm_res$s2)
#> $correlation
#> [1] 0.0091
#>
#> $chisq_pvalue
#> [1] 0.4218

    

      
      
    

    

  

  





A correlation of $0.009$ and a $\chi^2$ p-value near $0.42$ are exactly what Basu's theorem predicts. Under the normal model, $\bar{X}$ and $S^2$ are statistically independent, not merely uncorrelated. A scatter makes the independence visible.




  

    
RScatter: xbar vs s2 forms a structureless cloud

    
plot(norm_res$xbar, norm_res$s2,
     pch = 16, cex = 0.3, col = rgb(0.2, 0.3, 0.7, 0.4),
     xlab = "sample mean", ylab = "sample variance",
     main = "Basu in action: N(0, 1), n = 30")
abline(v = 0, col = "red", lty = 2)

    

      
      
    

    

  

  





The cloud has no visible trend, no fanning, and no curvature. If $\bar{X}$ and $S^2$ were dependent, the conditional variance of $S^2$ at the extremes of $\bar{X}$ would be different from its variance in the middle, and the cloud would bulge or pinch. It does not.



Tip
Basu's theorem gives the cleanest derivation of "sample mean and sample variance are independent under normality". The classical derivation relies on a change of variables and the orthogonal decomposition of the sample. With Basu, the argument is three lines: $\bar{X}$ is complete sufficient, $S^2$ is ancillary, therefore they are independent.





Try it: Re-run the Monte Carlo for $\mu = 10$ and $\mu = -5$. The correlation should still be close to zero, confirming that independence holds for every value of $\mu$.



  

    
RYour turn: independence for any mu

    
# Run mc_normal for mu = 10 and mu = -5, feed each result to is_independent
ex_res_10 <- mc_normal(mu = 10,  n = 30, reps = 4000)
ex_res_m5 <- mc_normal(mu = -5, n = 30, reps = 4000)
# your code here: call is_independent on each
#> Expected: correlations near 0, chi-square p-values well above 0.05

    

      
      
    

    

  

  





Click to reveal solution



  

    
RIndependence for any mu solution

    
is_independent(ex_res_10$xbar, ex_res_10$s2)
#> $correlation
#> [1] -0.0104
#>
#> $chisq_pvalue
#> [1] 0.2867

is_independent(ex_res_m5$xbar, ex_res_m5$s2)
#> $correlation
#> [1] 0.0046
#>
#> $chisq_pvalue
#> [1] 0.6711

    

      
      
    

    

  

  





Explanation: The theorem claims independence for every $\mu$, and the Monte Carlo at $\mu = 10$ and $\mu = -5$ both confirm it.








How does Basu's theorem apply to uniform and exponential distributions?



The normal is famous but far from the only model where Basu's theorem pays off. Two more canonical examples round out the picture.



First, the uniform on $(0, \theta)$. Here $X_{(n)}$, the sample maximum, is complete sufficient for $\theta$. The ratio $\bar{X} / X_{(n)}$ is ancillary because both numerator and denominator scale with $\theta$, and the ratio cancels it.




  

    
RUniform(0, theta): max and mean-over-max are independent

    
set.seed(202)
mc_unif <- function(theta, n, reps) {
  out <- replicate(reps, {
    x <- runif(n, min = 0, max = theta)
    c(x_max = max(x), ratio = mean(x) / max(x))
  })
  data.frame(x_max = out["x_max", ], ratio = out["ratio", ])
}

unif_res <- mc_unif(theta = 3, n = 20, reps = 5000)
is_independent(unif_res$x_max, unif_res$ratio)
#> $correlation
#> [1] 0.0075
#>
#> $chisq_pvalue
#> [1] 0.3054

    

      
      
    

    

  

  





The correlation is tiny and the $\chi^2$ test does not reject. The sample maximum tells you nothing about the mean-to-max ratio, just as Basu's theorem predicts.



Second, the exponential with rate parameter $\beta$. Here $\sum X_i$ is complete sufficient, and $X_1 / \sum X_i$ is ancillary because the $\beta$-scaling drops out of any ratio of coordinates of the sample.




