# Does Sqrt of confidence intervals require a correction factor?

#### dbooksta

##### New Member
I have the following formula for confidence intervals on samples from a Rayleigh process:

$$\frac{2(n-1)\overline{r^2}}{\chi_2^2} \leq \widehat{\sigma^2} \leq \frac{2(n-1)\overline{r^2}}{\chi_1^2}$$

I want to give confidence intervals in terms of $$\sigma$$, not $$\sigma^2$$.

If $$[x, y]$$ are the 95% confidence intervals for $$\sigma^2$$ can I just take the square root and say that $$[\sqrt{x}, \sqrt{y}]$$ are 95% confidence intervals for $$\sigma$$, or do I need to apply a correction factor for the concavity of the square root function?

If correction is required can you show me whether it applies only to the values, or does the confidence level itself require correction? (E.g., does the sqrt of 95% confidence intervals only represent 90% intervals?)

#### BGM

##### TS Contributor
Note that

$$0 \leq L(\mathbf{X}) \leq \theta \leq U(\mathbf{X}) \iff \sqrt{L(\mathbf{X})} \leq \sqrt{\theta} \leq \sqrt{U(\mathbf{X})}$$

(applying monotonic increasing function $$f(x) = \sqrt{x}$$ gives an equivalent inequality)

When $$[L(\mathbf{X}), U(\mathbf{X})]$$ is a confidence interval of $$\theta$$ with confidence level $$1 - \alpha$$,

$$\Pr\{L(\mathbf{X}) \leq \theta \leq U(\mathbf{X})\} = 1 - \alpha = \Pr\left\{\sqrt{L(\mathbf{X})} \leq \sqrt{\theta} \leq \sqrt{U(\mathbf{X})}\right\}$$

So we can see $$[\sqrt{L(\mathbf{X})}, \sqrt{U(\mathbf{X})}]$$ is a confidence interval of $$\sqrt{\theta}$$ with confidence level $$1 - \alpha$$ also. Your assertion is correct and there is no "correction" is needed.

In general you can obtain the CI for any parameter transformed by a monotonic increasing function (with appropriate range of course) in a similar way.

BTW it looks weird with $$n - 1$$ here if your sample size is $$n$$. Do not blindly borrow from the case of CI for normal variance with unknown mean.

#### dbooksta

##### New Member
BTW it looks weird with $$n - 1$$ here if your sample size is $$n$$. Do not blindly borrow from the case of CI for normal variance with unknown mean.
Yes, that's another question I have on my queue to address. In general, per Siddiqui 1961, 4.10, the $$\chi^2$$ limits should be based on n degrees of freedom. But I'm working with radii r computed based off the sample center of the distribution -- true center is unknown. Somebody else was convinced (albeit without proof) that you lose 2 degrees of freedom using the sample center instead of true center. Can you point me to a proof I could follow to derive that, and to confirm it's 2 degrees and not just 1?

Update: Actually it may be better to characterize this as the uncorrelated, symmetric, bivariate normal distribution, since I'm working off samples $$z_i$$ consisting of coordinates $$(x_i, y_i)$$ where x and y are i.i.d. normal with stdev $$\sigma$$.

Last edited:

#### dbooksta

##### New Member
So we can see $$[\sqrt{L(\mathbf{X})}, \sqrt{U(\mathbf{X})}]$$ is a confidence interval of $$\sqrt{\theta}$$ with confidence level $$1 - \alpha$$ also. Your assertion is correct and there is no "correction" is needed.
I was re-reading Siddiqui 1964 p. 1007 and now I'm doubting this: We estimate a parameter $$\hat{\gamma}$$, and we know that another parameterization $$\sigma = \sqrt{\gamma}$$ requires a correction factor cf(n) -- i.e., $$\hat{\sigma} = cf(n) \sqrt{\hat{\gamma}}$$.

The confidence intervals were derived for $$\gamma$$. But of course $$\hat{\gamma}$$ is an estimate from n samples, and now I'm interested in the confidence interval for $$\hat{\sigma}$$. But $$\hat{\sigma} \neq \sqrt{\hat{\gamma}}$$, rather $$\hat{\sigma} = cf(n) \sqrt{\hat{\gamma}}$$. So confidence intervals should be on the corrected estimate too, right? In which case I need to multiply the square root of the $$\gamma$$ confidence intervals by the same correction factor, right?

#### BGM

##### TS Contributor
If I have not understand wrongly, at 4.10 it is degrees of freedom of $$2N$$.

I am not sure which part you are referring to. My best guess is that you maybe talking about using chi-square goodness-of-fit test; but that is a general non-parametric method to test the multinomial counts so it should not be the related to your CI; unless you are inverting the test but I do not think you are trying to do that.

And for the second question, yes of course you can multiply a positive constant. The inequality will preserve.

#### dbooksta

##### New Member
No, I am trying to use the CI formula given in 4.10. But that was derived from true Rayleigh samples -- i.e., radius from the true center of the distribution. In my case I don't know the true center; I only have the sample center, and I compute the radius of each sample from that sample center.

So to address this question it's probably better to use the uncorrelated bivariate normal model of the variables, which is saying the same thing, but where I can say each sample value is a function of two normally distributed variables whose true mean is unknown. In that case I thought the confidence intervals would be given by the same formula 4.10, but with 2n - 2 degrees of freedom because we lose one degree in each dimension due to the unknown mean?

One more confusion: When I estimate $$\hat\gamma = s^2$$ I know from MC simulations that the correction factor for $$\hat\sigma$$ is $$c_4(2n - 1)$$ -- i.e., the one we use to correct for bias in standard deviation estimates for normal distributions. Since I see (2n -1) in that correction factor I am left doubting whether (2n - 2) is the correct value to use in the confidence interval equations.

Last edited:

#### BGM

##### TS Contributor
So now you are really switching the topic to normal inference.

Given observations

$$\begin{bmatrix} X_i \\ Y_i \end{bmatrix} \sim \mathcal{N} \left(\begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix}, \begin{bmatrix} \sigma^2 & 0 \\ 0 & \sigma^2 \end{bmatrix} \right), ~~ i = 1, 2, \ldots ,n$$

One can show that the MLE of $$\sigma^2$$ is

$$\hat{\sigma}^2 = \frac {1} {2n} \sum_{i=1}^n [(X_i - \bar{X})^2 + (Y_i - \bar{Y})^2]$$

And as usual we have the unbiased estimator

$$S^2 = \frac {1} {2n - 2} \sum_{i=1}^n [(X_i - \bar{X})^2 + (Y_i - \bar{Y})^2]$$

Therefore you are correct that we can make use of the fact that

$$\frac {(2n - 2)S^2} {\sigma^2} \sim \chi^2(2n-2)$$

to derive the confidence interval for $$\sigma^2$$

But I have to say for Rayleigh distribution should be restricted to known mean of $$(0, 0)$$ so the above arguments is just for general normal case.