Looking for some help on my statistics homework question!
The background to the question is: suppose that you toss a biased coin repeatedly (and independently) until you get a head. Let Y denote the number of tails observed before that happens. Then Y has a geometric distribution with parameter $\psi$, where $\psi$ is the probability of getting a head on any individual coin toss.
The probability mass function of the geometric distribution with the aforementioned parameter is : $f(y) = (1- \psi)^y * \psi$ where $y = 0, 1, 2 ….$
The mean of this distribution is $\mu = (1-\psi)/ \psi$ and the variance $\sigma^2 = (1-\psi)/\psi^2$.
The questions:
Show that the geometric distribution is a member of the natural exponential family.
I have attempted to manipulate the probability mass function to look like the natural exponential form and got that $f(y) = exp(y\log(1-\psi) + \log(\psi))$ but i'm not sure if this is correct?
Find the canonical link function for a GLM with geometric response variable.
So I have from above that $\theta$ is $\log(1-\psi)$, and so to get the link function, g, I find $\psi$ in terms of $\mu$ in the equation for the mean of the distribution above, which gives me $\psi = 1/(\mu+1)$. I then take theta, $y\log(1-\psi)$, and substitute $\psi$ for $\psi$ in terms of $\mu$ to get the canonical link function as $\log(1-(1/(\mu+1)))$. Again, I'm not entirely sure whether this is the correct way to answer the question.
For what I can find online about the geometric distribution, the probability mass function is written slightly differently, so any help on how I might be able to fix either of my answers would be much appreciated!
Best Answer
I know this is an old thread, but I rather put the answer here to help someone out in the future.
Yes, you have written it correctly. Furthermore, you can define $\theta=\ln(1-\psi)\implies \psi=1-e^{\theta}$, $a(\phi)=1$, $b(\theta)=-\ln(\psi)=-\ln(1-e^\theta)$, and $c(y;\phi)=0$ to show that it is a member of the exponential family, that is, $$f(y;\theta,\phi)=\exp\biggl\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y;\phi)\biggr\}.$$
You wrote the canonical link function properly. For a geometric distribution with pmf $(1-\psi)^y\psi$, $y=0,1,\ldots$, the canonical link function is given by $g(\mu)=\ln\bigl(\frac{\mu}{1+\mu}\bigr)$ using the $\mathbb{E}[Y]=\mu$ relationship that you described.