#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: #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: #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: #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: #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: #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: #hypothesis-testing #confidence-interval #mean #asymptotics #small-sample What "nice" property of a confidence …

Bounty: 50

Suppose that I have $X_{i} overset{i.i.d.}{sim} P$ with $E[X_{i}]=mu$ and $V[X_{i}^{2}] = sigma^{2}<infty$.

Then by the central limit theorem I know that:
begin{align}
sqrt{n} (bar{X}{n} – mu) overset{d}{to} N(0,sigma^{2})
end{align}

where $bar{X}
{n}$ is the sample average. Suppose for some silly reason I know the value of $sigma^{2}$. Then this asymptotic approximation allows me to justify confidence sets for $mu$ of the form:
begin{align}
bar{X}{n} pm q{alpha/2} sqrt{frac{sigma^{2}}{n}}
end{align}

where $q_{alpha/2}$ is the $alpha/2^{th}$ quantile of the standard normal. In particular:
begin{align}
lim_{n to infty} P left(q_{alpha/2} leq sqrt{n} frac{(bar{X}{n} – mu)}{sigma} leq -q{alpha/2} right) = 1-alpha\
implies lim_{n to infty} P left(bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}} leq mu leq bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} right) = 1-alpha\
end{align}

For simplicity, let:
$$CI_{1} = left[bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}} , bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} right]$$
Now suppose that I am a strange statistician, and that rather than the confidence interval constructed above, I prefer a confidence interval (for whatever reason) of my own making:
$$CI_{2} = left[bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}}+b_{n} , bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} -b_{n}right]$$
where $b_{n} = o(n^{-1/2})$ is some vanishing deterministic sequence. Note that $CI_{2}$ also provides $1-alpha$ coverage probability asymptotically.

My question: is there any reason to prefer $CI_{1}$ to $CI_{2}$? Asymptotically they are the same, so I suspect any reason would need to appeal to finite sample arguments. For example, I can always construct the sequence $b_{n}$ such that $CI_{1}$ and $CI_{2}$ are VERY different in finite sample. So what statistical justification would lead someone to use $CI_{1}$ versus $CI_{2}$? Is there a name for the desirable property $CI_{1}$ possesses that $CI_{2}$ does not?

Thanks so much!


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: #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: #confidence-interval #references #bootstrap #multiple-comparisons #bonferroni Multiple comparisons correction for depende…

Bounty: 300

In this blog post the authors discuss simultaneously estimating quantiles, and constructing a simultaneous confidence envelope for the estimation which covers the whole quantile function. They do this by bootstrapping and then computing pointwise bootstrap confidence intervals and applying a Bonferroni type correction for multiple comparisons. Since the comparisons are not independent, they compute something like an effective number of independent trials according to a formula

$$N_{eq}=frac{N^2}{sum_{i,j}r(b_i,b_j)}$$

where $N$ is the number of points to be estimated and $r(b_i,b_j)$ is the sample correlation between the $ith$ and $j$th bootstrap vectors.

My question is where this formula comes from. They provide a link to a source, but I don’t see this formula in the source. Is anyone aware of this particular correction being used in the literature?


Get this bounty!!!