# Relation between probability distribution for area and diameter of circle

If the area of a circle is normally distributed with mean $\mu$ and standard deviation $\sigma$, what distribution does the diameter follow?

Edit: I should clarify my question a bit. I found a reference suggesting the cross sectional area (of a reinforcement bar) follows a normal distribution with CoV=0.02. But the model I'm working with takes the bar diameter as input.

Thus I want to find what distribution to use for input, to have similar uncertainty. That is, when I run a large number of simulations of $\phi$, the resulting diameter $\pi\phi^2/4$ should be normally distributed with CoV 0.02. I have found that a CoV of 0.01 gives the intended result through trial and error, but would like to have a derivation.

Best regards, Mattias

• Doesn't seem like there's a nice distribution for this. But if you reverse the set up, you will have $\chi^2$! – Karn Watcharasupat Dec 14 '17 at 9:51
• An area cannot take negative values, so cannot be normally distributed. So if it is normally distributed then the diameter has uniform distribution (vacuously true). Btw, also it has exponential distribution (again vacuously true). Et cetera. – drhab Dec 14 '17 at 9:55

I think you are mostly on the right track with your analysis of this question. Perhaps the following will allow you to get a solution that meets your needs.

Distribution of areas. Technically, any normal distribution can take negative values, and so could not be used to model physical objects for which negative dimensions are impossible. However, you have a coefficient of variation (CoV) of $\sigma/\mu = .02.$ So if $\mu = 100,$ then $\sigma = 2$ and $A \sim \mathsf{Norm}(\mu=100,\,\sigma=2)$ has $P(A < 0) \approx 0,$ negligibly small for practical applications.

If you wanted a fussier model, you might use $A^\prime \sim \mathsf{Gamma}(shape=2500, rate=25)$ with $E(A^\prime) = 2500/25 = 100$ and $SD(A^\prime) = \sqrt{2500/625} = 2.$ The two density curves are almost indistinguishable in the interval $(90,110).$ In the figure below the thin blue curve is for the density of $A$ and the broken red curve is for $A^\prime.$

Simulated distribution of diameters. In your question, I think you have an incorrect relationship between area and diameter. It should be $A = \pi(D/2)^2$ so that $D = \sqrt{4A/\pi}.$ You are correct that $D$ should have a CoV of about $.01,$ as the following simulation in R shows.

a = rgamma(10^5, 2500, 25)  # gamma areas
d = sqrt(4*a/pi)
cva = sd(a)/mean(a); cva
## 0.02000077
cvd = sd(d)/mean(d); cvd
## 0.01000108
mean(d); sd(d)
## 11.28355
## 0.1128478


A histogram of the simulated values of $D$ shows that the distribution of $D$ is roughly, but not exactly $\mathsf{Norm}(11.3, 0.113),$ which has CoV $0.01.$ (A simulation based on normal areas, gives a similar result, except that the normal curve is offset a little to the left instead of to the right.)

Toward an exact distribution of diameters. If you look at the Wikipedia pages for 'gamma distribution' (under Related Distributions) and for 'generalized gamma distribution', you will find that the square root of a gamma distribution has a generalized gamma distribution. If you want an exact model for $D$ based on gamma areas $A^\prime,$ I will leave it to you to find the relevant generalized gamma distribution. (I think that will be less messy than trying to find the distribution of the square root of a normal distribution that is truncated to avoid any possibility of negative values.)

• Thank you very much for taking the time to put together this answer! It helped me a lot. – 09matblo Dec 18 '17 at 9:35