### 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.

Get this bounty!!!

### 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.

Get this bounty!!!

### 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.

Get this bounty!!!

### 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.

Get this bounty!!!

## #StackBounty: #self-study #confidence-interval #estimation #multivariate-normal Confidence interval for $rho$ when \$Xsim N_3(0,Sigma…

### 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.

Get this bounty!!!

## #StackBounty: #estimation #mutual-information Recommended Mutual Information Estimator for Continuous Variable

### Bounty: 50

The mutual information seems to be quite an interesting measure of the relationship between variables. As such I wanted to apply it to investigate the relationship of two continuous variables $$X$$ and $$Y$$ for which I only have a hundred observations. In particular, I would like to obtain a normed version of the mutual information such that it is $$1$$ in the case of perfect dependence. I guess this means that the entropy of $$X$$ and $$Y$$ also need to be estimated.

After doing some research, I realized that estimating the (unnormalized) mutual information of two continuous variables is highly nontrivial. As a result, multiple competing approaches exists. https://journals.aps.org/pre/pdf/10.1103/PhysRevE.76.026209 provides an overview of some of them. https://journals.aps.org/pre/pdf/10.1103/PhysRevE.76.026209 also compares multiple approaches under different settings and makes recommendations when to use which approach. However, this paper is already 12 years old and since then new estimators have been developed, for example, https://arxiv.org/pdf/1801.04062.pdf. So, is anybody active in this field and can provide a recommendation which estimator is currently to be preferred (in which situation)? Ideally, I would also like to obtain a confidence interval for the normed mutual information.

Get this bounty!!!