I came across the following probability theorem.
Let $X_1$ and $X_2$ be independent and identically distributed with density $f(x,\theta)$. If the random variable $Z=X_1+X_2$ is such that \begin{align} \log(f(X_1=x_1,X_2=x_2,\theta))&=\log{(f(X_1=x_1,\theta))}+\log{(f(X_2=x_2,\theta)}\\ &=\phantom{aaaaaa}g(Z,\theta)+h(x_1,x_2) \end{align} Then $f(x,\theta)$ can be written as $f(x,\theta)=exp(xc(\theta)+d(\theta)+S(x))$ for $c(\theta), d(\theta)$ and $S(x)$ real valued functions.
The book I am following has the following sketch for the proof
Define $r(x,\theta)=\log{(f(x,\theta))}-\log{(f(x,\theta_0))}$ for any fixed value of $\theta_0$ and let $q(x,\theta)=g(x,\theta)-g(x,\theta_0)$. Then it follows that $q(x_1+x_2,\theta)=r(x_1,\theta)+r(x_2,\theta)$.
Now, it follows that \begin{equation} r(x_1,\theta)-r(0,\theta)+r(x_2,\theta)-r(0,\theta)=r(x_1+x_2,\theta)-r(0,\theta) \quad (1) \end{equation}
The rest of the proof follows easily from the equation above and is left to the reader
I have some questions regarding the proof.
- I don't understand how equation (1) is obtained from $q(x_1+x_2,\theta)=r(x_1,\theta)+r(x_2,\theta)$
- Why the conclusion regarding the exponential follows from equation (1) above?
- What is the intuition behind the use of $q(x,\theta)$ and $r(x,\theta)$ in the proof?
You subtract a $\theta_0$ piece to get rid of $h(x_1, x_2)$, which allows for a simpler relation between $q$ and $r$ than between $f$ and $g$.
To get (1):
$$\begin{eqnarray} r(x_1,\theta)-r(0,\theta)+r(x_2,\theta)-r(0,\theta) &=& q(x_1+x_2,\theta) - 2r(0,\theta)\\ &=& q(x_1+x_2+0,\theta) - 2r(0,\theta)\\ &=& r(x_1+x_2,\theta) + r(0,\theta) - 2r(0,\theta)\\ &=& r(x_1+x_2,\theta) - r(0,\theta) \end{eqnarray} $$ As kimchi lover points out in the comments, then (assuming continuity) $r(x, \theta) - r(0,\theta)$ is linear in $x$, i.e. $$r(x, \theta) = c(\theta)x + d(\theta)$$ and the rest trivially follows.