I found in my intro to stats textbook that $t$-distribution approaches the standard normal as $n$ goes to infinity. The textbook gives the density for $t$-distribution as follows, $$f(t)=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\sqrt{n\pi}\Gamma\left(\frac{n}{2}\right)}\left(1+\frac{t^2}{n}\right)^{-\frac{n+1}{2}}$$

I think it might be possible to show that this density converges (uniformly) to the density of normal as $n$ goes to infinity. Given $$\lim_{n\to \infty}\left(1+\frac{t^2}{n}\right)^{-\frac{n+1}{2}}=e^{-\frac{t^2}{2}}$$, it would be great if we can show $$\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\to \sqrt{\frac{n}{2}}$$ as $n\to \infty$, yet I am stuck here. Can someone point out how to proceed or an alternative way to show that $t$-distribution converges to normal as $n\to \infty$.

  • $\begingroup$ why do you this holds true? $\endgroup$
    – Aksakal
    Feb 12, 2020 at 4:47
  • 2
    $\begingroup$ An alternative way begins with the observation that the Student t is a variance mixture of Gaussians. You needn't worry about the Gamma factors, because they simply Normalize the distribution, so you have already answered your question. (Use logs and Taylor's Theorem to demonstrate uniform convergence.) BTW, that Gamma ratio is incorrect. If it were right, then asymptotically $n/2=\Gamma(n/2+1)/\Gamma(n/2)$ would behave like $\sqrt{n+1}/2\times\sqrt{n} /2=\sqrt{n^2+n}/4\approx n/4.$ Evidently $\sqrt{n}/2$ should be $\sqrt{n/2}.$ You can apply Stirling's formula if you aren't convinced. $\endgroup$
    – whuber
    Feb 12, 2020 at 4:53
  • $\begingroup$ Here's the thread on the variance mixture expression: stats.stackexchange.com/questions/52906. $\endgroup$
    – whuber
    Feb 12, 2020 at 4:54
  • 1
    $\begingroup$ You can also show this using Slutsky's theorem: math.stackexchange.com/q/3240536/321264. $\endgroup$ Feb 12, 2020 at 15:28

5 Answers 5


Stirling's approximation gives $$\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right)$$ so

$$\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{e}\right)}^{\frac{n}{2}}}\left(1 + O\left(\tfrac{1}{n}\right)\right)\\= {\sqrt{\frac{\frac{n+1}{2}}{e}}}\left(1+\frac1n\right)^{\frac{n}{2}}\left(1 + O\left(\tfrac{1}{n}\right)\right) \\= \sqrt{\frac{n}{2}} \left(1 + O\left(\tfrac{1}{n}\right)\right)\\ \to \sqrt{\frac{n}{2}}$$ and you may have a slight typo in your question

In fact when considering limits as $n\to \infty$, you should not have $n$ in the solution; instead you can say the ratio tends to $1$ and it turns out here that the difference tends to $0$. Another point is that $\sqrt{\frac{n}{2}-\frac14}$ is a better approximation, in that not only does the difference tend to $0$, but so too does the difference of the squares.

  • $\begingroup$ as a note, $\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^{\frac{n}{2}} = e^{\frac{1}{2}}$. $\endgroup$
    – zyxue
    Sep 5, 2021 at 16:03

A generalization uncovers a fundamental idea. One nice thing about it is how it circumvents calculation altogether: the Gamma functions don't play any role and, in fact, neither do the specific expressions for the Normal and Chi-squared pdfs.

Recall that the Student $t$ distribution with $\nu$ degrees of freedom originates (both historically, pedagogically, and from a basic statistical standpoint) as the ratio

$$t_\nu = \frac{Z}{\sqrt{S_\nu^2/\nu}}$$

where $Z$ has a standard Normal distribution and $S^2$ is a random variable independent of $Z$ with a $\chi^2(\nu)$ distribution. (This characterization suffices to derive the probability density function proportional to

$$f_\nu(t) \propto \left(1 + \frac{t^2}{\nu}\right)^{-(\nu+1)/2}$$

for $\nu \in \{1,2,3,\ldots\};$ this is then generalized by allowing $\nu$ to be any positive real number. However, we will not need this detail; I present it only to make an explicit connection with how the question is framed.)

