# Why does the t-distribution become more normal as sample size increases?

As per Wikipedia, I understand that the t-distribution is the sampling distribution of the t-value when the samples are iid observations from a normally distributed population. However, I don't intuitively understand why that causes the shape of the t-distribution to change from fat-tailed to almost perfectly normal.

I get that if you're sampling from a normal distribution then if you take a big sample it will resemble that distribution, but I don't get why it starts out with the fat-tailed shape it does.

I'll try to give an intuitive explanation.

The t-statistic* has a numerator and a denominator. For example, the statistic in the one sample t-test is

$$\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$$

*(there are several, but this discussion should hopefully be general enough to cover the ones you are asking about)

Under the assumptions, the numerator has a normal distribution with mean 0 and some unknown standard deviation.

Under the same set of assumptions, the denominator is an estimate of the standard deviation of the distribution of the numerator (the standard error of the statistic on the numerator). It is independent of the numerator. Its square is a chi-square random variable divided by its degrees of freedom (which is also the d.f. of the t-distribution) times the square of $$\sigma_\text{numerator}$$.

When the degrees of freedom are small, the denominator tends to be fairly right-skew. It has a high chance of being less than its mean, and a relatively good chance of being quite small. At the same time, it also has some chance of being much, much larger than its mean.

Under the assumption of normality, the numerator and denominator are independent. So if we draw randomly from the distribution of this t-statistic we have a normal random number divided by a second randomly* chosen value from a right-skew distribution that's on average around 1.

* without regard to the normal term

Because it's on the denominator, the small values in the distribution of the denominator produce very large t-values. The right-skew in the denominator make the t-statistic heavy-tailed. The right tail of the distribution, when on the denominator makes the t-distribution more sharply peaked than a normal with the same standard deviation as the t.

However, as the degrees of freedom become large, the distribution becomes much more normal-looking and much more "tight" around its mean.

As such, the effect of dividing by the denominator on the shape of the distribution of the numerator reduces as the degrees of freedom increase.

Eventually - as Slutsky's theorem might suggest to us could happen - the effect of the denominator becomes more like dividing by a constant and the distribution of the t-statistic is very close to normal.

### Considered in terms of the reciprocal of the denominator

whuber suggested in comments that it might be more illuminating to look at the reciprocal of the denominator. That is, we could write our t-statistics as numerator (normal) times reciprocal-of-denominator (right-skew).

For example, our one-sample-t statistic above would become:

$${\sqrt{n}(\bar{x}-\mu_0)}\cdot{1/s}$$

Now consider the population standard deviation of the original $$X_i$$, $$\sigma_x$$. We can multiply and divide by it, like so:

$${\sqrt{n}(\bar{x}-\mu_0)/\sigma_x}\cdot{\sigma_x/s}$$

The first term is standard normal. The second term (the square root of a scaled inverse-chi-squared random variable) then scales that standard normal by values that are either larger or smaller than 1, "spreading it out".

Under the assumption of normality, the two terms in the product are independent. So if we draw randomly from the distribution of this t-statistic we have a normal random number (the first term in the product) times a second randomly-chosen value (without regard to the normal term) from a right-skew distribution that's 'typically' around 1.

When the d.f. are large, the value tends to be very close to 1, but when the df are small, it's quite skew and the spread is large, with the big right tail of this scaling factor making the tail quite fat:

• Thanks! This has clarified much, but I still was a little unsure about "Its square is a chi-square random variable divided by its degrees of freedom (which is also the d.f. of the t-distribution) times [the standard deviation of the] numerator". Did you mention that merely because it was a useful thing to know, or is it something of direct relevance to the answer to my question? I understand that it is the distribution of the denominator, as opposed to the distribution of the square of the denominator, that is depicted in your figure. Commented Aug 2, 2014 at 12:45
• The distribution of the statistic would be heavier-tailed than normal even if it weren't specifically the square root of a chi-square on its df; in that sense it wouldn't directly alter the answer to leave it out. But at the least it serves as an explanation for where the scaled-chi distributions in the diagram came from. Commented Aug 2, 2014 at 12:54
• I am thinking it might be a little more illuminating to conduct this analysis based on the reciprocal of the sample standard deviation. That, coupled with an argument that the sample SD is independent of the sample mean (a key idea that would benefit from a little more emphasis and explanation, IMHO), would help people see that division of the sample mean by the sample SD has to spread out what otherwise would be a Normal distribution. (This of course was the whole point of Gossett's discovery.)
– whuber
Commented Aug 2, 2014 at 19:15
• @whuber I've added a section discussing it in terms of the reciprocal, but also retained the original discussion (it seems to me to be more direct, but I appreciate that many people may get more out of it in terms of the reciprocal). I'll add a little on the independence as well Commented Aug 2, 2014 at 22:55
• The value of the denominator in the t-statistic ($s/\sqrt{n}$) is on average close to $\sigma/\sqrt{n}$. The thing that's approximately 1 (at least in larger samples) is $s/\sigma$ - which we see in the first plot - and correspondingly $\sigma/s$ (in the second plot). Hopefully it's clear why those things are close to 1 when the sample's large enough to give a reasonable estimate of $\sigma$. Commented Aug 3, 2014 at 4:55

@Glen_b gave you the intuition on why the t statistic looks more normal as the sample size increases. Now, I will give you a slightly more technical explanation for the case when you already got the distribution of the statistic.

It is well-known that the t-statistic is distributed as a student t dsitribution with $n-1$ degrees of freedom, where $n$ is the sample size. The corresponding density looks as follows:

$$\frac{\left(1+\frac{x^2}{n-1}\right)^{-n/2}}{\sqrt{n-1} B\left(\frac{n-1}{2},\frac{1}{2}\right)}.$$

It is possible to show that

$$\frac{1}{\sqrt{n-1} B\left(\frac{n-1}{2},\frac{1}{2}\right)}\rightarrow \frac{1}{\sqrt{2\pi}},$$

and

$$\left(1+\frac{x^2}{n-1}\right)^{-n/2}\rightarrow \exp(-x^2/2),$$

as $n\rightarrow \infty$. By taking the product of these two limits you can see that the Student-t density converges exactly to the standard normal density.

• The convergence of the PDFs does not seem to say much. For instance, you could mix in $1/n$ of a distribution with PDF proportional to $(1 + (x/n)^2)^{-1}$ with each $t_n$ distribution and still arrive at the same limiting PDF, but all the while the distributions in the sequence would be getting fatter tails. The possibility of subtle behavior like this makes arguments based on limits of PDFs less than satisfying. Besides, doesn't the question really ask about small degrees of freedom? It wants to know why the sequence "starts out with the fat-tailed shape that it does."
– whuber
Commented Aug 2, 2014 at 19:09
• @whuber The answer is straightforward: there is a $-n$ in the power, this makes the tails lighter as $n$ increases. We just need to worry about the case at hand, not about other hypothetical cases where weird things may happen. Commented Aug 2, 2014 at 19:41
• The problem is that you haven't demonstrated that your case isn't one of the weird ones!
– whuber
Commented Apr 19, 2021 at 23:21

I just wanted to share something that helped my intuition as a beginner (though it's less rigorous than the other answers).

If $$Z, Z_1, ..., Z_n$$ are iid standard normal RVs then the following RV,

$$\frac{Z}{\sqrt{\frac{Z_1^2+...+Z_n^2}{n}}}$$

has a t-distribution with $$n$$ degrees of freedom.

As $$n$$ gets really big, using the law of large numbers we can see that the denominator goes to $$1$$. So you're just left with $$Z$$ which is standard normal which is why the t-distribution looks normal as $$n$$ gets big.

To elaborate... note that $$E[Z^2] = 1$$ which says the expected value of a chi squared RV is one. The fraction in the square root is just the sample mean of $$n$$ iid $$Z_i^2$$ RVs. The sample mean as $$n$$ gets super big will equal the expected value of just one of the $$Z_i^2$$'s which is one.

So as $$n$$ gets really big you're just left with $$\frac{Z}{\sqrt{1}} = Z$$

• Nice point. With continuous mapping and Slutsky this can be turned into a proof as well. Commented Feb 27, 2020 at 15:30