  

    
RExponential(beta): sum and first-over-sum are independent

    
set.seed(303)
mc_exp <- function(beta, n, reps) {
  out <- replicate(reps, {
    x <- rexp(n, rate = beta)
    c(s = sum(x), ratio = x[1] / sum(x))
  })
  data.frame(s = out["s", ], ratio = out["ratio", ])
}

exp_res <- mc_exp(beta = 2, n = 20, reps = 5000)
is_independent(exp_res$s, exp_res$ratio)
#> $correlation
#> [1] -0.0021
#>
#> $chisq_pvalue
#> [1] 0.7392

    

      
      
    

    

  

  





Again the helper reports no evidence against independence. In both cases, the general pattern is the same: find a ratio that cancels the scale parameter, and you have found an ancillary statistic.



Key Insight
In any scale family, a ratio that cancels the scale parameter is ancillary. That recipe turns Basu's theorem into a practical tool: once you have identified a complete sufficient statistic $T$, construct a ratio of the sample that is invariant to rescaling, and Basu guarantees it is independent of $T$.





Try it: For iid $\text{Exp}(\beta)$, is the ratio $X_1 / X_2$ ancillary? Simulate and check with is_independent() against the total $\sum X_i$.



  

    
RYour turn: is X_1/X_2 ancillary for Exp(beta)?

    
set.seed(404)
ex_pair <- function(beta, n = 20, reps = 4000) {
  out <- replicate(reps, {
    x <- rexp(n, rate = beta)
    c(s = sum(x), ratio12 = x[1] / x[2])
  })
  data.frame(s = out["s", ], ratio12 = out["ratio12", ])
}

# Step 1: check that ratio12 has the same distribution for two betas
ex_b1 <- ex_pair(beta = 1)
ex_b5 <- ex_pair(beta = 5)
c(median_b1 = median(ex_b1$ratio12), median_b5 = median(ex_b5$ratio12))

# Step 2: check independence of s and ratio12 at, say, beta = 2
# your code here: use is_independent on ex_pair(beta = 2)
#> Expected: ratios have the same distribution across betas; correlation near 0; p-value not small

    

      
      
    

    

  

  





Click to reveal solution



  

    
RX_1/X_2 ancillary solution

    
ex_b2 <- ex_pair(beta = 2)
is_independent(ex_b2$s, ex_b2$ratio12)
#> $correlation
#> [1] 0.0022
#>
#> $chisq_pvalue
#> [1] 0.6105

    

      
      
    

    

  

  





Explanation: $X_1 / X_2$ is the ratio of two iid exponentials, which has a distribution that is free of $\beta$ (it is $F_{2,2}$-distributed with some rescaling). So the ratio is ancillary, and by Basu it must be independent of the complete sufficient $\sum X_i$.








Practice Exercises



Exercise 1: Verify $S^2$ is ancillary for $\mu$



Simulate $N(\mu, 1)$ samples of size $n = 50$ for $\mu = 0$ and $\mu = 5$, with $3000$ replicates each. Compute the empirical mean and sd of $S^2$ in both cases. Save the two summary vectors as my_s2_mu0 and my_s2_mu5.




  

    
RExercise 1: S^2 is ancillary for mu

    
# Exercise 1 starter
# Hint: replicate() + var() + compute mean and sd

# Write your code below:

    

      
      
    

    

  

  






Click to reveal solution



  

    
RExercise 1 solution

    
set.seed(505)
sim_s2 <- function(mu, n = 50, reps = 3000) {
  replicate(reps, var(rnorm(n, mean = mu, sd = 1)))
}

my_s2_mu0 <- c(mean = mean(sim_s2(0)), sd = sd(sim_s2(0)))
my_s2_mu5 <- c(mean = mean(sim_s2(5)), sd = sd(sim_s2(5)))
rbind(my_s2_mu0, my_s2_mu5)
#>               mean         sd
#> my_s2_mu0 1.001... 0.202...
#> my_s2_mu5 1.000... 0.201...

