# Find the expected value of “Y”, exponential family with lots of questions here

I have a problem I don't know how to approach. It is

A generalization of the 1-parameter exponential family, to allow 2-parameter distribution, is the family given by

$$f(y;\theta, \phi)=e^{\frac{y \theta - c(\theta)}{a(\phi)}+d(y,\phi)}$$

Show that $E(Y)=c'(\theta)$ and $var(Y)=c''(\theta) a(\phi)$

Well, first off, maybe this is common understanding for experienced people but is this exponential distribution a continuous one? A Discrete one?

I ask this because my only idea to approach the question is by first principles, so for the expected value of Y, I thought either integrating or summing $yf(y)$ would do (if that is possible at all) but from the information given, I don't know which to go for.

Is there a simpler way? I am confused, it would be great if someone would tell me how to solve this.

• The exponential distribution is continuous. – Em. Jan 30 '16 at 23:30
• Are you sure you are not missing a minus sign before $y\theta$? And what are $c,d,a$ supposed to be? (functions, which which properties?) – Clement C. Jan 30 '16 at 23:40
• I've typed it out as it is, no minus sign etc. $c$, from what it says which is all I know too, is a function of $\theta$, one of the 2 parameters and same for $a$ as a function of $\phi$. I take $\theta, \phi$ to be something like the "mean " "variance" etc. $y$ is the random variable. I tried integrating bu since $d$ is some unknown function of $y, \phi$, it's inconclusive.... – John Trail Jan 31 '16 at 2:52
• It seems to me that a simple special case is the exponential distribution. See Wikipedia, and parameterize using the mean $\mu$ instead of the rate $\lambda = 1/\mu,$ so that $f(x) = (1/\mu)e^{-x/\mu},$ for $x \ge 0.$ One way to show that $E(X) = \mu$ is to use integration by parts. That might get you started. – BruceET Jan 31 '16 at 4:34
• Can you please state what book this is from and please check to see if you made any typos on $f$? By the way, exponential families consist of both discrete and continuous distributions. – Clarinetist Jan 31 '16 at 4:43

Suppose $$Y$$ is continuous, and is of the two-parameter exponential family given above.
Without getting into technical details, assume that interchange of differentiation and integration is allowed. We know that $$\int_{\mathbb{R}}f(y; \theta, \phi)\text{ d}y = 1\text{.}$$
Differentiate both sides to get $$\dfrac{\partial}{\partial \theta}\int_{\mathbb{R}}f(y; \theta, \phi)\text{ d}y = \int_{\mathbb{R}}\dfrac{\partial}{\partial \theta}f(y; \theta, \phi)\text{ d}y = 0\text{.}\tag{1}$$ Now $$\dfrac{\partial}{\partial \theta}f(y; \theta, \phi) = e^{d(y, \phi)}\exp\left[\dfrac{y\theta-c(\theta)}{a(\phi)} \right](y-c^{\prime}(\theta)) = [y-c^{\prime}(\theta)]f(y;\theta, \phi)$$ so that $$(1)$$ implies that $$\int_{\mathbb{R}}[y-c^{\prime}(\theta)]f(y;\theta, \phi)\text{ d}y = \int_{\mathbb{R}}yf(y;\theta, \phi)\text{ d}y - \int_{\mathbb{R}}c^{\prime}(\theta)f(y;\theta, \phi)\text{ d}y = 0\text{.}\tag{2}$$ Notice two things: that $$c^{\prime}(\theta)$$ is not dependent on $$y$$, and do you recognize what $$\int_{\mathbb{R}}yf(y;\theta, \phi)\text{ d}y$$ is?
For a second run, it is shown similarly that$$\int_{\mathbb{R}}\dfrac{\partial^2}{\partial \theta^2}f(y; \theta, \phi)\text{ d}y = 0\text{.}\tag{3}$$ and I leave it as an exercise to you to show that \begin{align} \dfrac{\partial^2}{\partial \theta^2}f(y; \theta, \phi) &= [y-c^{\prime}(\theta)]^2 f(y;\theta, \phi) - c^{\prime\prime}(\theta)f(y;\theta, \phi) \\ &= \{y^2 - 2yc^{\prime}(\theta) + [c^{\prime}(\theta)]^2\}f(y; \theta, \phi)- c^{\prime\prime}(\theta)f(y;\theta, \phi)\text{.} \end{align} Integrate this over $$\mathbb{R}$$, set it equal to $$0$$ (i.e., do $$(3)$$), and proceed similarly to how $$(2)$$ is derived. Furthermore, use the result of $$(2)$$ to do $$(3)$$. Be sure to simplify this expression. This will finish your proof for the continuous case. The summation case is similar.