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I am working with a sum of variables $X_i$; they are all independent, but not identically distributed. For any $i$, I can show the bound $$\Lambda^*_{X_i}(t) := \sup_t \langle t, x \rangle - \Lambda_X(t) \ge f_i(t)$$ for some concave $f_i$, i.e. the Legendre transform of the log-moment generating function for each of the $X_i$ from below by some concave function (not sure about the convention on negative signs, but my bound is in the direction that says that tails of the $X_i$ are very light).

Can I then conclude that: $$\Lambda^*_{\sum_{i=1}^n X_i} (t) \ge n \min_{i} f_i(t/n)$$

intuitively, this seems like it should be trivial, e.g. by Gartner Ellis, but I am finding that the $^*$ operation does not respect sums nicely. (For now, I am not worried about convergence issues and such).

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$\newcommand{\La}{\Lambda}$ The Legendre transform $\La^*_X$ of the log-moment generating function of a random variable $X$ is given by the formula $$\La^*_X(x):=\inf_{t\ge0}(-tx+\La_X(t)), $$ where $\La_X(t):=\ln Ee^{tX}$. The function $\La^*_X$ is the pointwise infimum in $t\ge0$ of the concave functions $x\mapsto-tx+\La_X(t)$. So, $\La^*_X$ is concave.

So, we may take $f_i=\La^*_i:=\La^*_{X_i}$ for all $i$. Then for $\La_i:=\La_{X_i}$ we have \begin{align} \La^*_{\sum_{i=1}^n X_i}(x)&=\inf_{t\ge0}\sum_{i=1}^n(-t\tfrac xn+\La_i(t)) \\ &\ge\sum_{i=1}^n\inf_{t\ge0}(-t\tfrac xn+\La_i(t)) \\ &=\sum_{i=1}^n\La^*_i(\tfrac xn) \\ &=\sum_{i=1}^n f_i(\tfrac xn) \\ &\ge n\min_{i=1}^n f_i(\tfrac xn). \end{align}

So, we get the inequality opposite to what you suggested. Usually, this opposite inequality will be strict, if indeed the $X_i$'s are not identically distributed.

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  • $\begingroup$ Hi Iosef, thank you for the reply. I think we both had some signs flipped. My statement had some backwards inequalities that I have fixed. I think your definition of $\Lambda^*$ is also off by a negative sign. The bound you gave (i.e. that the inf of a sum is not smaller than the sum of infs) is of course true, but I am interested in the other direction. My guess after thinking more is that there should not be, in general, a good bound in the other direction. $\endgroup$
    – DJA
    Commented Dec 11, 2019 at 22:27
  • $\begingroup$ @DJA : You are still saying that you are unsure about signs, but what you seemed to be sure about was that "[your] bound [was] in the direction that says that tails of the $X_i$ are very light". The definition of $\Lambda^*_{X_i}(t)$ in my answer implies that the inequality $\Lambda^*_{X_i}(t)\le f_i(t)$ in your original post means exactly that the right tails are light enough. So, I think your original question has been answered. $\endgroup$
    – Iosif Pinelis
    Commented Dec 12, 2019 at 4:17
  • $\begingroup$ Previous comment continued: If you want to ask another question, I think you should do it in a separate post. Also, after you changed your post, it has become only more confusing. In particular, now your $\Lambda^*_{X_i}$ is convex, rather than concave. Also, the function $\Lambda_X$ is still undefined in your post. $\endgroup$
    – Iosif Pinelis
    Commented Dec 12, 2019 at 4:19

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