    

      
      
    

    

  

  





Explanation: The two rows agree within Monte Carlo error, confirming $S^2$ is ancillary for $\mu$.






Exercise 2: Use Basu's theorem to compute $E[X_1 / \sum X_i]$



For iid $X_i \sim \text{Exp}(\beta)$ with $n = 5$, show analytically that $E[X_1 / S] = 1/n$ where $S = \sum X_i$, using Basu's theorem. Then verify by simulation with $\beta = 2$.



Analytical hint: $X_1 / S$ is ancillary and $S$ is complete sufficient, so they are independent. Therefore $$E[X_1] = E\!\left[ \frac{X_1}{S} \cdot S \right] = E\!\left[ \frac{X_1}{S} \right] \cdot E[S].$$ Substitute $E[X_1] = 1/\beta$ and $E[S] = n/\beta$ and solve.




  

    
RExercise 2: E[X_1 / sum(X)] via Basu

    
# Exercise 2 starter
# Hint: simulate many samples, compute x[1] / sum(x), take the mean

# Write your code below:

    

      
      
    

    

  

  






Click to reveal solution



  

    
RExercise 2 solution

    
set.seed(606)
my_ratio <- replicate(10000, {
  x <- rexp(5, rate = 2)
  x[1] / sum(x)
})

mean(my_ratio)
#> [1] 0.2003
1 / 5
#> [1] 0.2

    

      
      
    

    

  

  





Explanation: The analytical answer is $1/n = 1/5 = 0.2$. The decomposition uses independence of $X_1/S$ and $S$, which is exactly Basu's theorem. No direct integration was required.






Exercise 3: Is the range ancillary under $U(0, \theta)$?



For iid $U(0, \theta)$, the range $R = X_{(n)} - X_{(1)}$ is not ancillary for $\theta$ (in contrast to the $U(\mu, \mu+1)$ case). Simulate $4000$ samples of size $n = 20$ at $\theta = 1$ and at $\theta = 3$. Compare the mean and sd of the range. Explain the result in one sentence.




  

    
RExercise 3: range under U(0, theta)

    
# Exercise 3 starter
# Hint: simulate the range at two thetas and compare

# Write your code below:

    

      
      
    

    

  

  






Click to reveal solution



  

    
RExercise 3 solution

    
set.seed(707)
sim_range_scale <- function(theta, n = 20, reps = 4000) {
  replicate(reps, {
    x <- runif(n, min = 0, max = theta)
    max(x) - min(x)
  })
}

my_r1 <- sim_range_scale(theta = 1)
my_r3 <- sim_range_scale(theta = 3)
rbind(theta1 = c(mean = mean(my_r1), sd = sd(my_r1)),
      theta3 = c(mean = mean(my_r3), sd = sd(my_r3)))
#>           mean         sd
#> theta1 0.906... 0.061...
#> theta3 2.717... 0.183...

    

      
      
    

    

  

  





Explanation: The mean and sd of the range at $\theta = 3$ are roughly three times those at $\theta = 1$. The distribution of the range is $\theta$-scaled, so it is not ancillary for $\theta$. The ancillary statistic in this scale family is a ratio that cancels $\theta$, not the range itself.






Complete Example



Let us bundle everything into one verification script: Basu's theorem across three canonical models in a single sweep.




  

    
RPutting it all together: Basu across three models

    
set.seed(909)

run_basu_suite <- function() {
  # 1. Normal: (xbar, s2)
  norm_pairs <- mc_normal(mu = 0, n = 30, reps = 4000)
  norm_check <- is_independent(norm_pairs$xbar, norm_pairs$s2)

  # 2. Uniform(0, theta): (x_max, mean/max)
  unif_pairs <- mc_unif(theta = 3, n = 20, reps = 4000)
  unif_check <- is_independent(unif_pairs$x_max, unif_pairs$ratio)

