Why the Square of a t Variable Is F

Why Squaring a t Variable Produces an F Variable

A Student t variable can be positive or negative, but its square is always nonnegative. Squaring also makes the two-sided nature of many t calculations explicit: a large squared value means that the original t value was far from zero in either direction. The resulting distribution is not another t distribution. It is an F distribution with numerator degrees of freedom 1 and denominator degrees of freedom equal to the original t degrees of freedom.

This connection is useful for translating probabilities and critical values, and for recognizing that a two-sided t test can be expressed through an F statistic. The identity is exact when the t variable has the stated Student t distribution; it does not require an approximation.

Definition: A standard Student t variable with $\nu$ degrees of freedom can be represented as $$ T=\frac{Z}{\sqrt{U/\nu}}, $$ where $Z\sim N(0,1)$, $U\sim\chi^2_\nu$, and $Z$ and $U$ are independent. An F variable with degrees of freedom $r$ and $s$ is a ratio $$ F=\frac{U_1/r}{U_2/s}, $$ where $U_1\sim\chi^2_r$ and $U_2\sim\chi^2_s$ are independent. We write its distribution as $F_{r,s}$.

The Distributional Identity

Start with the representation of $T$. Squaring gives $T^2=Z^2/(U/\nu)$. Because $Z$ is standard normal, its square has a chi-square distribution with 1 degree of freedom. It is independent of $U$, since $Z$ and $U$ are independent. The squared numerator and denominator therefore have exactly the form in the definition of an F variable.

Z^2\sim\chi^2_1,\qquad T^2=\frac{Z^2/1}{U/\nu} \sim F_{1,\nu}.

Theorem: If $T\sim t_\nu$, then $T^2\sim F_{1,\nu}$. The numerator degrees of freedom are 1, and the denominator degrees of freedom are $\nu$. This conclusion depends on $T$ having a Student t distribution; it does not hold for an arbitrary random variable merely because that variable is squared.

The degrees of freedom follow directly from the representation: the squared standard normal supplies the numerator chi-square with 1 degree of freedom, while the chi-square variable in the t denominator supplies the denominator with $\nu$ degrees of freedom. The order matters: $T^2$ has distribution $F_{1,\nu}$, not $F_{\nu,1}$.

Write the t representation.
Use $T=Z/\sqrt{U/\nu}$, with $Z$ standard normal, $U$ chi-square with $\nu$ degrees of freedom, and independence.

Square the numerator and denominator.
Then $T^2=Z^2/(U/\nu)$, and $Z^2$ is chi-square with 1 degree of freedom.

Match the F definition.
The ratio is an independent chi-square with 1 degree of freedom divided by 1, over a chi-square with $\nu$ degrees of freedom divided by $\nu$. Hence $T^2\sim F_{1,\nu}$.

Translating Events and Tails

The identity gives a direct way to translate an event involving $T^2$ into an F probability. For $x>0$, the event $T^2>x$ is the same event as $|T|>\sqrt{x}$. Since a Student t distribution is symmetric about zero and continuous, its positive and negative tails have equal probability. Consequently, an upper-tail probability for $T^2$ is twice the corresponding upper-tail probability for $T$.

P(T^2>x) =P(|T|>\sqrt{x}) =2P(T>\sqrt{x}) =P(F_{1,\nu}>x).

The same reasoning applies to intervals. If $0\leq a<b$, then $a<T^2<b$ means $\sqrt{a}<|T|<\sqrt{b}$. The probability is the probability in the positive interval plus the equal probability in the negative interval.

P(a<T^2<b) =2\left[P(T>\sqrt{a})-P(T>\sqrt{b})\right] =P(a<F_{1,\nu}<b).

These equalities are distributional statements. In a numerical calculation, a t table may give a one-sided tail, while an F table may give an upper-tail probability for the squared statistic. The two table entries correspond only after the threshold is squared and the tail convention is checked.

Worked Example: An Upper Tail for a Squared t Variable

Let $T\sim t_4$. Find $P(T^2>2.25)$. A t-distribution calculation gives $P(T>1.50)\approx0.1040$, rounded to four decimal places.

Method. Use the event equivalence and symmetry of the t distribution. First take the square root of the threshold: $$ \sqrt{2.25}=1.50. $$ Thus $T^2>2.25$ is equivalent to $T>1.50$ or $T<-1.50$. The two tail probabilities are equal, so $$ P(T^2>2.25)=2P(T>1.50)\approx2(0.1040)=0.2080. $$ Equivalently, $T^2\sim F_{1,4}$, and this is the upper-tail probability $P(F_{1,4}>2.25)$.

Check. The two disjoint tails each have probability approximately $0.1040$; their sum is $0.2080$. The common mistake is to report $0.1040$, counting only the positive t tail even though squaring includes both signs.

Worked Example: Probability Between Two Squared Thresholds

Let $T\sim t_5$. Find $P(1<T^2<4)$. Use the t tail probabilities $P(T>1)\approx0.1816$ and $P(T>2)\approx0.0510$, rounded to four decimal places.

Method. Taking square roots shows that the event is $1<|T|<2$. It consists of the positive interval from 1 to 2 and the corresponding negative interval from $-2$ to $-1$. By symmetry, the two intervals have equal probability. For the positive interval, subtract the tail beyond 2 from the tail beyond 1: $$ P(1<T<2)=P(T>1)-P(T>2) \approx0.1816-0.0510=0.1306. $$ Therefore, $$ P(1<T^2<4) =2(0.1306)=0.2612. $$ Using unrounded t tail probabilities before rounding the final result gives approximately $0.2613$, to four decimal places. In F notation this is $P(1<F_{1,5}<4)$.

