*Bounty: 50*

*Bounty: 50*

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

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 $X’Sigma^{-1}Xsim chi^2_3$.

So expanding the quadratic form, I get

$$x’Sigma^{-1}x=frac{1}{1-rho^2}left[x_1^2+(1+rho^2)x_2^2+x_3^2-2rho(x_1x_2+x_2x_3)right]$$

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

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

$$g_1(rho)=(x_2^2+chi^2_{alpha/2,3})rho^2-2(x_1x_2+x_2x_3)rho+x_1^2+x_2^2+x_3^2-chi^2_{alpha/2,3}$$

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 $$frac{3bar x}{sqrt{3+4rho+2rho^2}}sim N(0,1)$$

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

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

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

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.