Chi-squared, t, F distributions — definition, degrees of freedom: Question bank

This is a concept-trap page. No heavy arithmetic — every item hunts a misconception or a boundary case about how the three sampling distributions are built, why the degrees of freedom come out the way they do, and what happens at the edges (one degree of freedom, infinitely many, equal variances, zero deviation).

Before you begin, recall the three builders from the parent note:

Recall The one-line definitions you are being tested against

$χ_{k}^{2} = Z_{1}^{2} + \dots + Z_{k}^{2}$ — a sum of squared independent standard normals.
$t_{k} = \frac{Z}{V / k}$ with $Z \sim N (0, 1)$ , $V \sim χ_{k}^{2}$ independent — a normal over a scaled square root of a $χ^{2}$ .
$F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ — a ratio of two $χ^{2}$ 's, each first divided by its own df.

Cover the answer, commit to a reason, then reveal.

True or false — justify

A chi-squared variable can be negative.

False — it is a sum of squares

Z_{i}^{2}

, and squares are never negative; the whole distribution lives on

[0, \infty)

χ_{k}^{2}

is symmetric like the normal.

False — it is right-skewed (a long tail to the right), because squaring piles probability near zero and stretches large positive values; only as

k

grows does it slowly look bell-shaped.

The

t

distribution is symmetric about

0

True — the numerator

Z

is symmetric about

0

and the positive denominator

V / k

does not favour either sign, so

t

and

- t

are equally likely.

The

F

distribution is symmetric.

False — it is a ratio of two non-negative quantities, so it lives on

[0, \infty)

and is right-skewed; there is no reflection symmetry, only the reciprocal symmetry

1/ F_{d_{1}, d_{2}} \sim F_{d_{2}, d_{1}}

t_{k}

has larger variance than

N (0, 1)

for every finite

k > 2

True —

Var (t_{k}) = \frac{k}{k - 2} > 1

for all finite

k > 2

, because the random denominator (an estimated

σ

) injects extra spread; it shrinks to

1

only in the limit

k \to \infty

Adding two independent chi-squareds gives another chi-squared.

True —

χ_{a}^{2} + χ_{b}^{2} = χ_{a + b}^{2}

, since you are just appending more independent squared normals to the sum; the df simply add.

Adding two independent

t

variables gives another

t

False — there is no such closure rule; the numerators are normal but the random denominators do not combine into a single scaled

χ^{2}

, so the sum is not a

t

χ^{2}

with

k

df has variance

k

False — its mean is

k

; its variance is

2 k

, because each

Z_{i}^{2}

contributes variance

2

(from

E [Z^{4}] - 1 = 3 - 1

) and variances add over the

k

independent terms.

F_{d_{1}, d_{2}}

and

F_{d_{2}, d_{1}}

are the same distribution.

False — swapping which sample sits on top matters; they are equal only through the reciprocal relation

F_{d_{1}, d_{2}} = 1/ F_{d_{2}, d_{1}}

, not as identical distributions (unless

d_{1} = d_{2}

The mean of

F_{d_{1}, d_{2}}

is exactly

1

False — it is

\frac{d _{2}}{d _{2} - 2}

(for

d_{2} > 2

), slightly above 1; it only approaches 1 as

d_{2} \to \infty

because then the denominator

χ^{2}

per-df stabilises to exactly 1.

Spot the error

"For a sample of size

n

\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n}^{2}

Wrong df — it is

χ_{n - 1}^{2}

. Estimating

\overset{x}{ˉ}

forces

\sum (x_{i} - \overset{x}{ˉ}) = 0

, one constraint, so one degree of freedom is spent; free directions

= n - 1

F = \frac{U}{V}

where

U \sim χ_{d_{1}}^{2}

V \sim χ_{d_{2}}^{2}

The div-by-df is missing —

F = \frac{U / d _{1}}{V / d _{2}}

. Without dividing each by its own df, the ratio's typical size depends on

d_{1}, d_{2}

and would not sit near 1 under equal variances.

"Since

\overset{x}{ˉ}

is normal, always standardise it with

Z = \frac{x ˉ - μ}{s / n}

You wrote

s

(an estimate) but called it

Z

. Once

σ

is replaced by the random

s

, the ratio follows

t_{n - 1}

, not

N (0, 1)

; see Sample Variance and Bessel's Correction.

t = \frac{Z}{V / k}

where

Z

and

V

come from the same data, so they're dependent — but that's fine."

