I am struggling to find the expression that shows that some transformation of Y belongs to the exponential family. Were Y has the Pareto distribution. I am given the following pdf:
$$f(y;\theta)=\frac{\theta k^{\theta}}{y^{(\theta+1)}} \quad where \quad y>k,\theta>0,k>0,$$
I have done the following so far:
$$f_Y(y;\theta)=exp[log(\frac{\theta k^{\theta}}{y^{(\theta+1)}})]$$
$$ =exp[log(\theta k^{\theta})-log(y^{(\theta +1)})]$$
$$=exp[log(\theta k^{\theta})-(\theta + 1) log(y)]$$
$$=exp[(-\theta -1)\cdot log(y)+log(\theta k^{\theta})$$
I am confused about obtaining the mean and variance, since I have only learned how to do this when we have the following form of the exponential family:
But this seems to only work when you have three variables but I have two here. So I can't write it in this form. Could someone help me out?
Pareto distribution does not belong to the exponential family! It is easy to verify this because it support depends on the parameter.
Mean and variance can be calculated in the standard way, using the definition
$$\mathbb{E}[X]=\int xf(x)dx$$
$$\mathbb{E}[X^2]=\int x^2f(x)dx$$
$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$
In your specific case you have
$$\begin{align} \mathbb{E}[X]& = \theta k^{\theta}\int_k^{\infty}y^{-\theta}dy\\ & = \frac{\theta k^{\theta}}{1-\theta}\left[y^{1-\theta} \right]_{y=k}^{y=\infty} & = \begin{cases} \infty, & \text{if $\theta\le 1$ } \\ \frac{k\theta}{\theta-1}, & \text{if $\theta >1$ } \end{cases} \end{align}$$
... and so on