Check. The probability is positive and is twice the probability in one of the two equal intervals. Using the rounded tail entries gives $0.2612$; using the unrounded probabilities gives $0.2613$. The reported answer, based on unrounded values, is approximately $0.2613$. The calculation should not subtract the two tails and then omit the factor of 2.

Two-Sided t Tests and F Tests

Suppose that under a null hypothesis a test statistic $T$ has a $t_\nu$ distribution, and its observed value is $t_{\text{obs}}$. For a two-sided alternative, values at least as far from zero as the observed value count against the null in either direction. The two-sided p-value is $P(|T|\geq|t_{\text{obs}}|)$. Squaring both sides of the threshold expresses the same event as $T^2\geq t_{\text{obs}}^2$. Under the null, $T^2\sim F_{1,\nu}$, so the two-sided t p-value equals the upper-tail probability of the corresponding F statistic.

p_{\text{two-sided}} =P(|T|\geq|t_{\text{obs}}|) =P(T^2\geq t_{\text{obs}}^2) =P(F_{1,\nu}\geq t_{\text{obs}}^2).

This equivalence concerns a two-sided t test and its squared statistic. It does not mean that a one-sided t test has the same p-value as the upper tail of the squared statistic. Squaring removes the sign, so the F statistic cannot distinguish a large positive t value from a large negative one.

Worked Example: Converting a Two-Sided t p-Value to an F Tail

Under a null hypothesis, a test statistic has distribution $t_9$ and its observed value is $-1.80$. The t upper-tail probability at $1.80$ is approximately $0.0527$, rounded to four decimal places. Find the two-sided p-value and identify the equivalent F probability.

Method. The observed statistic is negative, but a two-sided test counts departures in either direction. By symmetry, the tail below $-1.80$ equals the tail above $1.80$. Therefore, $$ p_{\text{two-sided}} =2P(T>1.80) \approx2(0.0527)=0.1054. $$ Squaring the observed value gives $$ (-1.80)^2=3.24. $$ Since $T^2\sim F_{1,9}$ under the null, the same p-value is $P(F_{1,9}\geq3.24)$, approximately $0.1054$.

Check. The squared statistic is positive, and its F upper tail is the two-sided t probability, not just one t tail. The calculation uses the upper-tail probability $0.0527$, not $0.0523$; doubling $0.0527$ gives $0.1054$ to four decimal places.

Critical Values and a Common Exam Trap

The same tail distinction applies when converting critical values. Define $t_{\nu,\alpha}$ as the positive t value with upper-tail probability $\alpha$, so $P(T>t_{\nu,\alpha})=\alpha$. A two-sided t test with total significance level $\alpha$ allocates $\alpha/2$ to each t tail. Its positive critical value is therefore $t_{\nu,\alpha/2}$, and the equivalent upper-tail F critical value is its square.

P\left(F_{1,\nu}>t_{\nu,\alpha/2}^2\right) =\alpha.

A one-sided F upper-tail area of $\alpha$ corresponds to two t tails, each of area $\alpha/2$. This is why the t critical value for a two-sided level-$\alpha$ procedure is squared to obtain the corresponding F critical value. It would be incorrect to square $t_{\nu,\alpha}$ when converting a two-sided t procedure at level $\alpha$; that t quantile allocates area $\alpha$ to one tail, not $\alpha/2$.

Common exam trap: If $P(T>c)=0.05$, then $P(T^2>c^2)=0.10$, not $0.05$, because the squared event includes both $T>c$ and $T<-c$. For a two-sided t test with total level $0.05$, use the t upper-tail quantile with tail area $0.025$, then square it to obtain the corresponding F critical value.

A second frequent error is reversing the F degrees of freedom. The numerator comes from $Z^2$, which has 1 degree of freedom; the denominator comes from $U$, which has $\nu$ degrees of freedom. Thus the squared t variable is $F_{1,\nu}$ in that order.

When the Identity Applies

The identity relies on the exact Student t model, including the independence in its normal-over-square-root-chi-square representation. It is not a general rule that squaring any statistic produces an F distribution. In particular, if a statistic does not have a $t_\nu$ distribution under the model being used, the identity cannot be applied just by squaring its observed value.

For a statistic that does have a $t_\nu$ distribution, the identity can simplify a calculation or provide a check: square the observed t value, use numerator degrees of freedom 1 and denominator degrees of freedom $\nu$, and take the F upper tail for a two-sided t probability. The sign of the observed t value affects the direction of a one-sided test, but not this squared-statistic equivalence.

Key takeaway: If $T\sim t_\nu$, then $T^2\sim F_{1,\nu}$. For $x>0$, $P(T^2>x)=2P(T>\sqrt{x})$. The factor of 2 reflects the two t tails; the F tail already combines them.

Check Your Understanding

For each problem, use symmetry and the squared-t/F relationship where appropriate.

Let $T\sim t_1$. Find $P(T^2>1)$, using the fact that $P(T>1)=0.25$. (Answer: $0.5000$)
Let $T\sim t_1$. Given $P(T>1)=0.25$ and $P(T>\sqrt{3})=1/6$, find $P(1<T^2<3)$. (Answer: $0.1667$)
If $T\sim t_{12}$, state the distribution of $T^2$, including the numerator and denominator degrees of freedom. (Answer: $F_{1,12}$)
A two-sided t test has total significance level $0.10$ and $\nu=8$ degrees of freedom. State which upper-tail t quantile must be squared to obtain the equivalent F critical value, and give the F degrees of freedom. (Answer: square the t quantile with upper-tail area $0.05$; use $F_{1,8}$)
Under a null hypothesis, $T\sim t_\nu$ and the observed value is $-c$, where $c>0$. Express the two-sided p-value as an upper-tail probability for an F variable. (Answer: $P(F_{1,\nu}\geq c^2)$)

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