I have the following equation : $$ E[e^{(\theta +1) X_T} | \mathcal{F}_t] = e^{X_t} E[e^{\theta X_T} | \mathcal{F}_t] $$ where $X_t$ is a Levy process, $\mathcal{F_t}$ the filtration at time $t$, $T > t$, and $\theta$ some number. I am trying to find a solution for $\theta$ in closed form (it probably has to involve the charateristic function $E(e^{iuX_t})$ of the Levy process)
What I have tried:
Since $e^{(\theta+1)X_t}$ is $\mathcal{F}_t$-measurable, it can be written $$ e^{(\theta +1 )X_t} E[e^{(\theta +1)(X_T-X_t)} | \mathcal{F}_t] = e^{X_t} E[e^{\theta X_T} | \mathcal{F}_t]$$ By stationarity of increments of the Levy process we have $$ e^{\theta} E[e^{(\theta +1)X_{T-t}} | \mathcal{F}_t] = E[e^{\theta X_T} | \mathcal{F}_t] $$ Now by making the substitution $\theta +1 = iu$ we rewrite the equation in terms of characteristic functions: $$ e^{\theta} e^{(T-t)\psi(u)} = e^{t\psi(u)}E(e^{-X_T}|\mathcal{F}_t) $$ Where $\psi$ is the characteristic exponent (given in closed form by the Levy-Khintchine formula). This is almost what I need, except the extra expectation.
Let's start from the beginning. Using the independence and stationarity of the increments, we see that
$$\begin{align*} \mathbb{E}\bigg(e^{(\theta+1) X_T} \mid \mathcal{F}_t \bigg) &= e^{(\theta+1)X_t} \mathbb{E}\bigg(e^{(\theta+1) (X_T-X_t)} \mid \mathcal{F}_t \bigg) \\ &= e^{(\theta+1)X_t} e^{(T-t) \psi(-\imath (\theta+1))} \tag{1} \end{align*}$$
where $\psi$ denotes the characteristic exponent of the Lévy process. Similarly,
$$e^{X_t} \mathbb{E}\bigg(e^{\theta X_T} \mid \mathcal{F}_t \bigg) = e^{(\theta+1) X_t} \mathbb{E}\bigg(e^{\theta (X_T-X_t)} \mid \mathcal{F}_t \bigg) = e^{(\theta+1)X_t} e^{(T-t) \psi(-\imath \theta)}. \tag{2}$$
This means that $\theta$ has to solve the equation
$$\psi(-\imath (\theta+1)) = \psi(-\imath \theta).$$
For example, if $(X_t)_t$ is a Brownian motion, then the solution is given by $\theta = -1/2$.