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

Reminders you will lean on throughout (all proven in the parent note):

Level 1 — Recognition

Goal: name the distribution and its degrees of freedom. No arithmetic beyond counting.

Exercise 1.1. You add together the squares of $7$ independent standard normals: $W = Z_{1}^{2} + \dots + Z_{7}^{2}$ . What distribution does $W$ follow, and what is its degrees of freedom (df)?

Recall Solution 1.1

WHAT the object is: a sum of squared independent standard normals. That is the exact definition of a chi-squared. Counting df: df = how many independent squared normals we summed. We summed $7$ . $W \sim χ_{7}^{2}, df = 7.$ Sanity check via the mean: $E [W] = k = 7$ — each $Z_{i}^{2}$ contributes mean $1$ , and $7 \times 1 = 7$ . ✓

Exercise 1.2. A statistic is written as $\frac{Z}{V /9}$ where $Z \sim N (0, 1)$ and $V \sim χ_{9}^{2}$ are independent. Name the distribution and its df.

Recall Solution 1.2

Match the template. The $t$ definition is $t_{k} = \frac{Z}{V / k}$ : a normal on top, and on the bottom a chi-squared divided by its own df, then square-rooted. Here the bottom is $V /9$ with $V \sim χ_{9}^{2}$ , so $k = 9$ . $\frac{Z}{V /9} \sim t_{9}, df = 9.$

Exercise 1.3. Identify: $\frac{U /3}{V /8}$ with $U \sim χ_{3}^{2}$ , $V \sim χ_{8}^{2}$ independent. What is it, and which df is the numerator df?

Recall Solution 1.3

Match the template. $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ — each chi-squared divided by its own df, then a ratio. Top uses $U \sim χ_{3}^{2}$ so $d_{1} = 3$ ; bottom uses $V \sim χ_{8}^{2}$ so $d_{2} = 8$ . $\frac{U /3}{V /8} \sim F_{3, 8}, numerator df d_{1} = 3, denominator df d_{2} = 8.$

Level 2 — Application

Goal: substitute into $E$ and $Var$ formulas.

Exercise 2.1. Find $E [χ_{11}^{2}]$ and $Var (χ_{11}^{2})$ .

Recall Solution 2.1

Formulas: $E [χ_{k}^{2}] = k$ and $Var (χ_{k}^{2}) = 2 k$ . WHY these hold: each single $Z^{2}$ has $E [Z^{2}] = 1$ and $Var (Z^{2}) = 2$ ; independence lets means and variances add over the $k$ terms. $E [χ_{11}^{2}] = 11, Var (χ_{11}^{2}) = 2 \cdot 11 = 22.$

Exercise 2.2. For $t_{6}$ , compute $E [t_{6}]$ and $Var (t_{6})$ .

Recall Solution 2.2

Mean. $t$ is symmetric about $0$ (numerator $Z$ is symmetric, denominator is a positive number), so for $k > 1$ , $E [t_{k}] = 0$ . Here $k = 6 > 1$ : $E [t_{6}] = 0$ . Variance. $Var (t_{k}) = \frac{k}{k - 2}$ for $k > 2$ . With $k = 6$ : $Var (t_{6}) = \frac{6}{6 - 2} = \frac{6}{4} = 1.5.$ Note $1.5 > 1$ : heavier spread than $N (0, 1)$ (whose variance is $1$ ), because the random denominator adds uncertainty.

Exercise 2.3. For $F_{4, 10}$ , compute $E [F_{4, 10}]$ .

Recall Solution 2.3

Formula: $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2}$ for $d_{2} > 2$ . Only the denominator df matters for the mean. Here $d_{2} = 10$ : $E [F_{4, 10}] = \frac{10}{10 - 2} = \frac{10}{8} = 1.25.$ Interpretation: near $1$ , as expected — $F$ compares two per-df variances, and under equal variances each part has mean $\approx 1$ .

Level 3 — Analysis

Goal: choose the correct df yourself by tracking constraints and estimated parameters.

Exercise 3.1. A sample of size $n = 13$ is drawn from $N (μ, σ^{2})$ . The true $σ^{2} = 5$ and the sample gives $s^{2} = 6$ , where $s^{2} = \frac{1}{n - 1} \sum (x_{i} - \overset{x}{ˉ})^{2}$ . State the exact distribution of $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ and compute its observed value.

Recall Solution 3.1

Distribution. The key statistics result: for a sample from a normal, $\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n - 1}^{2} .$ WHY df $= n - 1$ , not $n$ : computing $\overset{x}{ˉ}$ forces the deviations to satisfy $\sum (x_{i} - \overset{x}{ˉ}) = 0$ . That is one constraint, so one of the $n$ deviations is determined by the other $n - 1$ — one df is lost. See Sample Variance and Bessel's Correction. With $n = 13$ : df $= 12$ , so the distribution is $χ_{12}^{2}$ . Observed value. $\frac{( n - 1 ) s ^{2}}{σ ^{2}} = \frac{12 \cdot 6}{5} = \frac{72}{5} = 14.4.$

Exercise 3.2. A one-sample $t$ -test uses $n = 25$ , $\overset{x}{ˉ} = 103$ , $s = 10$ , testing $μ_{0} = 100$ . Compute the $t$ -statistic and give its reference distribution. Should you use $Z$ or $t$ , and why?

