This is an example using the Poisson distribution illustrate exponential tilting.
Let \(p_0(x)\) denote the “standard” Poisson distribution (\(\mu=1\)):
\[p_0(x) = e^{-1}/x!\]and \(\tilde{p}(x\vert\theta)\) its exponentially tilted form:
\[\tilde{p}(x|\theta) = p_0(x)e^{\theta x}.\]This isn’t a proper distribution, hence the notation \(\tilde{p}(x\vert\theta)\), but it would be if we determined the normalizing constant, which I will denote \(\exp\{\psi(\theta)\}\):
\[\begin{alignat*}{2} \exp\{\psi(\theta)\} &= \sum_{x=0}^{\infty} e^{-1} e^{\theta x}/x! \\ &= e^{-1} \exp\{e^\theta\} & \hspace{8em}& e^x = \sum_{n=0}^\infty x^n / n! \\ &= \exp\{e^\theta - 1\} \end{alignat*}\]Thus \(\psi(\theta) = e^\theta - 1\) and
\[\as{ p(x|\theta) &= p_0(x)e^{\theta x - \psi(\theta)} \\ &= \exp\{-1 + \theta x - e^\theta + 1\} / x! \\ &= \exp\{\theta x - e^\theta \} / x!, }\]or, letting \(\mu = e^\theta\),
\[p(x|\theta) = e^{-\mu} \mu^x / x!,\]the familiar form of the Poisson distribution. Note the main idea, that we started with a base distribution and ended up with a family of derived distributions, arising from the process of exponential tilting.