Suppose $X\sim N_3(0,\Sigma)$, where $\Sigma=\begin{pmatrix}1&\rho&\rho^2\\\rho&1&\rho\\\rho^2&\rho&1\end{pmatrix}$.

On the basis of one observation $x=(x_1,x_2,x_3)'$, I have to obtain a confidence interval for $\rho$ with confidence coefficient $1-\alpha$.

We know that $X'\Sigma^{-1}X\sim \chi^2_3$.

So expanding the quadratic form, I get


To use this as a pivot for a two-sided C.I with confidence level $1-\alpha$, I setup $$\chi^2_{1-\alpha/2,3}\le x'\Sigma^{-1}x\le \chi^2_{\alpha/2,3}$$

I get two inequalities of the form $g_1(\rho)\le 0$ and $g_2(\rho)\ge 0$, where


and $$g_2(\rho)=(x_2^2+\chi^2_{1-\alpha/2,3})\rho^2-2(x_1x_2+x_2x_3)\rho+x_1^2+x_2^2+x_3^2-\chi^2_{1-\alpha/2,3}$$

Am I right in considering a both-sided C.I.? After solving the quadratics in $\rho$, I am guessing that the resulting C.I would be quite complicated.

Another suitable pivot is $$\frac{\mathbf1' x}{\sqrt{\mathbf1'\Sigma \mathbf 1}}\sim N(0,1)\quad\,,\,\mathbf1=(1,1,1)'$$

With $\bar x=\frac{1}{3}\sum x_i$, this is same as saying $$\frac{3\bar x}{\sqrt{3+4\rho+2\rho^2}}\sim N(0,1)$$

Using this, I start with the inequality $$\left|\frac{3\bar x}{\sqrt{3+4\rho+2\rho^2}}\right|\le z_{\alpha/2}$$

Therefore, $$\frac{9\bar x^2}{3+4\rho+2\rho^2}\le z^2_{\alpha/2}\implies 2(\rho+1)^2+1\ge \frac{9\bar x^2}{z^2_{\alpha/2}}$$

That is, $$\rho\ge \sqrt{\frac{9\bar x^2}{2z^2_{\alpha/2}}-\frac{1}{2}}-1$$

Since the question asks for any confidence interval, there are a number of options available here. I could have also squared the standard normal pivot to get a similar answer in terms of $\chi^2_1$ fractiles. I am quite sure that both methods I used are valid but I am not certain whether the resulting C.I. is a valid one. I am also interested in other ways to find a confidence interval here.

@JimB had commented that the intervals might not give real numbers within $[-1,1]$. Is there a way I can actually ensure that it does not happen?

  • $\begingroup$ I think you want to use $x' \Sigma^{-1} x \leq \chi^2_{1-\alpha,3}$ and then solve the quadratic for the lower and upper bounds for $\rho$. Try it with actual samples from the trivariate normal and you'll see that the quadratic solution when using the bound on the left-hand side of your equation typically results in imaginary numbers and occasionally real numbers but outside of the range -1 to 1. $\endgroup$ – JimB Mar 31 at 21:04
  • $\begingroup$ @JimB I had not considered that. Will check again. $\endgroup$ – StubbornAtom Mar 31 at 21:10
  • $\begingroup$ You should also check the other confidence interval approach with values for $\bar{x}$. With values close enough to zero you're going to get imaginary numbers and large values of $\bar{x}$ will get you bounds way outside -1 and +1. Estimating $\rho$ with just a single sample of 3 correlated values is not going to have very "tight" coverage. $\endgroup$ – JimB Mar 31 at 21:22
  • $\begingroup$ Even using the $x'\Sigma^{-1}x \leq \chi^2_{1-\alpha,3}$ approach results in between 1% and 3% of the estimates being imaginary numbers. (At least my klutzy implementation ends up with those figures.) $\endgroup$ – JimB Mar 31 at 21:45

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