Student’s t distribution

You will learn: Explain how an estimated scale changes standardized tails and moment conditions.

Start with: Normal distribution · Chi-squared distribution

How does estimating the standard deviation change a standardized sample mean? For independent normal observations, the resulting statistic follows Student’s t rather than a standard normal. Its heavier tails account for the random denominator.

f(x;\nu)=\frac{\Gamma(\tfrac{\nu+1}{2})}{\sqrt{\nu\pi}\,\Gamma(\tfrac{\nu}{2})}\left(1+\tfrac{x^{2}}{\nu}\right)^{-(\nu+1)/2}

t_3.0 Normal(0, 1)

Construct the random denominator

First three independent constructions used in the histogram
Z	V	√(V/ν)	T
0.9660	1.2965	0.6574	1.4694
1.3061	5.1278	1.3074	0.9990
1.0904	5.4330	1.3457	0.8102

Z is standard normal and independent of V, which is chi-squared with 3 degrees of freedom. A small denominator can produce a large T even when Z is ordinary. 5 of 1000 results lie beyond the plot; exact omitted mass is 0.9273%.

Absolute cutoff 2.00

P(|T|≥2) = 0.139326. Mean: 0. Variance: 3.0000. Unit scale is not unit standard deviation.

Quantile probability 0.975

The 97.5% quantile is 3.182446. The two-sided central 95.0% interval uses ±this cutoff.

Estimate an unknown standard deviation

A separate sample of 4 independent Normal(0,1) observations gives mean -0.2824, sample SD 0.7016, and T=(mean−0)/(SD/√n) = -0.8051. The SD uses n−1 in its variance denominator; the reference has ν=n−1=3.

Read the observations used for this statistic

-0.4951, 0.4647, -1.1539, 0.0547

ν (degrees of freedom) 3.0 n (samples) 1000

Bars: 1000 ratios Z/√(V/ν) with independent numerator and denominator. Solid: Student's t PDF. Dashed: Standard Normal for reference. As

\nu

grows, the two curves merge.

What to notice

Small $\nu$ means fat tails. At $\nu=1$ it’s the Cauchy distribution — no finite mean. At $\nu=2$ it has a mean but infinite variance.
Large $\nu$ means Gaussian. The density approaches Normal(0,1) as degrees of freedom grow. At ν=30, the 97.5% quantile is about 2.042 rather than the normal 1.960; tail calculations can still differ materially.
Algebraic tails. The density falls off like $|x|^{-(\nu+1)}$ — a polynomial decay, in contrast to the Normal’s exponential one.

Where it comes from

If you take a standard Normal $Z$ and divide by the square root of an independent chi-squared scaled by its degrees of freedom, you get a t:

T=\frac{Z}{\sqrt{V/\nu}}, \quad Z\sim\mathcal{N}(0,1),\ V\sim\chi^{2}_{\nu}

Plug in the sample mean and sample variance of n iid Normal draws and you get the classic t-statistic. The degrees of freedom $\nu$ = n − 1 capture how much the uncertainty about $\sigma$ inflates the distribution of the mean.

Compare the assumptions and event probabilities in normal vs Student’s t, or use the distribution chooser.

Follow the numerator and denominator

The histogram builds each result from independent Z and V. For example, Z=1 and V=2 at ν=4 give T=1/√(2/4)=√2≈1.4142. If V were 0.08 instead, the same Z would give T≈7.0711. A small denominator is enough to create an extreme ratio; it does not require an extreme normal numerator.

For an actual normal sample, T=(sample mean−μ₀)/(S/√n), with S² computed using n−1. The five illustrative observations 4,5,6,7,8 have mean 6 and S²=2.5. Against μ₀=5 the statistic is √2, with four degrees of freedom. Replacing S by a known population σ would give a different statistic and a standard-normal reference under the null.

A quantile is a cutoff for a specified event

At ν=4, the 97.5th percentile is approximately 2.776445. Symmetry puts 95% of this reference distribution between −2.776445 and +2.776445. At ν=30 that cutoff falls to approximately 2.042272, still above the standard-normal cutoff 1.959964. The probability panel computes the two-sided event |T|≥c, so its probability is twice the upper tail for c≥0.

The scale parameter in this standard t law is one, but its variance is ν/(ν−2) when ν>2. At ν=3, its standard deviation is √3, not one. At ν=2 the variance is infinite, and at ν=1 the mean is undefined. A scale-matched comparison must not silently claim matching variances.

Make a prediction

Will replacing σ with a sample SD produce an exact t distribution for any independent observations?

Explore the answer

No. The normal-sample construction gives an independent mean and sample variance with the required reference laws. Skewed or heavy-tailed observations need not satisfy that construction. The confidence-interval experiment shows the resulting small-sample coverage mismatch.

References

NIST: t distribution lists its moments and shape relationships. NIST: testing a process mean gives the estimated-SD statistic, and its critical-value table distinguishes one-sided and two-sided cutoffs. Continue to confidence intervals to check the repeated-sampling interpretation.

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