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I have the following questions (all related):

Suppose we have variables $\mathrm{x}^1, \mathrm{x}^2, \mathrm{y}$ and they are normalized to have mean 0 and variance 1 . It is known that the sample Pearson correlation coefficient between $\mathbf{x}_1$ and $\mathbf{x}_2$ are $0.7$.

  1. Consider a univariate model, that is, $$ y_i=x_{i 1} \beta_1+\epsilon_{i 1} \quad \text { and } \quad y_i=x_{i 2} \beta_2+\epsilon_{i 2} . $$

The sample Pearson correlation coefficients (i) between $\mathbf{x}_1$ and $\mathbf{y}$ and (ii) between $\mathbf{x}_2$ and $\mathbf{y}$ are $0.3$ and $0.75$. Find the $\hat{\beta}_1$ and $\hat{\beta}_2$ obtained by least squares.

Hint: What do $\mathrm{x}_1^{\prime} \mathrm{y}$ and $\mathrm{x}_2^{\prime} \mathrm{y}$ mean?

2) Consider a multiple linear regression model, that is, $$ y_i=x_{i 1} \alpha_1+x_{i 2} \alpha_2+\epsilon_i . $$ Compute $\hat{\alpha}_1$ and $\hat{\alpha}_2$ when they are estimated using the least squares.

3) In general, which condition(s) of $\mathrm{x}^1$ and $\mathrm{x}^2$ are sufficient to have $\hat{\alpha}_1=\beta_1^{(1)}$ and $\hat{\alpha}_2=\beta_1^{(2)}$ ?

For 1), I am using the formula $\hat{\beta}=r\left(\frac{S_y}{S_x}\right)$ but I'm not sure if it applies to a linear regression model without an intercept.

For 2), I am thinking of using the multiple correlation formula but to no avail since I'm not sure how to get the regression coeffificnets into the formula:

$R=\sqrt{\frac{r_{y x_1}^2+r_{y x_2}^2-2 r_{y x_1} \cdot r_{y x_2} \cdot r_{x_1 x_2}}{1-r_{x_1 x_2}^2}}$

For 3), I am thinking it is when $\mathrm{x}^1$ and $\mathrm{x}^2$ are independent.

Any help and guidance is much appreciated, thank you!

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  • $\begingroup$ The reason there is no intercept is that you standardised everything to have mean 0 (in fact you can try to find an intercept, but up to rounding precision it is guaranteed to be 0 in ordinary least squares). Also forcing the variances to 1 has the effect of making the covariances equal to the correlations $\endgroup$
    – Henry
    Commented Oct 20, 2022 at 18:13
  • $\begingroup$ Thank you, how would you approach the other questions? $\endgroup$ Commented Oct 20, 2022 at 18:34

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