# standardised random variable least square regression $X$ against $Y$, $Y$ against $X$ [duplicate]

Let $X$ and $Y$ be mean 0 and variance 1 random variables such that we choose $\alpha$ and $\beta$ to minimise

$$\mathbb{E}(X-\beta Y)^2$$

and

$$\mathbb{E}(Y-\alpha X)^2$$

after not so difficult derivation, I arrive at $\alpha = \mathbb{E}(XY)/\mathbb{E}X^2$ and $\beta = \mathbb{E}(XY)/\mathbb{E}Y^2$, so $\alpha = \beta$.

This seems very strange, because if $y=mx$ is regression line, then surely $x = \frac{1}{m }y$.

• (X - Y)^2 = (Y - X)^2 for each X and Y. Basically it's the same function (expand the expressions for prove). That's why you get alpha = beta – Ivan Oct 23 '15 at 12:12
• @Ivan sorry, you are going too fast for me. How are they the same function? The coefficient is in front of $X$ in one of the expression and $Y$ in the other. – Lost1 Oct 23 '15 at 12:15
• that's minimization problem. In best case you have E(X - \betaY)^2 = 0 expanding the expression X^2 - 2X\betaY + Y^2 = 0 \beta = (X^2 - 2xy + y^2) and E(Y-\alphaX)^2 = 0 \alpha = (Y^2 - 2xy + X^2) witch is basically the same – Ivan Oct 23 '15 at 12:19
• @Ivan when you expand you should get $X^2+2\beta XY +\beta^2 Y^2$? why has the beta's disappeared? I do not follow this. – Lost1 Oct 23 '15 at 12:21
• What do you precisely mean by "because if $y=mx$ is regression line, then surely $x = \frac{1}{m }y$". Is it to say that, if $\hat\beta$ is the regression coefficient of a regression of $y$ on $x$, $\hat\beta^{-1}$ is the coefficient of the regression of $x$ on $y$? – Christoph Hanck Oct 23 '15 at 14:07

The regression line of $Y$ against $X$ minimises the total squared distance of the $Y$ COORDINATES to the $y$ COORDINATES of the line, where as $X$ against $Y$ minimise the total squared distance of $X$ COORDINATES to the $x$ COORDINATES line.