The definition requires independence of

Z

and

V

. It works out because for a normal sample

\overset{x}{ˉ}

(giving

Z

) and

s^{2}

(giving

V

) are provably independent; without that fact the derivation collapses.

t^{2} \sim F_{k, 1}

Reversed —

t^{2} = \frac{Z ^{2} /1}{V / k} = F_{1, k}

. The squared normal

Z^{2} \sim χ_{1}^{2}

is the numerator (df

= 1

), and

V \sim χ_{k}^{2}

is the denominator (df

= k

"Chi-squared per degree of freedom,

χ_{k}^{2} / k

, has mean

k

Its mean is

1

, not

k

— dividing a mean-

k

variable by

k

gives mean

1

. This normalisation is exactly why each part of

F

is put on a per-df footing.

"Degrees of freedom must be a whole number equal to the sample size."

Df is data minus estimated parameters, not the sample size; and in general (e.g. Welch's

t

, Gamma Distribution-based definitions) df need not even be an integer.

Why questions

Why do we square the

Z_{i}

instead of just summing them?

A plain sum

\sum Z_{i}

is again normal and averages to zero, so positive and negative misses cancel; squaring measures total magnitude of deviation regardless of sign — a genuine "spread" score.

Why does

t_{k} \to N (0, 1)

k \to \infty

As df grow,

V / k \to 1

by the law of large numbers, so the random denominator stops wobbling and freezes at 1, leaving

t = Z /1 = Z

Why does the

t

have fatter tails than the normal?

The denominator

V / k

is random; occasionally it is small by chance, which inflates

t

to extreme values more often than a fixed denominator would — heavier tails.

Why divide each

χ^{2}

by its own df in

F

A raw

χ_{k}^{2}

has mean

k

, so a bare ratio would depend on the df sizes; per-df normalisation makes each part average

\approx 1

, so

F \approx 1

signals equal variances — the interpretation used in ANOVA.

Why is the df

n - 1

and not

n

for the sample variance?

Computing

\overset{x}{ˉ}

from the same data imposes the linear constraint

\sum (x_{i} - \overset{x}{ˉ}) = 0

; once

n - 1

deviations are known the last is forced, so only

n - 1

pieces are truly free.

Why must the two

χ^{2}

's in

F

be independent?

The clean ratio distribution only holds if numerator and denominator vary independently; if they shared randomness (e.g. same sample), the ratio's cancellations would distort its shape and the

F

tables would not apply.

Why is

F

the natural tool for comparing two variances rather than a difference

s_{1}^{2} - s_{2}^{2}

A ratio is scale-free — under equal variances it centres near 1 regardless of the common

σ^{2}

— whereas a difference still depends on the unknown units and scale.

Why does

E [Z^{2}] = 1

and not

0

Because

E [Z^{2}] = Var (Z) + (E [Z])^{2} = 1 + 0 = 1

; the mean of the squared value picks up the variance, even though

E [Z] = 0

Edge cases

What is

χ_{1}^{2}

— the smallest interesting case?

It is just

Z^{2}

, the square of a single standard normal: heavily right-skewed with a spike near

0

(small

Z

's are common) and mean

1

, variance

2

Does

t_{k}

have a mean when

k = 1

No —

t_{1}

is the Cauchy distribution; its tails are so heavy that the mean integral does not converge, which is why

E [t_{k}] = 0

is stated only for

k > 1

Does

t_{k}

have a finite variance when

k = 2

No —

Var (t_{k}) = \frac{k}{k - 2}

blows up as

k \to 2^{+}

, so the variance is finite only for

k > 2

; at

k = 1, 2

the tails are too fat.

What does

F_{1, 1}

look like at the extreme?

Both df equal 1, so it is

\frac{Z _{1}^{2}}{Z _{2}^{2}}

— a ratio of two squared normals; extremely heavy-tailed with no finite mean, since

d_{2} = 1

fails the

d_{2} > 2

requirement.

What happens to

F_{d_{1}, d_{2}}

when the two population variances are truly equal?

The two normalised variance estimates are each about 1, so

F

concentrates near 1; this is precisely the null-hypothesis behaviour exploited in Hypothesis Testing and ANOVA.

If all sample values are identical, what is

s^{2}

and the

χ^{2}

statistic?

Then every

x_{i} - \overset{x}{ˉ} = 0

, so

s^{2} = 0

and

\frac{( n - 1 ) s ^{2}}{σ ^{2}} = 0

— a legitimate boundary value at the very left edge of the

χ^{2}

support, though of probability zero for continuous data.

k \to \infty

, what shape does a standardised

χ_{k}^{2}

approach?

Since

χ_{k}^{2}

is a sum of

k

i.i.d. terms, the Central Limit Theorem applies:

\frac{χ _{k}^{2} - k}{2 k} \to N (0, 1)

, so the skew slowly washes out and it looks normal.

What is the largest tail a

t

can have, and when?

The heaviest is at

k = 1

(Cauchy); tails get progressively lighter as

k

increases, converging to the thin normal tails as

k \to \infty