Student’s t distribution

The normal distribution page introduced standardization when the population scale is fixed. Let X_1,\ldots,X_n be independent variables with common distribution \mathcal N(\mu,\sigma^2), and define

\overline X\equiv\frac1n\sum_{i=1}^nX_i

and

S^2\equiv\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline X)^2.

Then

\frac{\overline X-\mu}{S/\sqrt n}\sim t_{n-1}.

Student’s t distribution can be defined from two independent random variables. Let Z\sim\mathcal N(0,1) and U\sim\chi^2_\nu, with Z\perp\!\!\!\perp U, and define

T\equiv\frac{Z}{\sqrt{U/\nu}}.

Then

T\sim t_\nu, \qquad \nu>0.

The parameter \nu is the degrees of freedom. Because U/\nu can be close to zero, the t_\nu distribution assigns more probability to extreme values than the standard normal distribution when \nu is finite.

The distribution is symmetric about zero:

f_T(-t)=f_T(t).

Its mean is zero only when \nu>1, and its variance is finite only when \nu>2.

Comparing central intervals

The middle 95 percent of a standard normal distribution lies between approximately -1.96 and 1.96. For t_5, the middle 95 percent lies between approximately

-2.57\text{ and }2.57.

The wider interval reflects heavier tails. A value of 2.2 is outside the central normal interval but inside the central t_5 interval.

As the degrees of freedom increase, the t family approaches the standard normal family.

reference	lower .025 quantile	upper .975 quantile
t_5	about -2.57	about 2.57
t_{10}	about -2.23	about 2.23
t_{30}	about -2.04	about 2.04
standard normal	about -1.96	about 1.96

The table makes the limiting pattern visible without treating degrees of freedom as a generic scale parameter.

Calculating a two-sided tail probability

Suppose the observed standardized value is |t|=2.5 with 5 degrees of freedom. Symmetry gives

\mathbb{P}(|T|\geq2.5) =2\mathbb{P}(T\geq2.5).

In R,

Code

2 * pt(2.5, df = 5, lower.tail = FALSE)

The probability is approximately .0545. Under the standard normal reference, the corresponding two-sided tail probability is about .0124.

Interpreting degrees of freedom

In the sample-mean construction, the centered deviations satisfy the linear constraint

\sum_{i=1}^n(X_i-\overline X)=0.

Thus only n-1 centered deviations can vary freely, giving

\nu=n-1.

Other statistics impose different constraints. Thus \nu should not be treated as another name for sample size.

Smaller \nu represents greater uncertainty about the scale used in the standardized quantity. The reference distribution then places more probability in the tails. Larger \nu reduces that extra tail mass.

Inspecting several references in base R

Code

degrees_freedom <- c(5, 10, 30)

data.frame(
  df = degrees_freedom,
  lower = qt(.025, df = degrees_freedom),
  upper = qt(.975, df = degrees_freedom)
)

These quantiles are properties of the reference distributions, not estimates of the lexical decision effect itself.

Keeping two uses separate

The t family may be used as a reference distribution for a standardized estimator. It may also be used as a model for response deviations with heavier tails. These are different statistical roles.

Choosing a t reference for an estimator does not change the observed response values or make the sample mean resistant to an extreme measurement. A robust response model requires an explicit likelihood choice.

Check your understanding

Which central 95 percent interval is wider, t_5 or t_{30}? Explain using the defining random denominator.
Why does estimating one mean from n values leave n-1 freely varying deviations?
Would |t|=2.5 have more upper tail probability under t_5 or the standard normal reference?
What must be specified before claiming that a t family makes a response model robust?

The t family represents a signed standardized quantity. The next page turns cumulative probability into simulated values.