Generalization Part 1

Let $Z$ instead be a random variable with any distribution. Later I will want to work with its logarithm, so for this purpose use the indicator function $\mathcal I$ to split $Z$ into its negative, zero, and positive parts:

$$Z = -\mathcal{I}(Z\lt 0)(-Z) + \mathcal{I}(Z=0)Z + \mathcal{I}(Z\gt 0)Z = -Z_{-} + Z_0 + Z_{+}.$$

The fraction $t_\nu$ analogously splits into three parts by dividing each term by $\sqrt{S_\nu^2/\nu}.$ The part with numerator $Z_0$ is identically $0$ and the other parts are expressed as ratios with strictly positive random variables $Z_{-}$ and $Z_{+}$ in their numerators. These are the ratios we need to analyze.

Generalization Part 2

Let us suppose $S_\nu^2$ is a sequence of positive random variables that, for sufficiently large $\nu,$ have finite variances $v^2_\nu$ and (therefore) have finite means $m_\nu$ such that

$$\lim_{\nu\to\infty} \frac{m_\nu}{\nu}=1$$


$$\lim_{\nu\to\infty} \frac{v^2_\nu}{\nu^2} = 0.$$

(Both are well-known, easily-established properties of Chi-squared distributions.) This is just a specific way of stipulating that $S_\nu^2$ tends to get more and more concentrated (relative to its location) around the value $\nu$ as $\nu$ increases, but equivalently it shows that $S_\nu^2/\nu$ tends to $1$ while its variance tends to $0.$ Chebyshev's Inequality then implies an arbitrarily large amount of the probability of $S_\nu^2/\nu$ eventually becomes concentrated in arbitrarily small neighborhoods of $1.$ That in turn implies an arbitrarily large amount of the probability of $\varphi_\nu=\log\left(S_\nu^2/\nu\right)$ becomes concentrated in arbitrarily small neighborhoods of $0.$

In mathematical analysis, a sequence like $(\varphi_\nu)$ is sometimes called a "mollifier" (provided $\varphi_\nu$ is smooth and compactly supported). The key idea is that adding a mollifier to another random variable has less and less of an effect, converging (almost surely) to that other variable in the limit. That result does not depend on the smoothness of the mollifying functions and it only really requires that their supports constrict down to zero. However, since our $\varphi_\nu$ do not have compact support, the usual conclusion that convergence occurs almost everywhere (with respect to Lebesgue measure) has to be weakened to convergence in probability.


Let $W$ represent either $Z_{+}$ or $Z_{-}$ and let $T_\nu = S_\nu^2/\nu.$ Because $W$ and $T_\nu$ are both positive, we may take logarithms:

$$\log\left(\frac{W}{\sqrt{T_\nu}}\right) = \log(W) + \left(- \frac{1}{2}\log(T_\nu)\right).$$

The factor of $-1/2$ does not affect the mollifying properties of the sequence of $\varphi_\nu = \log(T_\nu).$ Thus, the sequence $\log(W/\sqrt{T_\nu})$ converges in probability to $\log(W).$ Since the $\log$ is continuous, we see that $W/\sqrt{T_\nu}$ converges to $W.$

Obviously when $W$ is an atom at $0,$ the sequence $W/\sqrt{T_\nu}$ is constantly $0.$

Finally, now that we have seen that all three components of $Z/\sqrt{T_\nu}$ converge to the corresponding components of $Z,$ we conclude

In the generalized setting, $t_\nu=\frac{Z}{\sqrt{S_\nu^2/\nu}}$ converges in probability to $Z.$

If, in addition, $Z$ and $S_\nu^2$ (for each $\nu,$ at least eventually for large $\nu$) have continuous distributions with bounded densities (as in the case of Normal and Chi-squared distributions in the Student $t$ setting), it is now straightforward to show the sequence of distribution functions of $t_\nu$ converges uniformly to the distribution function of $Z.$ (The boundedness allows us to conclude that the convergence is uniform.)

