# 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 $X_2$?

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Is there a typo on the right hand side of your first inequality? – Yemon Choi Aug 3 '10 at 6:58
What is the typo? – PV1707 Aug 3 '10 at 7:00
You've fixed it now, I think. (In the old version you had Tr($e^{AB}$) which can't be right.) – Yemon Choi Aug 3 '10 at 7:06
You also forgot to state that $X_1$ and $X_2$ must be independent for the expectation identity to hold. – Per Vognsen Aug 3 '10 at 7:07
A quantum observable $A$ is a Hermitian operator on the state space, the operator $e^{itA}$ is time evolution by $t$ when $A$ is used as the Hamiltonian (the energy observable), when a state is represented by a density operator $\rho$ then $tr(A \rho)$ is the expectation of the observable values of $A$ in that state, etc. So the setup for the Golden-Thompson inequality seem to be a 'Wick rotation' away from the quantum mechanical setup that would allow you to draw conclusions. It seems to be about characteristic functions, not moment-generating functions. – Per Vognsen Aug 3 '10 at 8:06

## 1 Answer

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 ${\mathbb E}(e^{2tB}) > [{\mathbb E}(e^{tB})]^2$

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Jensen's inequality gives you $\mathbb{E} (e^{2tX} ) > [\mathbb{E} (e^{tX})]^2$ for any random variable $X$ with exponential moments. – Jeff Schenker Aug 3 '10 at 20:39
Unless $X$ is constant, that is. – Jeff Schenker Aug 3 '10 at 20:40
Good point, Jeff: I vaguely realized that Jensen's inequality applied while I was typing this, but couldn't remember the conditions for strict inequality to hold. – Yemon Choi Aug 3 '10 at 21:21