Recall Solution 3.2

Statistic. $t = \frac{x ˉ - μ _{0}}{s / n} = \frac{103 - 100}{10/ 25} = \frac{3}{10/5} = \frac{3}{2} = 1.5.$ Reference distribution. $t_{n - 1} = t_{24}$ . The df comes from the chi-squared hidden inside $s^{2}$ , which carries $n - 1 = 24$ df (same constraint as 3.1). $Z$ or $t$ ? Use $t$ . We did not know the true $σ$ ; we replaced it with the random estimate $s$ . That extra randomness in the denominator demands the fatter-tailed $t_{24}$ , not $Z$ . See Hypothesis Testing.

Exercise 3.3. Two independent samples: sample A has $n_{1} = 7$ , sample B has $n_{2} = 16$ . Under the assumption of equal true variances, you form $F = s_{1}^{2} / s_{2}^{2}$ . What are the numerator and denominator df, and what distribution does $F$ follow?

Recall Solution 3.3

Each sample's spread is a chi-squared. From 3.1's logic, $\frac{( n _{1} - 1 ) s _{1}^{2}}{σ ^{2}} \sim χ_{n_{1} - 1}^{2}$ and $\frac{( n _{2} - 1 ) s _{2}^{2}}{σ ^{2}} \sim χ_{n_{2} - 1}^{2}$ . Form the ratio per df. Under equal $σ^{2}$ , the common $σ^{2}$ cancels, and $\frac{s _{1}^{2}}{s _{2}^{2}} = \frac{[ ( n _{1} - 1 ) s _{1}^{2} / σ ^{2} ] / ( n _{1} - 1 )}{[ ( n _{2} - 1 ) s _{2}^{2} / σ ^{2} ] / ( n _{2} - 1 )} = \frac{U / d _{1}}{V / d _{2}} \sim F_{d_{1}, d_{2}} .$ Df. $d_{1} = n_{1} - 1 = 6$ (numerator, from sample A), $d_{2} = n_{2} - 1 = 15$ (denominator, from sample B). $F = \frac{s _{1}^{2}}{s _{2}^{2}} \sim F_{6, 15} .$

Level 4 — Synthesis

Goal: combine two facts — additivity, the $t^{2} = F$ link, reciprocal symmetry.

Exercise 4.1. $A \sim χ_{4}^{2}$ and $B \sim χ_{9}^{2}$ are independent. What is the distribution of $A + B$ , and what are its mean and variance?

Recall Solution 4.1

Additivity. A $χ_{4}^{2}$ is $4$ squared normals; a $χ_{9}^{2}$ is $9$ more independent squared normals. Appending them gives $4 + 9 = 13$ squared independent normals: $A + B \sim χ_{13}^{2} .$ Mean and variance. $E [A + B] = 13, Var (A + B) = 2 \cdot 13 = 26.$ Cross-check without additivity: means add ( $4 + 9 = 13$ ✓) and, by independence, variances add ( $8 + 18 = 26$ ✓).

Exercise 4.2. A random variable $T \sim t_{20}$ . Find the distribution of $T^{2}$ , and hence compute $E [T^{2}]$ two different ways to confirm they agree.

Recall Solution 4.2

The $t^{2} = F$ link. Write $t_{k} = \frac{Z}{V / k}$ , so $T^{2} = \frac{Z ^{2}}{V / k} = \frac{Z ^{2} /1}{V / k} .$ Since $Z^{2} \sim χ_{1}^{2}$ (one squared normal, df $1$ ), the top is a $χ_{1}^{2}$ divided by its df $1$ , and the bottom is $χ_{k}^{2}$ divided by $k$ . That is exactly an $F$ with $d_{1} = 1$ , $d_{2} = k$ : $T^{2} \sim F_{1, 20} .$ $E [T^{2}]$ way 1 — via $F$ : $E [F_{1, 20}] = \frac{d _{2}}{d _{2} - 2} = \frac{20}{18} = \frac{10}{9} \approx 1.1111$ . $E [T^{2}]$ way 2 — via $t$ : $E [T^{2}] = Var (T) + (E [T])^{2} = \frac{k}{k - 2} + 0 = \frac{20}{18} = \frac{10}{9}$ . Both give $\frac{10}{9}$ . ✓ (They must agree — $T^{2}$ is the same random variable either way.)

Exercise 4.3. You computed $F_{5, 12} = 0.4$ for the ratio $s_{A}^{2} / s_{B}^{2}$ , but a table only lists upper-tail values of $F_{12, 5}$ . Use reciprocal symmetry to convert.

