## #StackBounty: #mathematical-statistics #estimation The difference of normal means is also minimax?

### Bounty: 50

Let $$X_i sim N(xi, sigma^2)$$ and $$Y_i sim N(eta, tau^2)$$ for known $$sigma^2$$ and $$tau^2$$.

I know that $$bar{X}$$ and $$bar{Y}$$ are minimax under squared error loss since their variance is fixed, and a sequence of Bayes estimators can be constructed such that their Bayes risk converge to the maximum risk of $$bar{X}$$ and $$bar{Y}$$.

I am wondering how I can show $$delta(X,Y) =bar{Y}-bar{X}$$ is also minimax for $$eta-xi$$?

Essentially I need to show that for any $$T(X,Y)$$, we have

$$sup_{xi,eta} E(T(X,Y)-(eta-xi))^2 leq sup_{eta,xi} E((bar{Y}-bar{X})-(eta-xi))^2 = frac{sigma^2+tau^2}{n}$$

The only thing I can think of doing is constructing another sequence of priors whose Bayes risks converge to the RHS. However, this seems kind of tedious now that we’re in 2 dimensions. I feel like there’s a “trick” here that I should be using?

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## #StackBounty: #estimation #causality #identifiability What does it mean to "non-parametrically" identify a causal effect with…

### Bounty: 100

I am wondering, within the context of causal inference, what it means to “non-parametrically” identify a causal effect within the super-population perspective. For example, in Hernan/Robins Causal Inference Book Draft:

It defines non-parametric identification on pg. 43 and 123 as:

…identification that does not require any modeling assumptions when
the size of the study population is quasi-infinite. By acting as if we
could obtain an unlimited number of individuals for our studies, we
could ignore random fluctuations and could focus our attention on
systematic biases due to confounding, selection, and measurement.
Statisticians have a name for problems in which we can assume the size
of the study population is effectively infinite: identification
problems.

I understand the identification part to mean that under the strong ignorability assumption, there is only ONE way for the observed data to correspond to a causal effect estimand. What confuses me is why we need to assume the size of the study is quasi-infinite.

For example, in the book it gives an example of a 20 person study where each subject was representative of 1 billion identical subjects, and to view the hypothetical super-population as that of 20 billion people. Specifically, on pg. 13 it states that:

… we will assume that counterfactual outcomes are
deterministic and that we have recorded data on every subject in a
very large (perhaps hypothetical) super-population. This is equivalent
to viewing our population of 20 subjects as a population of 20 billion
subjects in which 1 billion subjects are identical to the 1st subject, 1 billion
subjects are identical to the 2nd subject, and so on.

My confusion here is what it means to assume a single person is representative of 1 billion identical individuals. Is it assuming that each of the 1 billion are identical with respect to their outcomes and treatment only, but differ with respect to the covariates? Or is it assuming the individual is a summary measure of the 1 billion? My instinct is that the notion of the 1 billion is entertaining the fact we may draw many times without having a case where we have a lack of samples. I.e., small sample sizes result in more unstable estimates.

Essentially, what is so crucial about assuming there are many identical individuals in the “background”, if they are just going to be the same as a patient you observe? What happens or breaks down if instead of the 1 billion, we only had 2 identical individuals?

Thank you for any insight.

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### 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!!!