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 $X_2$?

  • $\begingroup$ Is there a typo on the right hand side of your first inequality? $\endgroup$ – Yemon Choi Aug 3 '10 at 6:58
  • $\begingroup$ What is the typo? $\endgroup$ – PV1707 Aug 3 '10 at 7:00
  • $\begingroup$ You've fixed it now, I think. (In the old version you had Tr($e^{AB}$) which can't be right.) $\endgroup$ – Yemon Choi Aug 3 '10 at 7:06
  • $\begingroup$ You also forgot to state that $X_1$ and $X_2$ must be independent for the expectation identity to hold. $\endgroup$ – Per Vognsen Aug 3 '10 at 7:07
  • $\begingroup$ 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. $\endgroup$ – Per Vognsen Aug 3 '10 at 8:06

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

  • 1
    $\begingroup$ Jensen's inequality gives you $\mathbb{E} (e^{2tX} ) > [\mathbb{E} (e^{tX})]^2$ for any random variable $X$ with exponential moments. $\endgroup$ – Jeff Schenker Aug 3 '10 at 20:39
  • $\begingroup$ Unless $X$ is constant, that is. $\endgroup$ – Jeff Schenker Aug 3 '10 at 20:40
  • $\begingroup$ 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. $\endgroup$ – Yemon Choi Aug 3 '10 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.