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This cute little problem gives us the wisdom that when we minimize two functions at a single point uniquely, then their sum is also minimized at the same point. This Invariant Regression Estimate is applied to calculate the least square estimates of two group regression from ISI MStat 2016 Problem 7.

Suppose \({(y_{i}, x_{1 i}, x_{2 i}, \ldots, x_{k i}): i=1,2, \ldots, n_{1}+n_{2}}\) represents a set of multivariate observations. It is found that the least squares linear regression fit of \(y\) on \(\left(x_{1}, \ldots, x_{k}\right)\) based on the first \(n_{1}\) observations is the same as that based on the remaining \(n_{2}\) observations, and is given by

\(y=\hat{\beta}_{0}+\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}\)

If the regression is now performed using all \(\left(n_{1}+n_{2}\right)\) observations, will the regression equation remain the same? Justify your answer.

- \(f(\tilde{x})\) and \(g(\tilde{x})\) are both uniquely minimized at \( \tilde{x} = \tilde{x_0}\), then \(f(\tilde{x}) + g(\tilde{x})\) is uniquely minimized at \( \tilde{x} = \tilde{x_0}\).

Observe that we need to find the OLS estimates of \({\beta}{i} \forall i \).

\(f(\tilde{\beta}) = \sum_{i = 1}^{n_1} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \)

\(g(\tilde{\beta}) = \sum_{i = n_1}^{n_1+n_2} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \)

\(\hat{\tilde{\beta}} = (\hat{{\beta}_0}, \hat{{\beta}_1}, ..., \hat{{\beta}_k )} \)

\( h(\tilde{\beta}) = f(\tilde{\beta}) + g(\tilde{\beta}) = \sum_{i = 1}^{n_1+n_2} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \).

Now, \( h(\tilde{\beta})\) is the loss squared erorr under the grouped regression, which needs to be minimized with respect to \(\tilde{\beta} \).

Now, by the given conditions, \(f(\tilde{\beta})\) and \(g(\tilde{\beta})\) are both uniquely minimized at \( \hat{\tilde{\beta}}\), therefore \(h(\tilde{\beta}) = f(\tilde{\beta}) + g(\tilde{\beta})\) will be uniquely minimized at \(\hat{\tilde{\beta}}\) by the prerequisite.

Hence, the final estimate of \(\tilde{\beta}\) will be \( \hat{\tilde{\beta}}\).

This cute little problem gives us the wisdom that when we minimize two functions at a single point uniquely, then their sum is also minimized at the same point. This Invariant Regression Estimate is applied to calculate the least square estimates of two group regression from ISI MStat 2016 Problem 7.

Suppose \({(y_{i}, x_{1 i}, x_{2 i}, \ldots, x_{k i}): i=1,2, \ldots, n_{1}+n_{2}}\) represents a set of multivariate observations. It is found that the least squares linear regression fit of \(y\) on \(\left(x_{1}, \ldots, x_{k}\right)\) based on the first \(n_{1}\) observations is the same as that based on the remaining \(n_{2}\) observations, and is given by

\(y=\hat{\beta}_{0}+\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}\)

If the regression is now performed using all \(\left(n_{1}+n_{2}\right)\) observations, will the regression equation remain the same? Justify your answer.

- \(f(\tilde{x})\) and \(g(\tilde{x})\) are both uniquely minimized at \( \tilde{x} = \tilde{x_0}\), then \(f(\tilde{x}) + g(\tilde{x})\) is uniquely minimized at \( \tilde{x} = \tilde{x_0}\).

Observe that we need to find the OLS estimates of \({\beta}{i} \forall i \).

\(f(\tilde{\beta}) = \sum_{i = 1}^{n_1} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \)

\(g(\tilde{\beta}) = \sum_{i = n_1}^{n_1+n_2} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \)

\(\hat{\tilde{\beta}} = (\hat{{\beta}_0}, \hat{{\beta}_1}, ..., \hat{{\beta}_k )} \)

\( h(\tilde{\beta}) = f(\tilde{\beta}) + g(\tilde{\beta}) = \sum_{i = 1}^{n_1+n_2} (y - {\beta}_{0} - (\sum_{j=1}^{k} \hat{\beta}_{j} x_{j}))^2 \), where \(\tilde{\beta} = ({\beta}_0, {\beta}_1, ..., {\beta}_k ) \).

Now, \( h(\tilde{\beta})\) is the loss squared erorr under the grouped regression, which needs to be minimized with respect to \(\tilde{\beta} \).

Now, by the given conditions, \(f(\tilde{\beta})\) and \(g(\tilde{\beta})\) are both uniquely minimized at \( \hat{\tilde{\beta}}\), therefore \(h(\tilde{\beta}) = f(\tilde{\beta}) + g(\tilde{\beta})\) will be uniquely minimized at \(\hat{\tilde{\beta}}\) by the prerequisite.

Hence, the final estimate of \(\tilde{\beta}\) will be \( \hat{\tilde{\beta}}\).

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interesting problem, probably explanation/ solution could be more elaborate, still, a very good effort

Thanks for supporting us. We will try to be more elaborate. But, we will try to elaborate at such a point, when a student well equipped with the prerequisites can understand thoroughly. If someone can't then, it is an indication, that person must go back and give a quick revision.

Very interesting to read this article.I would like to thank you for the efforts you had made for writing this awesome article. This article inspired me to read more. keep it up.