# Assuming $u\sim N(0,\sigma^2)$ when y is highly skewed

does it make sense to assume $u\sim N(0,\sigma^2)$ when I know from a histogram that $y$ is highly skewed. Because from the assumption $u\sim N(0,\sigma^2)$ it follows that $y\sim N(x\beta,\sigma^2)$ and I'm absoluteley not sure if the assumption $u\sim N(0,\sigma^2)$ makes sense in a case where I know that the distribution of $y$ is not bell shaped. The alternative would be just to make OLS without any assumption about the error term, but in this case I can't analyze outliers and leverages (what I'd really like to do, because otherwise I can't present much more than a line which minimizes the sqaured sum of the residuals). [addendum: I can't make an outlier analysis because I can not define "outlier" in a context where I don't assume a normal distribution, because there is no outlying without a distribution] Besides your answers I'd really like to have a recommendation for a good book, where I can find some thoughts about what assumptions should we make when y is obviously not normal distributed.

Regards

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There is Taleb's book, although it will merely point the risk of this assumption. –  Wok Jul 20 '11 at 15:41
You need to put some more background into this. In the linear model $y_i = \alpha + \beta x_i + \epsilon_i$, it is not unusual to assume the $\epsilon_i$ are iid with a normal distribution of mean $0$, but there is no requirement for the $x_i$ or $y_i$ to be normally distributed. –  Henry Jul 20 '11 at 15:41
Well, just because $y|x \sim N(x \beta, \sigma^2)$ doesn't mean that a histogram of the marginal distribution of $y$ will look bell shaped; I believe that will only happen if $x$ is also normally distributed. –  Macro Jul 20 '11 at 16:37
@Henry For background you could read the almost identical series of questions here, here, and here. "What we've got here is failure to communicate." –  whuber Jul 20 '11 at 17:17
@Mark Pick any $\beta$ and $\sigma$, generate 10 iid draws $\epsilon_1$, ..., $\epsilon_{10}$ from a normal(0, $\sigma$) distribution, and create the dataset $((2^i, \beta 2^i + \epsilon_i), i=1,\ldots,10)$. When $|\beta|$ and $\sigma$ are near $1$, the y's will be highly positively skewed but the data are perfectly linear with beautifully normal errors. In short, the skewness of the y's comes from the skewness of the x's but (of itself) reveals nothing at all about the distribution of the residuals. You check distributional assumptions by studying the residuals, not the y's. –  whuber Jul 20 '11 at 19:44