Quantile function of Student's T-distribution and Normal distribution 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?
 A: 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.)
A: Let's compare any Student's distribution $t_{\nu}$ and any normal distribution $N(a,\sigma^2)$.
Student's distribution $t_{\nu}$ has density $$f_{\nu}(t) \sim C_{
\nu} t^{- \nu -1}, t \to +\infty$$
(see en.wikipedia.org/wiki/Student%27s_t-distribution) and normal distribution $N(a,\sigma^2) $has density
$$f_{a, \sigma^2}(t) = C_{a, \sigma^2} e^{-\frac{(x- a)^2}{2\sigma^2}}.$$
Hence $$P(t_{\nu} > x) >   P(N(a,\sigma^2)  > x) \text{ ${}$(1)}$$
for $x \ge x_0(\mu, a, \sigma)$.
But (1) may be false for $x$ which are not big enough.
For example, $(n+100)^{100}$ grows slowlier, than $e^n$, but $(n+100)^{100}  > e^n$ for $n \le 100$.
A: Your focus on variance equality could be smart most of the time, but here, it leads to some errors.
First, when you calculate confidence intervals in the Gaussian framework, knowing or not the population variance, you will have the quantile of the standard normal or the quantile of the student with df given by the sample size minus 1. Nothing to do with the variance equality you require.
Second, you are right about the comparison, but you do not compare a normal and a student. You are competing a normal with a scaled student :) Y is not a student in your example.
