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I have $N=1500$ sampled data points $\mathbf{y} = \{y_1, ... ,y_N\}$ which I would like to relate to some explanatory variables $\mathbf{X}$ and a parameter vector $\mathbf{\beta}$ i.e.

$\mathbb{E}(Y_i) = \mathbf{x_i}^T\mathbf{\beta}$

The first thing I would like to do is fit a distribution to $\mathbf{y}$.The R function fitdistr() suggests that a Weibull distribution be used. Unfortunately, some of the data is negative valued i.e. $y_i < 0$, and thus I cannot fit a Weibull distribution properly.

I want a distribution that acknowledges that my sample can be negative. Do I fit the Weibull to a shifted distribution or do I create a new distribution that fits my data? This new distribution would belong to the exponential family of distributions.

Vykta Wakandigara
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    Fitting a distribution to $y$ is meaningless for your model, because it posits that the distribution changes according to the values of $x.$ – whuber Jun 08 '19 at 18:39
  • So why do we specify distributions to a response before we apply a glm? Are you saying that the step of estimating a distribution for y is unnecessary when we have $\mathbb{E}(Y_i) = \mathbf{x}_i^T\mathbf{\beta}$. See https://stats.stackexchange.com/questions/240455/fitting-exponential-regression-model-by-mle – Vykta Wakandigara Jun 08 '19 at 21:32
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    First thing to do is to visualise the data. Plot histograms of the data to get a general impression of the distribution. Then you might choose a distributional assumption based on that impression combined with what you know about the data generating process. – Michael Lew Jun 08 '19 at 22:32
  • Is there a distribution that is usually used for data skewed to the right and with a support that includes negative values? – Vykta Wakandigara Jun 08 '19 at 22:49
  • Please ask with details about your real problem! Maybe look nto skew-normal distribution, at least if the skew is not too large – kjetil b halvorsen Jun 07 '21 at 23:43

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