#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.


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#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!


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#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.


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#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?


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#StackBounty: #time-series #confidence-interval #standard-error #differences Standard Error of the cumulative value for time series

Bounty: 50

I have two time series, as in the picture below. The data was gathered experimentally. A practical example could be a measured mass flow rate, where I measure the mass flow rate over a certain time period with changing boundary conditions and I am interested in the total mass of fluid consumed during this period.
Comparison of two time series

The depicted intervals represent the Standard error of the mean and were determined through ten repeated measurements of the entire cycle. For each time step, the interval equals:
$ SEM= frac{sigma}{sqrt{10}}$
The ten replications naturally constitute ten identifiable time series.

I now have three questions :

(1). How can I assess, at a specific time step, if the difference between the means of the time series is statistically significant ? t-test ?

(2). How can I determine the SEM of the cumulated value for each time series ?

(3). How can I assess if the difference between the cumulated values of both time series is statistically significant ?

Does it make sense to create an “interval” of the cumulative value of both time series by cumulating the upper bound values and lower bound values ?


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#StackBounty: #confidence-interval #binomial #pdf #proportion #cdf Calculating a Confidence Interval for a Proportion for a Sample of D…

Bounty: 50

I’m interested in a (preferably analytic) solution or approximation to the following problem:

Let $s_1$ be a sample from an unknown distribution of size $N_1$ and with proportion of successes $p_1$. Let $s_2$ be an independent sample from the same distribution of size $N_2$ with proportion $p_2$. Given $N_1$, $p_1$, and $N_2$, can we calculate a Confidence Interval for $p_2$?

I would love a general purpose analytic solution if anyone has one, but for simplicity I am fine with considering the case where both $s_1$ and $s_2$ satisfy the conditions for their sampling distributions to be approximated by a Gaussian distribution.

Now, my approaches to solving this have led me to 2 options:

  1. Find upper and lower bounds for the confidence interval of $p$ (the population proportion of “successes”), and plug these back into confidence intervals for $p_2$ using the sampling distribution for $p$ with size $N_2$. Then take the max and min of those intervals. Or
  2. Treat $p$ as a normally distributed random variable with $mu=p_1$ and $sigma=sqrt{frac{p_1(1-p_1)}{N_1}}$, which would imply the CDF for $p_2$ can be found by:

    $CDF(x) = int_0^1{NormPDF(frac{y-p_1}{sqrt{frac{p_1(1-p_1)}{N_1}}})cdot NormCDF(frac{x-y}{sqrt{frac{y(1-y)}{N_2}}})dy}$

    where $NormPDF$ and $NormCDF$ are the PDF and CDF functions for the standard normal distribution.

The problem with 1 is that the interval found will be much wider than I would ideally want (this is what I am currently using in my equations). The problem with 2 is that I have no idea how to convert this into an analytic function (through approximation with $erf$ since I assume there is no analytic solution to the integral). My goal is to graph these intervals as a function of $p_1$ in desmos along with other sampling/prediction strategies for comparison – this is why I would really like an analytic solution or approximation.

If someone can solve this, or point me in the right direction to finding a solution that would be greatly appreciated!


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