  # 3. Exponential(beta): (sum, x_1/sum)
  exp_pairs <- mc_exp(beta = 2, n = 20, reps = 4000)
  exp_check <- is_independent(exp_pairs$s, exp_pairs$ratio)

  data.frame(
    model        = c("Normal",     "Uniform(0,theta)", "Exponential"),
    sufficient_T = c("xbar",       "x_max",            "sum(x)"),
    ancillary_V  = c("s^2",        "mean/max",         "x_1/sum"),
    correlation  = c(norm_check$correlation, unif_check$correlation, exp_check$correlation),
    chisq_p      = c(norm_check$chisq_pvalue, unif_check$chisq_pvalue, exp_check$chisq_pvalue),
    verdict      = c(
      ifelse(norm_check$chisq_pvalue > 0.05, "PASS", "FAIL"),
      ifelse(unif_check$chisq_pvalue > 0.05, "PASS", "FAIL"),
      ifelse(exp_check$chisq_pvalue  > 0.05, "PASS", "FAIL")
    )
  )
}

final_table <- run_basu_suite()
print(final_table)
#>               model sufficient_T ancillary_V correlation chisq_p verdict
#> 1            Normal         xbar         s^2      0.0083  0.3921    PASS
#> 2  Uniform(0,theta)        x_max    mean/max     -0.0044  0.6127    PASS
#> 3       Exponential       sum(x)     x_1/sum      0.0018  0.8204    PASS

    

      
      
    

    

  

  





All three rows come back as PASS, and the correlations cluster near zero as theory demands. With a single R script we have verified Basu's theorem across three textbook cases and built a reusable pattern for any future model: find the complete sufficient statistic, find an ancillary ratio, and run the independence helper.



Summary



A one-page takeaway of everything we covered.



    


  • Ancillary: distribution does not depend on $\theta$. It lives orthogonal to the parameter.
    • 


  • Sufficient: condenses all $\theta$-information from the sample into a single statistic (or vector).
    • 


  • Complete sufficient: sufficient, and no non-zero function of the statistic has mean zero for every $\theta$.
    • 


  • Basu's theorem: complete sufficient and ancillary statistics are independent, for every $\theta$, with no further calculation required.
    • 









Model
Complete sufficient $T$
Ancillary $V$
Basu says






$N(\mu, \sigma^2)$, $\sigma$ known
$\bar{X}$
$S^2$
$\bar{X} \perp S^2$




$U(\mu, \mu+1)$
order statistic
range $R$
$T \perp R$




$U(0, \theta)$
$X_{(n)}$
$\bar{X} / X_{(n)}$
$X_{(n)} \perp \bar{X}/X_{(n)}$




$\text{Exp}(\beta)$
$\sum X_i$
$X_1 / \sum X_i$
$\sum X_i \perp X_1/\sum X_i$







The practical payoff: whenever you need independence of two statistics under a parametric model, check whether one is complete sufficient and the other is ancillary. If yes, Basu's theorem gives you the result in one line.



References



    


  1. Basu, D. (1955). On statistics independent of a complete sufficient statistic. Sankhyā 15: 377-380. [Original paper]
    2. 


  2. Lehmann, E. L. (1981). An interpretation of completeness and Basu's theorem. JASA 76(374): 335-340. Link
    3. 


  3. Casella, G. & Berger, R. L. (2002). Statistical Inference, 2nd Edition. Duxbury. Chapter 6: Principles of Data Reduction.
    4. 


  4. Wikipedia contributors. Basu's theorem. Link
    5. 


  5. Mackey, L. (2015). STATS 300A Theory of Statistics, Lecture 4: Completeness and Ancillarity. Stanford University. Link
    6. 


  6. Berkeley Statistics (2024). Stat210A Reader: Completeness, Ancillarity, and Basu's Theorem. Link
    7. 


  7. Ghosh, M. (2002). Basu's theorem with applications: a personalistic review. Sankhyā A, 64(3, Part 1): 509-531.
    8. 


  8. Watkins, J. (n.d.). F4: Completeness (lecture notes). University of Arizona. Link
    9. 





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