Recall Solution 4.3

Reciprocal symmetry. Swapping numerator and denominator inverts the value and swaps the df: $\frac{1}{F _{d_{1}, d_{2}}} \sim F_{d_{2}, d_{1}} .$ So if $F_{5, 12} = 0.4$ , then $\frac{1}{F _{5, 12}} = \frac{1}{0.4} = 2.5 \sim F_{12, 5} .$ Interpretation: a small $F_{5, 12}$ (sample B more variable) becomes a large $F_{12, 5} = 2.5$ once you flip which variance is on top — the same evidence, re-expressed for the table you have.

Level 5 — Mastery

Goal: reason about limiting behaviour and structural links.

Exercise 5.1. Explain, then compute the limit: what happens to $Var (t_{k})$ as $k \to \infty$ , and what distribution does $t_{k}$ approach? Give the numerical variances for $k = 3, 10, 100$ to show the trend.

Recall Solution 5.1

WHY there is a limit. In $t_{k} = \frac{Z}{V / k}$ , the term $V / k$ is the average of $k$ independent $Z_{i}^{2}$ 's (since $V = \sum Z_{i}^{2}$ ). By the law of large numbers, that average converges to its mean $E [Z^{2}] = 1$ . So the random denominator stops wobbling and locks to $1$ , leaving just $Z$ . Hence $t_{k} \to N (0, 1)$ . Variance limit. $lim_{k \to \infty} Var (t_{k}) = lim_{k \to \infty} \frac{k}{k - 2} = 1,$ matching $Var (Z) = 1$ . Trend (see figure): $Var (t_{3}) = \frac{3}{1} = 3, Var (t_{10}) = \frac{10}{8} = 1.25, Var (t_{100}) = \frac{100}{98} \approx 1.0204.$ The excess over $1$ shrinks: $3 \to 1.25 \to 1.02$ — the tails thin out toward the normal.

Figure — Chi-squared, t, F distributions — definition, degrees of freedom

Exercise 5.2. Consider $F_{d_{1}, d_{2}}$ with fixed $d_{1}$ and $d_{2} \to \infty$ . Show $E [F] \to 1$ and explain structurally what the object becomes. Then state what a single $χ_{d_{1}}^{2} / d_{1}$ is doing.

Recall Solution 5.2

Mean limit. $lim_{d_{2} \to \infty} E [F_{d_{1}, d_{2}}] = lim_{d_{2} \to \infty} \frac{d _{2}}{d _{2} - 2} = 1.$ Structural reason. $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}}$ . As $d_{2} \to \infty$ , the denominator $V / d_{2}$ is an average of $d_{2}$ squared normals → converges to $E [Z^{2}] = 1$ (law of large numbers again). The ratio collapses to just the numerator: $F_{d_{1}, d_{2}} d_{2} \to \infty \frac{U}{d _{1}}, U \sim χ_{d_{1}}^{2} .$ So $F$ with a huge denominator df is essentially a single chi-squared divided by its df — a per-df measure of spread with mean $1$ . This is why, with a large second sample, comparing variances via $F$ reduces to just weighing the first sample's spread. Foundational for ANOVA.

Exercise 5.3. A $t$ -test at $n = 2$ (so $k = 1$ ) gives $t_{1}$ . Its variance formula $\frac{k}{k - 2}$ gives $\frac{1}{- 1} = - 1$ . A variance cannot be negative — explain the degenerate case honestly and say what is true instead.

Recall Solution 5.3

The formula does not apply. $Var (t_{k}) = \frac{k}{k - 2}$ is stated only for $k > 2$ . For $k = 1$ the formula is out of its domain — plugging in gives a meaningless $- 1$ , which is a signal that the variance does not exist, not a real value. What is actually true. For $k = 1$ , the random denominator $V /1 = ∣ Z_{2} ∣$ can be arbitrarily close to $0$ , so $t_{1} = \frac{Z _{1}}{∣ Z _{2} ∣}$ has tails so heavy that neither its mean nor its variance exists (both integrals diverge). In fact $t_{1}$ is the standard Cauchy distribution. Honest boundary summary:

Mean of $t_{k}$ exists only for $k > 1$ .
Variance of $t_{k}$ exists only for $k > 2$ ; for $k = 2$ , $\frac{k}{k - 2} = \frac{2}{0}$ is undefined (infinite variance); for $k = 1$ even the mean fails. Lesson: always check the domain condition attached to a moment formula before substituting.

Recall One-line self-test

Which df controls the mean of an $F$ ? ::: The denominator df $d_{2}$ , via $E = \frac{d _{2}}{d _{2} - 2}$ ( $d_{2} > 2$ ). Why is $t^{2} \sim F_{1, k}$ ? ::: Because $Z^{2} \sim χ_{1}^{2}$ , so squaring puts a df- $1$ chi-squared over $χ_{k}^{2} / k$ — the $F$ template. Why does $t_{k} \to N (0, 1)$ ? ::: The denominator $V / k$ averages $k$ squared normals to its mean $1$ , freezing to a constant.