Studentized Sample Means And Slutsky’s Theorem

Why Studentize a Sample Mean?

The central limit theorem describes a sample mean after scaling by the population standard deviation, $\sigma$. In applications, however, $\sigma$ is usually unknown. A natural alternative is to estimate it with the sample standard deviation, $s$, and use the resulting statistic. The key question is whether this substitution changes the limiting distribution.

Under standard iid assumptions, it does not: the studentized sample mean converges in distribution to a standard normal random variable. The reasoning combines two results already central to asymptotic analysis: the central limit theorem for the numerator and consistency of the sample variance in the denominator.

Definition: For an iid sample $X_1,\ldots,X_n$ with sample mean $\bar{x}=n^{-1}\sum_{i=1}^n X_i$, define the sample variance and standard deviation by $$ s^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{x})^2, \qquad s=\sqrt{s^2}. $$ When $s>0$, the studentized sample mean for a population mean $\mu$ is $$ T_n=\frac{\sqrt{n}(\bar{x}-\mu)}{s}. $$ It uses an estimated standard deviation; the corresponding statistic with known population standard deviation is $\sqrt{n}(\bar{x}-\mu)/\sigma$.

The factor $\sqrt{n}$ reflects the sampling variability of a mean: when the observations have variance $\sigma^2$, the variance of $\bar{x}$ is $\sigma^2/n$. Thus $\sigma/\sqrt{n}$ is the standard deviation of the sample mean, and $s/\sqrt{n}$ estimates it. The studentized statistic is the difference between the sample mean and its target, measured in estimated standard errors.

Assumptions and the Limiting Result

Suppose $X_1,X_2,\ldots$ are independent and identically distributed, with $E[X_i]=\mu$ and $0<\operatorname{Var}(X_i)=\sigma^2<\infty$. These assumptions ensure that the iid central limit theorem applies and that the sample variance estimates a finite, nonzero population variance. No normality assumption is required for the asymptotic result.

Theorem: Under the iid, finite-variance, and positive-variance assumptions above, $$ \frac{\sqrt{n}(\bar{x}-\mu)}{s}\ \xrightarrow{d}\ N(0,1). $$ The statistic is defined when $s>0$. Since $s^2$ converges in probability to $\sigma^2>0$, the probability that $s=0$ tends to zero, so this qualification does not affect the limiting distribution.

The proof has two parts. First, the central limit theorem gives a standard normal limit when the numerator is divided by the fixed population standard deviation. Second, consistency means that $s$ is close to $\sigma$ with probability approaching one. Slutsky’s theorem then allows the fixed scaling to be replaced by this consistent estimate.

Why the Sample Variance Is Consistent

Write the sample variance in terms of the first and second raw sample moments:

s^2 =\frac{n}{n-1}\left(\frac{1}{n}\sum_{i=1}^n X_i^2-\bar{x}^{\,2}\right).

The finite-variance assumption gives $E[X_i^2]<\infty$. By the weak law of large numbers, the average of the $X_i^2$ converges in probability to $E[X_i^2]$, while $\bar{x}$ converges in probability to $\mu$. The continuous mapping theorem gives $\bar{x}^{\,2}\xrightarrow{p}\mu^2$, and $n/(n-1)$ converges to 1. Combining these limits yields

s^2 \xrightarrow{p} E[X_i^2]-\mu^2 =\operatorname{Var}(X_i)=\sigma^2.

Because $\sigma^2>0$, the square-root function is continuous at the limit. Therefore $s\xrightarrow{p}\sigma$, and, since $\sigma>0$, $\sigma/s\xrightarrow{p}1$ wherever the ratio is defined. The probability of $s=0$ tends to zero: for any sufficiently small $\epsilon>0$, the event $s=0$ implies $|s-\sigma|>\epsilon$, whose probability tends to zero.

Slutsky’s Theorem and Studentization

Theorem: If $A_n\xrightarrow{d}A$ and $B_n\xrightarrow{p}b$, where $b$ is a constant, then $A_nB_n\xrightarrow{d}bA$. In particular, if $b\ne0$, then $A_n/B_n\xrightarrow{d}A/b$ when the ratio is defined with probability tending to one.

Apply the iid central limit theorem to the mean:

\frac{\sqrt{n}(\bar{x}-\mu)}{\sigma}\xrightarrow{d}N(0,1).

The studentized statistic can be written as the product of this term and a factor converging in probability to one:

\frac{\sqrt{n}(\bar{x}-\mu)}{s} = \left(\frac{\sqrt{n}(\bar{x}-\mu)}{\sigma}\right) \left(\frac{\sigma}{s}\right), \qquad \frac{\sigma}{s}\xrightarrow{p}1.

Slutsky’s theorem now gives the stated standard normal limit. This is a distributional approximation as $n$ grows; it does not say that the statistic is exactly standard normal for a particular finite sample. Nor does consistency of $s$ mean that $s$ equals $\sigma$ in the observed sample. It means that the estimation error becomes small in probability as the sample size increases.

Worked Example: An Approximate Test of a Population Mean

Suppose a hypothetical iid sample of $n=100$ measurements has sample mean $\bar{x}=52.4$ units and sample standard deviation $s=12$ units. Consider $H_0:\mu=50$ units against $H_a:\mu\ne50$ units. Assume the population has finite, positive variance. The sample is consistent with the theorem’s sampling assumptions, and the observed $s$ is positive.

