Analogues of the Golden-Thompson inequality Are there any analogues of the Golden-Thompson Inequality for moment generating functions? The Golden-Thompson Inequality asserts the following: If $A$ and $B$ are two $n \times n$ Hermitian matrices, then $\text{tr}(e^{A+B}) \leq \text{tr}(e^{A}e^{B})$ with no commutativity hypotheses on $A$ and $B$. 
If $A$ and $B$ commute, then $e^{A+B} = e^{A}e^{B}$. If not, then $e^{A+B} = e^{A}e^{B}e^{-\frac{1}{2}[A,B]} \dots$ 
So the trace operator gets rid of the "messiness." In a similar way, suppose we are given $X_1$ and $X_2$ as independent random variables. Then $E[e^{t(X_1+X_2)}] = E[e^{tX_1}] E[e^{tX_2}]$. Can we obtain a similar inequality to the Golden-Thompson with no hypotheses on $X_1$ and $X_2$?  
 A: Let $X_1=X_2=Z$ be a standard Gaussian (i.e. normally distributed with mean zero and variance 1). Then
$$ {\mathbb E}(e^{t(X_1+X_2)}) = {\mathbb E}(e^{2tZ}) = \frac{1}{\sqrt{2\pi}} \exp(2t^2) $$
while
$$ {\mathbb E}(e^{tX_1}) {\mathbb E}(e^{tX_2}) = [{\mathbb E}(e^{tZ})]^2 = \frac{1}{2\pi} \exp(t^2) $$
If I've got these calculations correct, they suggest to me that you are not going to get an analogue of Golden-Thompson, at least not the most naive analogue. Intuitively, one expects the product of the expectations to be smaller than the expectation of the product, in general, because of positive correlation effects.
[One does always have the easy Cauchy-Schwarz bound, but this seem to be of a very different flavour to the kind of inequality you describe in your question.]
EDIT a much simpler example: take $X_1=X_2=B$ to be a Bernoulli random variable which takes the values $0$ and $1$, each with probability 1/2. Then
$$ {\mathbb E}(e^{2tB}) = \frac{1}{2}(1+e^{2t}) $$
while
$$ [{\mathbb E}(e^{tB})]^2 = \frac{1}{4}(1+e^t)^2 $$
so that for all sufficiently large $t$ we have ${\mathbb E}(e^{2tB}) > [{\mathbb E}(e^{tB})]^2$
