Exponential tilting

exponential tilting reweights a probability law in favour of larger or smaller values. Let $K(\theta )=\log {\mathbb{E}}_P{e}^{\theta X}$ be finite at the chosen parameter. Define the tilted law by its Radon-Nikodym derivativeIts total mass is one by construction. Positive $\theta$ favours larger $x$, while negative $\theta$ favours smaller $x$; it changes probabilities without creating values outside the original support.

For $\theta$ in the interior of the finite moment-generating-function domain, differentiation under the integral givesThe interior-domain condition supplies the integrable exponential bounds needed for these derivatives. Endpoint tilts can exist without a finite tilted mean, so existence of a normalized tilt alone is insufficient for the mean formula.

For a Bernoulli distribution with success probability $p\in (0,1)$, the tilted success probability is $p_{\theta }=p{e}^{\theta }/(1-p+p{e}^{\theta })$: tilting adds $\theta$ to its log odds. A Normal distribution $N(\mu ,{\sigma }^2)$ tilts to $N(\mu +{\sigma }^2\theta ,{\sigma }^2)$ by completing the square. These calculations explain why exponential tilting is useful for making a rare Sample mean typical while retaining an explicit Likelihood ratio.

Suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X}<\infty$ for all $\lambda \ge 0$. Under the exponential tilting $d{\mu }^{\lambda }=e^{\lambda x}d\mu /\varphi (\lambda )$, $X$ is an integrable random variable whenever $\lambda >0$: its negative exponential tail is bounded, and its positive contribution is dominated by a larger positive exponential moment. Differentiation under the integral sign gives the displayed mean. At the endpoint $\lambda =0$, the positive part of $X$ is integrable, but the negative part need not be. The right derivative ${\varphi }_{+}^{\prime }(0)=\mathbb{E}X$ is valid as an extended value, possibly $-\infty$; a finite mean at zero requires an extra negative-tail integrability assumption.