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Quant Interview Question: Given $y$ = $a_1$$x_1$, $y$ = $b_1$$x_1$+$b_2$$x_2$, $cov(x_1,y)$ != 0, $cov(x_1,x_2)$ > 0, $cov(x_2,y)$ = 0. Compare $abs(a_1)$ and $abs(b_1)$. Which one is larger and why?

My intuition tells me that this has something to do with linear regression since covariance is given. The OLS coefficient estimates consist of the covariance, but after writing them out, I'm not sure how to proceed. Specifically, I'm not sure how to move towards the absolute of the coefficients and how to move from the estimates to the true values.

Any help would be greatly appreciated!

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    $\begingroup$ Although this might look like OLS, it isn't really, because the $a_i$ and $b_i$ are not estimates. Compute the variance of $y$ in two ways and use the crucial inequality $\operatorname{Cov}(x_1,x_2)\gt 0.$ $\endgroup$
    – whuber
    Commented Aug 28, 2022 at 18:21
  • $\begingroup$ I got $(a^2-b^2)$$(Var(x_1))$ = ${b_2}^2$$Var(x_2)$ + $2b_1b_2Cov(x_1,x_2)$. Now, if we know that $b_1$ and $b_2$ have the same sign, we know abs(a) > abs(b_1) but idk how to proceed. I listed out all the variance I can get from rearranging the variables. $\endgroup$
    – wakuwaku
    Commented Aug 28, 2022 at 19:50
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    $\begingroup$ @whuber I get $a_1=b_1$ without using $\operatorname{Cov}(x_1,x_2)\gt 0$ at all. Am I missing something? $\endgroup$
    – statmerkur
    Commented Aug 28, 2022 at 22:54
  • $\begingroup$ @statmerkur I realized there's a flaw in the conditions. Not sure if this is intended. See my comment under your answer. $\endgroup$
    – wakuwaku
    Commented Aug 29, 2022 at 0:36

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Using the bilinearity of the covariance, the definition of the variance of a random variable, and $\mathrm{Cov}\left(x_2, y\right)=0$, we get $$ \begin{align} 0 \leq \mathbb V\left(y\right) &= \mathrm{Cov}\left(y, y\right) \\ &= \mathrm{Cov}\left(b_1 x_1 + b_2 x_2, y\right) \\ &= \mathrm{Cov}\left(b_1 x_1, y\right) + \mathrm{Cov}\left(b_2 x_2, y\right) \\ &= b_1\mathrm{Cov}\left(x_1, y\right) + b_2\mathrm{Cov}\left(x_2, y\right) \\ &= b_1\mathrm{Cov}\left(x_1, a_1 x_1\right) + 0 \\ &= a_1b_1\mathbb V\left(x_1\right). \end{align} $$ From $\mathrm{Cov}\left(x_1, y\right) \neq 0$ we now see that $a_1 \neq 0,\mathbb V\left(x_1\right) \neq 0,$ and either $\mathrm{sgn}\left(a_1\right) = \mathrm{sgn}\left(b_1\right)$ or $b_1=0$.
We also have $$ \mathbb V\left(y\right) = \mathbb V\left(a_1 x_1 \right) = a_1^2\mathbb V\left(x_1 \right) $$ and thus $$ a_1b_1\mathbb V\left(x_1\right)= a_1^2\mathbb V\left(x_1 \right) \iff a_1 b_1 = a_1^2 \iff b_1 = a_1. $$

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  • $\begingroup$ If you apply $Cov(x_2, y)$ = $Cov(x_2, a_1x_1)$ = $a_1Cov(x_2, x_1)$ = 0. However, $Cov(x_1, x_2)$ > 0. Doesn't this mean that $a_1$ = 0 and therefore contradicts your conclusion that $a_1$ != 0? @whuber $\endgroup$
    – wakuwaku
    Commented Aug 29, 2022 at 0:39
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    $\begingroup$ @wakuwaku I agree, trying to satisfy all three (instead of just two) covariance conditions simultaneously seems to yield such contradictions. $\endgroup$
    – statmerkur
    Commented Aug 29, 2022 at 8:35

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