Does the normal distribution converge to a certain distribution if the standard deviation grows without bounds? it appears to me that the pdf starts looking like a uniform distribution with bounds given by $[-2 \sigma, 2 \sigma]$. Is this true?

# Solved – Does the normal distribution converge to a uniform distribution when the standard deviation grows to infinity

convergencenormal distribution

#### Related Solutions

Actually $s$ doesn't need to systematically underestimate $\sigma$; this could happen even if that weren't true.

As it is, $s$ is biased for $\sigma$ (the fact that $s^2$ is unbiased for $\sigma^2$ means that $s$ will be biased for $\sigma$, due to Jensen's inequality*, but that's not the central thing going on there.

* *Jensen's inequality*

If $g$ is a convex function, $g\left(\text{E}[X]\right) \leq \text{E}\left[g(X)\right]$ with equality only if $X$ is constant or $g$ is linear.

Now $g(X)=-\sqrt{X}$ is convex,

so $-\sqrt{\text{E}[X]} < \text{E}(-\sqrt{X})$, i.e. $\sqrt{\text{E}[X]} > \text{E}(\sqrt{X})\,$, implying $\sigma>E(s)$ if the random variable $s$ is not a fixed constant.

Edit: a simpler demonstration not invoking Jensen --

Assume that the distribution of the underlying variable has $\sigma>0$.

Note that $\text{Var}(s) = E(s^2)-E(s)^2$ this variance will always be positive for $\sigma>0$.

Hence $E(s)^2 = E(s^2)-\text{Var}(s) < \sigma^2$, so $E(s)<\sigma$.

*So what is the main issue?*

Let $Z=\frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

Note that you're dealing with $t=Z\cdot\frac{\sigma}{s}$.

That inversion of $s$ is important. So the effect on the variance it's not whether $s$ is smaller than $\sigma$ on average (though it is, very slightly), but whether $1/s$ is larger than $1/\sigma$ on average (and those two things are NOT the same thing).

And it is larger, to a greater extent than its inverse is smaller.

Which is to say $E(1/X)\neq 1/E(X)$; in fact, from Jensen's inequality:

$g(X) = 1/x$ is convex, so if $X$ is not constant,

$1/\left(\text{E}[X]\right) < \text{E}\left[1/X\right]$

So consider, for example, normal samples of size 10; $s$ is about 2.7% smaller than $\sigma$ on average, but $1/s$ is about 9.4% *larger* than $1/\sigma$ on average. So even if at n=10 we made our estimate of $\sigma$ 2.7-something percent larger** so that $E(\widehat\sigma)=\sigma$, the corresponding $t=Z\cdot\frac{\sigma}{\widehat\sigma}$ would not have unit variance - it would still be a fair bit larger than 1.

**(at other $n$ the adjustment would be different of course)

Since the t-distribution is like the standard normal distribution but with a higher variance (smaller peak and fatter tails)

If you adjust for the difference in spread, the peak is higher.

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

You need to find out, which function in Python gives you the inverse of the cumulative distribution function of the normal distribution: https://en.wikipedia.org/wiki/Normal_distribution#Cumulative_distribution_function The cummulative distribution function of the normal distribution is often called $\Phi$ and it gives you the percentage of normally distributed random numbers up to a certain z-value. The inverse should give you the z-value to a certain percentage.

See the red curve in the above linked Wikipedia article: https://upload.wikimedia.org/wikipedia/commons/c/ca/Normal_Distribution_CDF.svg

I don't know about Python, but in R you'd use qnorm(). Say you had a mean of 0 and a Standard deviation of 1 and you wanted to find the intervall were 1% of the data is to the left an 1% to the right of the intervall (that is 99% to the left) than you'd call

```
qnorm(c(0.01, 0.99), mean = 0, sd = 1)
```

and get

```
[1] -2.326348 2.326348
```

So 2.326 is the exact number of what you estimated to 2.333.

Edit: Apparently the corresponding Python function is norm.ppf() in scipy.stats : https://stackoverflow.com/questions/24695174/python-equivalent-of-qnorm-qf-and-qchi2-of-r

## Best Answer

The other answers already here do a great job of explaining why Gaussian RVs don't converge to anything as the variance increases without bound, but I want to point out a seemingly-uniform property that such a collection of Gaussians

doessatisfy that I think might be enough for someone to guess that they are becoming uniform, but that turns out to not be strong enough to conclude that. $\newcommand{\len}{\text{len}}$Consider a collection of random variables $\{X_1,X_2,\dots\}$ where $X_n \sim \mathcal N(0, n^2)$. Let $A = [a_1,a_2]$ be a fixed interval of finite length, and for some $c \in \mathbb R$ define $B = A +c$, i.e. $B$ is $A$ but just shifted over by $c$. For an interval $I = [i_1,i_2]$ define $\len (I) = i_2-i_1$ to be the length of $I$, and note that $\len(A) = \len(B)$.

I'll now prove the following result:

Result: $\vert P(X_n \in A) - P(x_n\in B)\vert \to 0$ as $n \to \infty$.I call this uniform-like because it says that the distribution of $X_n$ increasingly has two fixed intervals of equal length having equal probability, no matter how far apart they may be. That's definitely a very uniform feature, but as we'll see this doesn't say anything about the actual distribution of the $X_n$ converging to a uniform one.

Pf: note that $X_n = n X_1$ where $X_1 \sim \mathcal N(0, 1)$ so $$ P(X_n \in A) = P(a_1 \leq n X_1 \leq a_2) = P\left(\frac{a_1}{n} \leq X_1 \leq \frac{a_2}n\right) $$ $$ = \frac{1}{\sqrt{2\pi}}\int_{a_1/n}^{a_2/n} e^{-x^2/2}\,\text dx. $$ I can use the (very rough) bound that $e^{-x^2/2} \leq 1$ to get $$ \frac{1}{\sqrt{2\pi}}\int_{a_1/n}^{a_2/n} e^{-x^2/2}\,\text dx \leq \frac{1}{\sqrt{2\pi}}\int_{a_1/n}^{a_2/n} 1\,\text dx $$ $$ = \frac{\text{len}(A)}{n\sqrt{2\pi}}. $$

I can do the same thing for $B$ to get $$ P(X_n \in B) \leq \frac{\text{len}(B)}{n\sqrt{2\pi}}. $$

Putting these together I have $$ \left\vert P(X_n \in A) - P(X_n \in B)\right\vert \leq \frac{\sqrt 2 \text{len}(A) }{n\sqrt{\pi}} \to 0 $$ as $n\to\infty$ (I'm using the triangle inequality here).

$\square$

How is this different from $X_n$ converging on a uniform distribution? I just proved that the probabilities given to any two fixed intervals of the same finite length get closer and closer, and intuitively that makes sense that as the densities are "flattening out" from $A$ and $B$'s perspectives.

But in order for $X_n$ to converge on a uniform distribution, I'd need $P(X_n \in I)$ to head towards being proportional to $\text{len}(I)$ for

anyinterval $I$, and that is a very different thing because this needs to apply to any $I$, not just one fixed in advance (and as mentioned elsewhere, this is also not even possible for a distribution with unbounded support).