Is the marginal distribution of an exponential tilted measure also an exponential tilted measure? Given a random variable $X \in \mathbb{R}^2$ with probability distribution $P$ and moment generating function $M_X(\lambda) = E[e^{\langle \lambda, X \rangle} ] < \infty$, we define the exponentially tilted measure $P_\lambda$ as follows:
$$
  P_\lambda(dx) = \frac{e^{\langle \lambda,  x \rangle}}{E[e^{\langle \lambda, X \rangle}]} P(dx)
$$
Let $Y$ be a random variable with probability distribution $P_\lambda$. My question is whether $Y_1$ has the distribution of an exponentially tilted measure? That is if $P^{(1)}$ is the probability distribution of $X_1$, does there then exist a $\theta$ such that $Y_1$ has probability distribution $P^{(1)}_\theta$?
 A: You have
$$
p'_{y'}
= \int \frac{e^{\lambda_1 x + \lambda_2 y'}}{E[e^{\lambda_1 X + \lambda_2 Y}]}p_{x,y'}dx
= e^{\lambda_2 y'}p_{y'}\int \frac{
e^{\lambda_1 x}}{\int e^{\lambda_1 x + \lambda_2 y}p_{x,y}dxdy}\frac{p_{x,y'}}{p_{y'}}dx
=e^{\lambda_2 y'}\frac{\phi_{X|Y=y'}(\lambda_1)}{\phi_{X,Y}(\lambda_1,\lambda_2)}p_{y'}
$$
So you want to show that
$
\frac{\phi_{X,Y}(\lambda_1,\lambda_2)}{\phi_{X|Y=y'}(\lambda_1)} = \phi_Y(\lambda_2).
$
or at least that it equals $\phi_Y(\lambda_2+c) e^{-c y'}$ for some constant $c$.
This seems a bit too strong.
If we have a distribution over $\{0,1,2\}\times\{0,1\}$ with probability matrix $$\begin{pmatrix}p_{00}&p_{01}\\p_{10}&p_{11}\\p_{20}&p_{21}\end{pmatrix}.$$
Then the $\theta$ tilted distribution is $$\begin{pmatrix}p_{00}&e^{\theta_1}p_{01}\\e^{\theta_2}p_{10}&e^{\theta_1+\theta_2}p_{11}\\e^{2\theta_2}p_{20}&e^{\theta_1+2\theta_2}p_{21}\end{pmatrix}\big/C$$ for some normalization constant $C$. The marginals are
$\begin{pmatrix}p_{00}+e^{\theta_1}p_{01}\\e^{\theta_2}(p_{10}+e^{\theta_1}p_{11})\\e^{2\theta_2}(p_{20}+e^{\theta_1}p_{21})\end{pmatrix}\big/C$
and in particular $$\frac{P[X'=0]}{P[X'=1]} = \frac{p_{00}+e^{\theta_1}p_{01}}{e^{\theta_2}(p_{10}+e^{\theta_1}p_{11})}
,\quad
\frac{P[X'=0]}{P[X'=2]} = \frac{p_{00}+e^{\theta_1}p_{01}}{e^{2\theta_2}(p_{20}+e^{\theta_1}p_{21})}
.$$
Meanwhile, simply tilting the marginal distribution gives
$\begin{pmatrix}p_{00}+p_{01}\\e^{\theta}(p_{10}+p_{11})\\e^{2\theta}(p_{20}+p_{21})\end{pmatrix}\big/C'$.
So here $$\frac{P[X'=0]}{P[X'=1]} = \frac{p_{00}+p_{01}}{e^{\theta}(p_{10}+p_{11})}
,\quad
\frac{P[X'=0]}{P[X'=2]} = \frac{p_{00}+p_{01}}{e^{2\theta}(p_{20}+p_{21})}
.$$
We would need to find a value for $\theta$ that made those two fractions equal the two above, but that seems impossible for general $\theta_1, \theta_2$.
