We all know that the t-distribution has a heavier tail than normal distribution (this means for a given percentile level $\alpha$, the quantile of student's T-distribution $q_{t_v}^\alpha$ is greater than the quantile of normal distribution $q_{N(0,1)}^\alpha$, $v$ is the degree of freedom of the t-distribution). But it seems to me that we have this conclusion just because we are comparing apples and oranges, the two distribution don't have the same variance. Why don't compare a normal distribution with a variance of $1$ with a Student's T-distribution with a variance of $10000$ ?
Now, if we compare the quantile of a normal distribution with a Student's T-distribution of same variance, we will have a paradoxical conclusion.
Let $X$ and $Y$ two random variables with mean of $0$ and variance of $1$. $X$ follows the normal distribution $N(0,1)$ while $Y$ follows the Student's T-distribution $t_v$.
As the variance of $Y$ is equal to $1$, $Y$ must equal in distribution to $\sqrt{\frac{v-2}{v}} t_v$ (because the variance of $t_v$ is equal to $\frac{v}{v-2}$ ):
$$Y \overset {d}{=} \sqrt{\frac{v-2}{v}} t_v$$
Now, we calculate the quantile at the level $\alpha$ for both $X$ and $Y$:
$$
P(X \leq q_{X}^\alpha) = P(N(0,1) \leq q_{X}^\alpha)=\alpha \Rightarrow q_{X}^\alpha=F_{N(0,1)}^{-1}(\alpha)
$$
and
$$
P(Y \leq q_{}^\alpha) = P(\sqrt{\frac{v-2}{v}} t_v \leq q_{Y}^\alpha)=\alpha
$$
$$
\Rightarrow \sqrt{\frac{v}{v-2}} q_{Y}^\alpha = F_{t_v}^{-1}(\alpha)
$$
$$
\Rightarrow q_{Y}^\alpha = \sqrt{\frac{v-2}{v}} F_{t_v}^{-1}(\alpha)
$$
We take $\alpha = 0.95$ for example, the quantile $q_{X}^\alpha \approx 1.64 $ is always greater than $q_{Y}^\alpha$, $\forall v$ (for example, if $v = 5$, $q_{Y}^\alpha = \sqrt{\frac{5-2}{2}} F_{t_5}^{-1}(0.95) \approx 1.56$.
In practice, we usually have observations from a random variable $Z$ and we can easily compute the variance and mean of this variable, so the variance and mean are known. From my demonstration, the quantile of $Z$ is smaller if we suppose $Z$ follows the Student's T-distribution rather than Normal distribution.
This conclusion seems contradictory to what we read about the heavier tail of Student's T distribution, doesn't it?
Best Answer
Further to the calculations in @BotnakovN's answer, you can always compare a $t$-distribution to the Normal distribution with the same mean and variance, or even compare both distributions' z-scores and obviate any "different variance" objection. You then find the $t$ tails are heavier in the sense their decay is subexponential (indeed, they're fat), while the Gaussian tails decay superexponentially. (See also all definitions here.)