Estimating standard deviation in Poisson regression

Question

I'm interested in an estimator of the standard deviation in a Poisson regression. So the variance is

$$Var(y)=\phi\cdot V(\mu)$$

where $\phi=1$ and $V(\mu)=\mu$. So the variance should be $Var(y)=V(\mu)=\mu$. (I'm just interested in how the variance should be, so if overdispersion occurs ($\widehat{\phi}\neq 1$), I don't care about it). Thus an estimator of the variance should be

$$\widehat{Var}(y)=V(\widehat{\mu})=\widehat{\mu}$$

and an estimator of the standard deviation should be

$$\sqrt{\widehat{Var}(y)}=\sqrt{V(\widehat{\mu})}=\sqrt{\widehat{\mu}}.$$

Is this correct? I haven't found a discussion about standard deviation in the context with Poisson regression yet, that's why I'm asking.

Example:

So here is an easy example (which makes no sense btw) of what I'm talking about.

data1 <- function(x) {x^(2)}
numberofdrugs <- data1(1:84)
data2 <- function(x) {x}   
healthvalue <- data2(1:84)
plot(healthvalue, numberofdrugs)
test <- glm(numberofdrugs ~ healthvalue, family=poisson)
summary(test) #beta0=5.5 beta1=0.042
mu <- function(x) {exp(5.5+0.042*x)}
plot(healthvalue, numberofdrugs)
curve(mu,  add=TRUE, col="purple", lwd=2)
# the purple curve is the estimator for mu and it's also 
# the estimator of the variance,but if I'd like to plot 
# the (not constant) standard deviation I just take the 
# square root of the variance. So it is var(y)=mu=exp(Xb) 
# and thus the standard deviation is sqrt(exp(Xb))
sd <- function(x) {sqrt(exp(5.5+0.042*x))}
curve(sd, col="green", lwd=2)

Is the the green curve the correct estimator of the standard deviation in a Poisson regression? It should be, no?

Answer 1

Poisson regression finds a value $\hat{\beta}$ maximizing the likelihood of the data. For any value of $x$, you would then suppose $Y$ has a Poisson($\exp(x \hat{\beta})$) distribution. The standard deviation of that distribution equals $\exp(x \hat{\beta}/2)$. This appears to be what you mean by $\sqrt{\widehat{\mu}}$.

There are, of course, other ways to estimate the standard deviation of $Y|x$. However, staying within the context of Poisson regression, $\exp(x \hat{\beta}/2)$ is the ML estimator of SD($Y|x$) for the simple reason that the ML estimator of a function of the parameters is the same function of the ML estimator of those parameters. The function in this case is the one sending $\hat{\beta}$ to $\exp(x \hat{\beta}/2)$ (for any fixed value of $x$). This theorem will appear in any full account of maximum likelihood estimation. Its proof is straightforward. Conceptually, the function is a way to re-express the parameters, but re-expressing them doesn't change the fact that they maximize (or fail to maximize, depending on their values) the likelihood.

Answer 2

You are thinking too much in terms of "normally distributed" here. For a normal distribution, you have two parameters then mean $\mu$ and the variance, $\sigma^2$. So you require two pieces of information to characterize the probability distribution for the normal case.

However, in the Poisson distributed case, there is only one parameter, and that is the rate $\lambda$ (I relabeled to avoid confusion with normal). This characterizes the Poisson distribution, and so there is no need to refer to other quantities.

This is why probably why don't hear standard deviation "estimation" mentioned in Poisson regression. Asking for a standard deviation estimator for a Poisson random variable is analogous to asking for a kurtosis estimator for a normally distributed random variable. You can get one, but why bother? By estimating the rate parameter $\lambda$, you have all the information you need.