I have seen references to the "tilting method" in the theory of large deviations. Is there a simple explanation of what this is, exactly?

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    $\begingroup$ There is a brief discussion of tilting in section 23.4 of Klenke's book Probability Theory. A Comprehensive Course, Springer Verlag, 2008. $\endgroup$ Apr 15, 2014 at 15:51

1 Answer 1


Tilting refers to a change of measure of the form $e^{\lambda \cdot x}/C(\lambda)$ where $C(\lambda)=E(e^{\lambda\cdot X})$. (The setup is that the measure for which you are proving large deviations is a topological vector space $X$, and $\lambda$ is in the dual space to $X$). The name tilting comes because the effect of the change of measure translates to modifying the ``typical behavior'' under the tilted measure; more technically, the tilting adds a linear term to the log-moment generating function, and that translates to a linear shift in the corresponding Legendre transform that will represent the rate function.

EDIT (in response to Igor's further question in comment): The reason you change measure is the following. You are trying to compute $P_n(A)$ which is small. Write now $$P_n(A)=\int_A dQ_n (dP_n/dQ_n)$$ and choose $Q_n$ such that $Q_n(A)$ is large (i.e., make $A$ "typical", so that $n^{-1}\log Q_n(A)\sim 0$). Now if $(dP_n/dQ_n)$ is almost constant on $A$, say equal $e^{-nJ}$, then you are happy because then $P_n(A)\sim e^{-nJ} Q_n(A)$. The devil is in the details of the words ``almost constant'' and $\sim$ but this is roughly the idea.

  • $\begingroup$ Thanks, but why would one do such a thing? Presumably you want to prove your large deviation result with a given measure.... $\endgroup$
    – Igor Rivin
    Apr 15, 2014 at 18:26

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