# Distribution of the ratio of sample range to sample standard deviation for normal when n=3

Let $$X_{1},X_{2},X_{3}$$ be i.i.d samples from $$N(\mu,\sigma^2)$$. Let $$u$$ denote $$u=\frac{X_{(3)}-X_{(1)}}{S_{3}}\,,$$

where $$X_{(i)}$$ denotes the $$i$$th order statistic, and$$S_{3}=\sqrt{\frac{\sum_{i=1}^3(X_i-\bar{X})^2}{2}}$$

Now I know the density function of u is$$f(u)=\frac{3}{\pi}\left(1-\frac{u^2}{4}\right)^{-\frac{1}{2}},\quad(\sqrt{3}\le u\le2)$$ But how to get that?Does anyone have any ideas? Would appreciate some help.

EDITED:

Thank you guys for help!!! I get the formula from a paper by E.S. Pearson, published in 1964. For $$n>3$$, it's not possible to find the exact distribution of $$u$$. For $$n=3$$, Pearson quoted this relation from Thomson(1955):

$$u=2\cos\left({\frac{1}{6}\pi(1-F(u))}\right)\quad (\text{it's easy to find }\sqrt{3}\le u\le2 \text{ here})$$

Then we can just take arccos and take the derivative to get the formula. I read the paper by Thomson in 1955, only to find

Finally, I find the paper by Lieblein in 1952,

But I can't see the relation between this paper and the problem we talk, let alone EASILY solve the problem from Lieblein's results. That's all the information I have now.

• This question needs more detail, including the distribution of $X_i$, and on what basis you assert that you know the answer. Commented May 29, 2023 at 15:10
• @Wolfies That would help -- but the distribution is named in the title; and the reference to order statistics indicates the sample is iid.
– whuber
Commented May 29, 2023 at 15:55
• I don't see why a reference to order statistics implies the sample is iid. One can derive order statistics for non-identical distributions, as well as for dependent samples (for example, with copulas) and still have tractable results. Commented May 29, 2023 at 16:11
• Where does this result come from? The formula for $f(u)$ indicates positive probability density for $-\sqrt{3}<u<0$, but how can $(X_{(3)}-X_{(1)})/S_3$ be negative?
– Ute
Commented May 29, 2023 at 16:53
• A quick calculation shows that $f(u) =0$ for $u\notin [\sqrt{3},2]$ - it is just a typo :-)
– Ute
Commented May 29, 2023 at 17:12

This is an amazing result!

Here is how you could derive a similar one for uniform distributions; it could be useful at least for similar questions. Here we can use that if $$X_1, X_2, X_3$$ are i.i.d. uniform random variables, then the distribution of $$X_{(2)}$$ conditioned on $$X_{(1)}=a$$ and $$X_{(3)}=a+b$$ is uniform on $$[a, a+b]$$. Therefore, we can write $$X_{(2)}=a +b*T$$, where $$T$$ is uniform on $$[0,1]$$.

Express the sample variance in terms of $$b$$ and $$T$$, after some calculation: $$S_3^2 = \frac{1}{3}b^2(T^2-T+1),$$ and get the ratio $$U =\frac{b}{S_3} = \left(\frac{T^2-T+1}{3}\right)^{-\frac{1}{2}} = \sqrt{3}(T^2-T+1)^{-1/2}.$$

Since $$T\in[0,1]$$, we have $$T^2-T+1 \in [3/4, 1]$$, thus the ratio $$U$$ takes only values in $$[\sqrt{3}, 2]$$. This particular result does also hold if the $$X_i$$ are not uniform distributed, as $$X_{(2)}$$ always lies between $$X_{(1)}$$ and $$X_{(3)}$$.

To get the distribution of $$U$$, we need to calculate \begin{aligned} F_U(u) = P(U \leq u) &= P\left(\sqrt{3}\left({T^2-T+1}\right)^{-\frac{1}{2}}\leq u\right)\\ &= P\left((T-1/2)^2 \geq \frac{3}{u^2}-\frac{3}{4}\right) \\&= 2 F_T\left(\frac{1}{2}-\sqrt{\frac{3}{u^2}-\frac{3}{4}}\right) \end{aligned} Taking the derivative, we find \begin{align}\label{eq:udens}\tag{\ast} f_U(u) &= 2\sqrt{3}\left(\frac{1}{u^2}-\frac{1}{4}\right)^{-1/2}u^{-3}f_T\left(\frac{1}{2}-\sqrt{3}\left(\frac{1}{u^2}-\frac{1}{4}\right)^{1/2}\right) \\&= 2\sqrt{3}\left(1-\frac{u^2}{4}\right)^{-1/2}u^{-2}f_T\left(\frac{1}{2}-\sqrt{3}\left(\frac{1}{u^2}-\frac{1}{4}\right)^{1/2}\right). \end{align}

#### Uniform r.v. $$X_i$$

Using that $$T$$ is uniform when the $$X_i$$ are uniform, we obtain, for $$u\in[\sqrt{3}, 2]$$, that $$F_U(u)=1-2\sqrt{3}\left(\frac{1}{u^2}-\frac{1}{4}\right)^{1/2}$$ and $$f_U(u) = \frac{2\sqrt{3}}{u^2\sqrt{1-\frac{u^2}{4}}}.$$

#### Normal r.v. $$X_i$$

For normal distributed variables, the distribution of $$T=(X_{(2)}-X_{(1)})/(X_{(3)}-X_{(1)})$$ is not uniform. The resulting density of $$U$$ looks however nicer than the one for uniform $$X$$. There must be an ingenious trick to derive it. If someone knows it, please post.see Kinfinity's edit: the result is known since the 1950's, and was apparently due to hard work solving integrals.

The result for uniform variables is not too different from the result for normals, though. Due to the symmetry and smoothness of the normal density, $$X_{(2)}$$ is approximately uniform given $$X_{(1)}$$ and $$X_{(3)}$$

The paper by Lieblein (1952) gives the distribution of a closely related statistic, $$Y_1=\min(Y_{11}, 1-Y_{11})$$, where $$Y_{11}=(X_{(2)}-X_{(1)})/(X_{(3)}-X_{(1)})$$ (called $$T$$ above), by heroic integration (no Wolfram Alpha at hand that time). The density of $$Y_1$$'s distribution is calculated as Eq. (9) in Lieblein, and from that we find $$f_T(t) = \frac{3\sqrt{3}}{2\pi(1-t+t^2)}.$$ Now we can go on and plug this density in to \eqref{eq:udens}

• Thank you Ute!!! Commented May 30, 2023 at 4:45
• @Kinfinity, if you think this answers your query, upvote it and accept the same by clicking the tick symbol alongside the post. Commented May 30, 2023 at 6:02
• Thank you for the references, @Kinfinity ! It just looks like hard work, but the result for normal distributed variables is surprisingly nice. I still believe that there is a more ingeneous way to prove it :-)
– Ute
Commented May 30, 2023 at 9:01
• This is a very neat adaptation of the Lieblein paper, though Lieblein was solving for the closest pair (here in a sample of 3), rather than the widest pair. So this approach does not come across as the natural approach for the given problem. I am also hopeful that an elegant approach exists. Commented May 30, 2023 at 18:06
• @wolfies: yeah, shamelessly made use of symmetry...
– Ute
Commented May 30, 2023 at 21:46