--- title: "Student's t distribution" --- The [normal distribution page](normal-distribution.qmd) introduced standardization when the population scale is fixed. Let $X_1,\ldots,X_n$ be independent variables with common distribution $\mathcal N(\mu,\sigma^2)$, and define $$ \overline X\equiv\frac1n\sum_{i=1}^nX_i $$ and $$ S^2\equiv\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline X)^2. $$ Then $$ \frac{\overline X-\mu}{S/\sqrt n}\sim t_{n-1}. $$ [**Student's t distribution**](https://online.stat.psu.edu/stat414/Lesson26) can be defined from two independent random variables. Let $Z\sim\mathcal N(0,1)$ and $U\sim\chi^2_\nu$, with $Z\perp\!\!\!\perp U$, and define $$ T\equiv\frac{Z}{\sqrt{U/\nu}}. $$ Then $$ T\sim t_\nu, \qquad \nu>0. $$ The parameter $\nu$ is the degrees of freedom. Because $U/\nu$ can be close to zero, the $t_\nu$ distribution assigns more probability to extreme values than the standard normal distribution when $\nu$ is finite. The distribution is symmetric about zero: $$ f_T(-t)=f_T(t). $$ Its mean is zero only when $\nu>1$, and its variance is finite only when $\nu>2$. ## Comparing central intervals The middle 95 percent of a standard normal distribution lies between approximately $-1.96$ and $1.96$. For $t_5$, the middle 95 percent lies between approximately $$ -2.57\text{ and }2.57. $$ The wider interval reflects heavier tails. A value of 2.2 is outside the central normal interval but inside the central $t_5$ interval. As the degrees of freedom increase, the t family approaches the standard normal family. | reference | lower $.025$ quantile | upper $.975$ quantile | |---|---:|---:| | $t_5$ | about $-2.57$ | about $2.57$ | | $t_{10}$ | about $-2.23$ | about $2.23$ | | $t_{30}$ | about $-2.04$ | about $2.04$ | | standard normal | about $-1.96$ | about $1.96$ | The table makes the limiting pattern visible without treating degrees of freedom as a generic scale parameter. ## Calculating a two-sided tail probability Suppose the observed standardized value is $|t|=2.5$ with 5 degrees of freedom. Symmetry gives $$ \mathbb{P}(|T|\geq2.5) =2\mathbb{P}(T\geq2.5). $$ In R, ```{r} #| eval: false 2 * pt(2.5, df = 5, lower.tail = FALSE) ``` The probability is approximately $.0545$. Under the standard normal reference, the corresponding two-sided tail probability is about $.0124$. ## Interpreting degrees of freedom In the sample-mean construction, the centered deviations satisfy the linear constraint $$ \sum_{i=1}^n(X_i-\overline X)=0. $$ Thus only $n-1$ centered deviations can vary freely, giving $$ \nu=n-1. $$ Other statistics impose different constraints. Thus $\nu$ should not be treated as another name for sample size. Smaller $\nu$ represents greater uncertainty about the scale used in the standardized quantity. The reference distribution then places more probability in the tails. Larger $\nu$ reduces that extra tail mass. ## Inspecting several references in base R ```{r} #| eval: false degrees_freedom <- c(5, 10, 30) data.frame( df = degrees_freedom, lower = qt(.025, df = degrees_freedom), upper = qt(.975, df = degrees_freedom) ) ``` These quantiles are properties of the reference distributions, not estimates of the lexical decision effect itself. ## Keeping two uses separate The t family may be used as a reference distribution for a standardized estimator. It may also be used as a model for response deviations with heavier tails. These are different statistical roles. Choosing a t reference for an estimator does not change the observed response values or make the sample mean resistant to an extreme measurement. A robust response model requires an explicit likelihood choice. ## Check your understanding 1. Which central 95 percent interval is wider, $t_5$ or $t_{30}$? Explain using the defining random denominator. 2. Why does estimating one mean from $n$ values leave $n-1$ freely varying deviations? 3. Would $|t|=2.5$ have more upper tail probability under $t_5$ or the standard normal reference? 4. What must be specified before claiming that a t family makes a response model robust? The t family represents a signed standardized quantity. The [next page](inverse-transform-sampling.qmd) turns cumulative probability into simulated values.