I have understood the basic knowledge of GLM. I know why a GLM consist of a predictor, a link function and a distribution. But I don't know how does the conditional mean connect to the distribution.

Using Poisson regression as example, we have $\log(\mu_i) = \beta_0 + \beta_1x_i$, where $\mu_i$ is the conditional mean.

How is the $\mu_i$ computed from the observed values $y_i$ and the Poisson distribution?

And also, how is the parameter $\lambda$ in the Poisson distribution computed?

  • $\begingroup$ Do you mean how the $\beta_0$ and $\beta_1$ are estimated? If you have those, then you get from $\log(\mu_i)$ to $\mu_i$ itself by $\mu_i = \exp(\hat\beta_0 + \hat\beta_1x_i)$. $\endgroup$
    – Dave
    Nov 19, 2021 at 18:48
  • $\begingroup$ In general that is a way to estimate the median but not the mean. I'm thinking of a log-normal distribution; need to check if this also applies to Poisson. $\endgroup$ Nov 19, 2021 at 18:56
  • $\begingroup$ @Dave, i do not quite understand how the reponse variable like count data contributed to the μi. $\endgroup$ Nov 20, 2021 at 16:11

1 Answer 1


A parametric distribution is defined by its family and its parameters. Consider the Poisson family of distributions. This distribution has a single parameter $\lambda$. In GLMs, we wish to model the mean of the parametric distribution as a function of a linear model of covariates. The mean should be called, as you say, or is treated as "the conditional mean (with respect to the covariates)." Therefore, for any parametric distribution, we must reparametrize it such that one parameter represents the mean ($\mu$). Furthermore, each observation is independent with its own mean $\mu_i$, but these means depend on a fixed number of shared parameters: $g(\mu_i) = \beta_0 + \sum_{j=1}^p \beta_p x_{ij}$, with $x_{ij}$ denoting the measurement of covariate $j$ for subject $i$.

For the Poisson family, let $y_i \sim Poi(\lambda_i)$ for $i=1,\cdots,n$ denote a set of independent count data. Suppose we also observe a vector of $p$ covariates $\boldsymbol{x}_i$ for each count datum. We believe that the logarithm of the mean (the canonical link) is linear related to these covariates (i.e. $\log(\mu_i) = \beta_0 + \sum_{j=1}^p \beta_p x_{ij}$). Since the mean of $y_i$, $\mu_i=\lambda_i$, we have that $y_i \sim Poi\left(\lambda_i = \exp \left( \beta_0 + \sum_{j=1}^p \beta_p x_{ij} \right)\right)$. Thus we merely replace $\lambda_i$ in the Likelihood with the model we think is correct.

  • $\begingroup$ If "each observation is independent with its own mean μi", then there are a number of i distribution, where each distribution has one mean. How can one single observation can generate one distribution? It is not a "distribution" but a data point... Sorry if I make silly statement $\endgroup$ Nov 20, 2021 at 16:10

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