  • $\begingroup$ Very nice answer. Is there any way to get at the convergence of the density functions without considering the densities explicitly? I can't seem to come up with anything constructive. $\endgroup$
    – Danny
    Apr 8, 2021 at 0:07
  • $\begingroup$ @Danny Since $$\log f_n(t)=-\frac{n+1}{2}\log(1+t^2/n)$$plus other stuff and the expression shown equals $-(n+1)t^2/(2n)+o(1/n),$ which converges to $-t^2/2,$ $f_n(t)$ must converge to some multiple of $\exp(-t^2/2).$ The multiple $\sqrt{1/(2\pi)}$ is determined by the normalization to unity. Thus, you never have to compute the limit of the Gamma functions -- indeed, you can now infer from this result that $$\lim_{n\to\infty} \frac{\Gamma\left(\frac{n+1}{2}\right)}{\sqrt{\pi n}\,\Gamma\left(\frac{n}{2}\right)}=\frac{1}{\sqrt{2\pi}}.$$ $\endgroup$
    – whuber
    May 20, 2021 at 2:01

While this is not as elementary as Stirling's approximation, the pointwise convergence of the density can be shown using dominated convergence theorem.

The density of a t-distribution with $n$ degrees of freedom is of the form $$f_n(x)=c_n\cdot\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}\quad,\,x\in\mathbb R$$

Let $g_n(x)=\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}$, so that $g_n(x)\to e^{-x^2/2}$ as $n\to \infty$.

So just remains to show that $c_n\to \frac1{\sqrt{2\pi}}$ as $n\to\infty$.

Now, $$\left(1+\frac{x^2}{n}\right)^{(n+1)/2}\ge \left(1+(n+1)\frac{x^2}{n}+\frac{n+1}{2n}x^4\right)^{1/2}\ge \left(1+\frac{x^4}{2}\right)^{1/2}$$

This implies $$|g_n(x)|\le \left(1+\frac{x^4}{2}\right)^{-1/2}\,,$$

where $$\int_{-\infty}^\infty \left(1+\frac{x^4}{2}\right)^{-1/2}\,dx<\infty$$

So by dominated convergence theorem,

$$\lim_{n\to\infty}\int_{-\infty}^\infty g_n(x)\,dx=\int_{-\infty}^\infty \lim_{n\to\infty}g_n(x)\,dx=\int_{-\infty}^\infty e^{-x^2/2}\,dx=\sqrt{2\pi}$$

Finally, as $\int_{-\infty}^\infty f_n(x)\,dx=c_n\int_{-\infty}^\infty g_n(x)\,dx=1$, taking limit on both sides yields


The nice thing about this approach is that we don't need to know what $c_n$ is to determine its limit.

Yet another way to derive this result is by using Slutsky's theorem, as shown here.


There are many ways you can establish that the T-distribution approaches the normal distribution in the limit. For the direct method you are using, you can find asymptotic expansions for the ratio of gamma functions are analysed in detail in Tricomi and Erdélyi (1951). The simplest expansion comes through application of Stirling's inequality, to obtain the general result (p. 133):

$$\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)} = z^{\alpha-\beta} \Big[ 1 + \frac{(\alpha - \beta) (\alpha + \beta - 1)}{2z} + \mathcal{O}(|z|^{-2}) \Big].$$

Taking $\beta = 0$ gives the simplified asymptotic form:

$$\frac{\Gamma(z+\alpha)}{\Gamma(z)} = z^{\alpha} \Big[ 1 + \frac{\alpha (\alpha - 1)}{2z} + \mathcal{O}(|z|^{-2}) \Big].$$

To obtain a form for the function of interest we can take $z = \tfrac{n}{2}$ and $\alpha = \tfrac{1}{2}$ to obtain:

$$H(n) \equiv \frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}{2})} = \sqrt{\frac{n}{2}} \Big[ 1 - \frac{1}{4n} + \mathcal{O}(n^{-2}) \Big].$$

We therefore have the desire limit:

$$\lim_{n \rightarrow \infty} \frac{\Gamma(\tfrac{n+1}{2})}{\sqrt{n \pi} \ \Gamma(\tfrac{n}{2})} = \lim_{n \rightarrow \infty} \frac{1}{\sqrt{2 \pi}} \Big[ 1 - \frac{1}{4n} + \mathcal{O}(n^{-2}) \Big] = \frac{1}{\sqrt{2 \pi}}.$$


An easy, intuitive way is to recognize that the noncentral scaled t-distribution with n degrees of freedom is the posterior predictive of the normal model based on n data points. (I think this is essentially its origin, and gives a common sense interpretation of the t-test.) As n goes to infinity, the model becomes "perfect", and must converge to a normal distribution.


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