Let's say we are performing simple linear regression of $y =b_1x + b_0$. Assume we estimate $b_1 = c$, we can use $c = 2$ for this example. When we consider the flipped regression of $x =b_1'y + b_0'$ can we establish any upper or lower bound on $b_1'$? Other than it being greater than 0?? This is the only thing I can think of because we know if $b_1 = 2, cov(x,y) > 0$
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The interpretation of this question, and therefore what the correct answer is, depend on what you mean by "know $b_1=c:$" are you referring to the actual parameters (which in practice are unknown) or to estimates based on data? – whuber Aug 11 '23 at 12:41
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1You might find the answers to Effect of switching... to be helpful. – Dilip Sarwate Aug 11 '23 at 18:21
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https://stats.stackexchange.com/questions/22718 is a very highly voted thread that shows the answer. For more like this, please search our site for switch x y regression. – whuber Aug 12 '23 at 14:46
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The equation of Regression Line, $y$ on $x$, in its explicit form reads
$$ y= r \frac{SD_y}{SD_x} x + \bar{y} - r\frac{SD_x}{SD_x} \bar{x}$$
And the question has provided us the coefficient of $x$ in the above equation to be equal to $c$, and that is given to be positive. That implies, $$ r\frac{SD_y}{SD_x} =c$$ $$\frac{SD_x}{SD_y} = r/c$$ $$r \frac{SD_x}{SD_y} = r^2/c$$
The equation of Regression Line, x on y, is simply obtained by replacing all y’s with x’s, thus, $$ x= r \frac{SD_x}{SD_y} y + \bar{x} - r\frac{SD_x}{SD_y} \bar{y}$$
Using the notations given in the question, we have: $$ b_1’ = r\frac{SD_x}{SD_y} = r^2/c$$ Since, $$|r| \leq 1 \implies r^2 \leq 1$$ $$ r^2/c \leq 1/c$$ Thus, $$ b_1’ \leq 1/c$$
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