Edit: Sorry yeah not too clear, probably posted this too late at night...
Essentially I have data which appears to be in exponential form - a log-log graph put it close to a straight line. Using R, I fit it using lm(log(y)~log(x)) which gave good estimates for $a$ and $b$ with a high $R^2$ value and the estimated equation following the shape of the data points.
My question is would Y still follow a normal distribution as if I had used lm(y~x)? Unless I'm seriously mixing things up, in this case $Y\sim N(aX+b,\sigma^2)$ so would this still be the case after log-transforming x and y?
Thanks
I am assuming you have estimated the coefficients of the regression by minimizing the squared distances. The error would be normal as usual.
Think like this, you have transformed the data by taking logarithm (assuming all data points are positive here), and you presumed that error would be log-normal. But if this is the case the error would always be positive. Which is not possible. Because as you know from the one of the assumptions that you take when doing linear regression (OLS), is all the errors sum up to zero.
Distribution of $Y$ would change. Without much thinking you can say that because you have changed the domain of $Y$ as well as the mean and the variance of the distribution. Intuitively speaking, by taking the $log$, you have squeezed the domain (e.g. what was $Y$= 100 then, is $log(100)=2$ now. So you can see you changed the distribution function to somewhat having "peakedness", "spiky"! This happens because the integration over the entire domain must sum to 1, as you have contracted the domain, so you have to lump your probability mass over smaller length compared to before.
Hope this helps!