# Linear regression with prior on $\arctan \beta_1$

Suppose we have $\hat{y} = \beta_1 x + \beta_0$ (I ask only for the univariate case.)

A typical Bayesian approach might involve Normal priors on both parameters. I was thinking today about a different approach. I imagined that the slope parameter should have a noninformative prior that uniformly sweeps the range of possible angles, so $\gamma = \frac{\arctan \beta_1 + \pi/2}{\pi/2}$; $\gamma \sim \text{Beta}(1,1)$ or whatever informative Beta we choose. Then $\beta_0\mid\beta_1$ is Normal or something.

Alternatively, define parameter $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\hat{y} = \left(\tan{\theta}\right)x + \beta_0$.

Has this been done before? Is this useful?

• It could only be useful in circumstances where it's ok to get different answers depending on the units of measurement chosen for $y$ and $x$, because a change in one unit of measurement can change that "noninformative" prior into a strongly informative one. The problems that would permit such behavior would primarily be those where $y$ and $x$ are the same kind of quantity expressed using the same units of measurement and the same origin. – whuber Jul 30 '15 at 20:13
• @whuber, how come? – Simon Kuang Jul 30 '15 at 20:15
• Look at an example. Suppose $y$ measures mean happiness on a $0-10$ scale and $x$ is mean income per country. If you measure $x$ in dollars per annum, then your prior will put half its probability on exceeding a happiness of $10$ for annual incomes above $10$, which is the entire world. If you measure $x$ in millions of dollars per day, your prior will put most of its probability on billionaire-level incomes, far exceeding anything possible.There aren't enough countries in the world to make the posterior distributions come close to matching--the results depend on how you measure income. – whuber Jul 30 '15 at 20:23
• In addition to whuber's noteworthy remarks, speaking of an angle for a unidimensional parameter is not particularly pertinent. – Xi'an Jul 31 '15 at 7:06
• @whuber I still don't get it. Of course you can incorporate a scale parameter into the prior like we always do. Isn't the prior centered at a slope of 0? – Simon Kuang Aug 2 '15 at 16:09