Reversing the order of conditioning in a sum to compare conditional variances Suppose $Z=X+Y$ where $X$ is independent of $Y$ and $Y\sim N(0,1)$. I would like to compare $\text{var}(E(X|Z))$ to $\text{var}(E(Z|X))$. Obviously, $\text{var}(E(Z|X))=\text{var}(X)$.
In particular, my guess is that $\text{var}(E(Z|X)) > \text{var}(E(X|Z))$. If $Z$ is normal, then this is easy to prove directly from known formulas for conditional distributions of a multivariate Gaussian. Furthermore, if $Y$ is non-normal, then this appears to be false as well. The difficulty, as I see it, is in getting a handle on $E(X|Z)$ while accounting for the fact that $Y$ is Gaussian. Another approach is to compute (or control somehow) the marginal $P(Z)$, which would help understand $P(X|Z)=P(X,Z)/P(Z)$.
 A: EDIT: It seems that one can remove the integrability condition.
Suppose that $\text{Var}[X]<\infty$ and $Z=X+Y$, where $Y\sim\mathcal{N}(0,1)$ and $X$ and $Y$ are independent.
Because of the law of total variance,
$$\text{Var}\{E[Z|X]\}=\text{Var}\{X\}\\=E[\text{Var}\{X|Z\}]+ \text{Var}\{E[X|Z]\}\geq \text{Var}\{E[X|Z]\},$$
with equality iff $ \text{Var}\{X|Z\}=0$. So to prove your conjecture we only need to show $ \text{Var}\{X|Z\}>0$.
Theorem $\text{Var}\{X|Z\}=0$ if and only if $X$ is constant.
Proof: If $X$ is constant then $\text{Var}[X]=0$ and therefore also $\text{Var}\{X|Z\}=0$. For the converse, suppose that $\text{Var}\{X|Z\}=0$. Let $\mathbb{P}$ be the measure under which $Z=X+Y$, where $X$ has law $P_X$, $Y$ is standard Gaussian and independent of $X$, and $\mathbb{\tilde P}$ be the measure under which $Z$ is standard Gaussian and independent of $X$, which has the same law as under $\mathbb{P}$. Let
$$ L := \exp\left[X Z-\frac12 X^2\right].$$
Since $L$ is nonnegative, we have by Tonelli's theorem and the MGF of $\mathcal{N}(0,1)$ that
$$\mathbb{\tilde{E}}[L]=\int_{\mathbb{R}^2}e^{xz-\frac12 x^2}\mathbb{\tilde P}_X(dx)\otimes \mathbb{\tilde P}_Z(dz)=\int_{\mathbb{R}}e^{-\frac12 x^2}\left(\int_{\mathbb{R}}e^{xz}\mathbb{\tilde P}_Z(dz)\right)\mathbb{\tilde P}_X(dx)\\=\int_{\mathbb{R}}e^{-\frac12 x^2}e^{\frac12 x^2}\mathbb{\tilde P}_X(dx) = 1.$$
Thus $\mathbb{P}\ll\mathbb{\tilde P}$, $ \frac{d\mathbb{P}}{d\mathbb{\tilde P}} = L$, and we have an abstract Bayes formula
$$\mathbb{E}[\varphi(X)|Z]=\frac{\mathbb{\tilde E}[\varphi(X)L|Z]}{\mathbb{\tilde E}[L|Z]}.$$
Let $M(Z):=\mathbb{\tilde E}[L|Z]$, which is finite by assumption. Then the conditional variance of $X$ given $Z$ can be written as
$$\text{Var}\{X|Z\}=\frac{M''(Z)}{M(Z)}-\left(\frac{M'(Z)}{M(Z)}\right)^2 = \frac{d}{dz}\frac{M'(z)}{M(z)} \Bigg|_{z=Z}.$$
Since $\text{Var}\{X|Z\}=0$, we have
$$\frac{M'(Z)}{M(Z)} = \frac{M'(0)}{M(0)} = \frac{\mathbb{E}[Xe^{-\frac12 X^2}]}{\mathbb{E}[e^{-\frac12 X^2}]} . $$
But we also have
$$ X = \mathbb{E}\left[ X|Z\right] = \frac{M'(Z)}{M(Z)}.$$
It therefore follows that $X$ is constant.
