Let $X \sim \chi^2_n$ random variable. I am looking for a large deviation upper bound for $X$. The answer here, says that

Since you said that you're looking for an upper bound, it should also be noted that an examination of the proof of Cramer's theorem shows that you can actually get uniform exponential upper bounds (and not just asymptotics as stated above). In fact, $$ P(X_n < cn) \leq e^{-n I(c)}, \quad \forall n\geq 1, \text{ and } 0 < c < 1, $$ where $I(c) = \frac{c - 1 - \log c}2.$

This bound will be quite useful to me, but I don't quite see how the proof follow's from Cramer's theorem. I am looking at the proof of Cramer's Theorem here, and I don't see how I can get rid of the asymptotics.

The answer provided here is unsatisfactory, since that mention the bounds $$P(X_n \ge n + 2\sqrt{nx}+2x) \le e^{-x}.$$ $$P(X_n \le n - 2\sqrt{nx}) \le e^{-x}.$$

which could correspond to the kind of bound I mention above if I set $x = nI(c)$, in which case, the left hand sides are not particularly useful. I do the following simulation to indicate what I mean. I let $n = 5$, and for $0 < c < 1$, I produce plots of what $nc$ looks like, and what $n - 2\sqrt{n*nI(c)}$ looks like. According to the inequalities, the upper bound on the probability of even $X_n < nc$ and $X_n < n - 2\sqrt{n*nI(c)}$ is the same, $e^{-nI(c)}$. (This is R code)

> c <- seq(0.001, .999, length = 100)
> Ic <- (c - 1 - log(c))/2
> n <- 5
> plot(c, n - 2*sqrt(n*n*Ic), type = "l")
> lines(c, n*c, type = "l", col = "red")
> legend("bottomright", legend = c("n - 2sqrt(n*n*I(c))", "nc"),lty = 1, col = c("black", "red") )

enter image description here

This is not useful for me, since I am looking for bounds of the form $e^{-nI(c)}$, and would like a reasonable expression for $t$ in $Pr(X_n < t) < e^{-nI(c)}$. The value of $t = n - 2\sqrt{n*nI(c)}$, although a stronger result, is of no help analytically, because $t$ is less than 0.

  • $\begingroup$ Also, asked here, but no satisfactory answers. $\endgroup$ – Greenparker Aug 6 '18 at 8:38
  • $\begingroup$ why you find the answer in your provided link unsatisfactory? $\endgroup$ – pointguard0 Aug 6 '18 at 8:42
  • $\begingroup$ @pointguard0 Updated the question to indicate reasons. Thanks $\endgroup$ – Greenparker Aug 6 '18 at 8:47
  • $\begingroup$ all right, can you explain why l.h.s. are not useful? $\endgroup$ – pointguard0 Aug 6 '18 at 9:13
  • $\begingroup$ in fact, i suspect that provided inequalities are stronger than that of yours $\endgroup$ – pointguard0 Aug 6 '18 at 9:14

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