Under $H_0$, the studentized statistic is $$ T_n=\frac{\sqrt{100}(52.4-50)}{12} =\frac{10(2.4)}{12} =2.00. $$

For a large-sample, two-sided comparison, the approximate p-value is $$ 2[1-\Phi(2.00)]\approx 0.0455, $$ where $\Phi$ is the standard normal cumulative distribution function. At significance level $\alpha=0.05$, this asymptotic test rejects $H_0$. The p-value is the probability, under the null and the stated approximation, of a statistic at least as far from zero as the observed statistic. It is not the probability that $H_0$ is true.

Worked Example: A Large-Sample Confidence Interval

In a hypothetical iid sample of $n=64$ measurements, suppose $\bar{x}=18.6$ minutes and $s=4.8$ minutes. Assume finite, positive population variance and independent sampling from the same population. Studentization gives the approximate standard error $$ \frac{s}{\sqrt{n}}=\frac{4.8}{8}=0.600\text{ minutes}. $$

Using the standard normal critical value 1.96, the approximate 95% interval is $$ 18.6\pm1.96(0.600) =18.6\pm1.176 =(17.424,\ 19.776)\text{ minutes}. $$

The interval is an asymptotic normal approximation. Its frequentist interpretation is that, under repeated sampling from a population satisfying the assumptions, intervals constructed by this procedure have coverage approaching 95% as sample size increases. It does not assign a 95% probability to the fixed population mean being inside this particular interval.

Worked Example: Studentizing a Sample Proportion

Let $X_i$ indicate whether person $i$ in a hypothetical iid sample has a specified characteristic: $X_i=1$ if yes and $X_i=0$ otherwise. Suppose $n=400$ observations include 130 yes responses, so $\bar{x}=\hat{p}=130/400=0.325$. Consider testing $H_0:p=0.30$ against a two-sided alternative. The observations are iid Bernoulli with finite variance, and the null variance is positive because $0<0.30<1$.

The sample variance of these binary observations is $$ s^2=\frac{400}{399}\hat{p}(1-\hat{p}) =\frac{400}{399}(0.325)(0.675) \approx0.219925, \qquad s\approx0.468961. $$ The studentized statistic is $$ T_n=\frac{\sqrt{400}(0.325-0.30)}{0.468961} \approx1.066. $$

The two-sided standard normal approximation gives $2[1-\Phi(1.066)]\approx0.286$. This is not evidence against $H_0$ at $\alpha=0.05$. The calculation illustrates studentization for a mean of binary observations; it does not make the finite-sample statistic exactly standard normal.

When the Result Matters—and What It Does Not Say

Studentization is useful because it turns an asymptotic result involving an unknown population scale into a procedure that can be calculated from the sample. It underlies large-sample inference for means, including the approximate tests and intervals above. The same logic is also a basic pattern in econometrics: a statistic with a limiting distribution can often retain that limit after a consistently estimated scale is substituted.

The conclusion depends on the assumptions and the limit argument. If observations are dependent or not identically distributed, the iid central limit theorem and the variance-consistency argument used here may not apply as stated. If the population variance is zero, division by its standard deviation is impossible. If the variance is infinite, the finite-variance central limit theorem invoked here does not supply the result. Other conditions and methods may be appropriate in those settings, but they are not established by this theorem.

A further distinction matters in practice: the theorem gives a limiting distribution, not a guarantee of good accuracy at every sample size. A highly skewed population or extreme observations can make a normal approximation inaccurate in a moderate sample. The result alone does not provide an error bound for the approximation.

Common Mistakes

Treating an asymptotic limit as an exact finite-sample law. The theorem says the distribution approaches standard normal under the assumptions; it does not say every finite-sample studentized mean is standard normal.
Replacing $s$ with $\sigma$ without explanation. They are not interchangeable in a realized sample. The justification is that $s\xrightarrow{p}\sigma$, followed by Slutsky’s theorem.
Ignoring the positive-variance condition. If $\sigma^2=0$, the standardization is undefined. The theorem requires $0<\sigma^2<\infty$.
Rounding a statistic before computing a tail probability. Use the unrounded statistic in the probability calculation when possible; otherwise the reported p-value may not agree with the displayed statistic.
Reading a p-value as a probability that the null is true. It is a probability about data or a test statistic calculated under the null, not a posterior probability for the hypothesis.

Key takeaway: With iid observations of finite, positive variance, the central limit theorem standardizes the sample mean using $\sigma$, the sample variance is consistent for $\sigma^2$, and Slutsky’s theorem justifies replacing $\sigma$ with $s$. The resulting studentized mean has a standard normal limiting distribution.

Check Your Understanding

Answer each question using the assumptions and limiting argument developed above.

State the assumptions needed for the studentized sample mean to converge to a standard normal distribution.
Why does consistency of $s^2$ follow from the weak law applied to the first and second raw sample moments?
In the Slutsky argument, which factor converges in distribution, and which factor converges in probability to one?
Does the theorem imply that the studentized mean is exactly standard normal in a sample of size 20? Explain.
What is the correct frequentist interpretation of an approximate 95% confidence interval obtained using this result?