Let $X_n$ be a chi-squared random variable with $n$ degrees of freedom. What are the sharpest known lower bounds on the tails of its distribution? Specifically, I am looking for the lower bounds in the form:

$$P(X_n-n\geq f_1(x)\sqrt{n}+g_1(x))\geq \exp(-x)$$ $$P(n-X_n\geq f_2(x)\sqrt{n}+g_2(x))\geq \exp(-x)$$

for some $f_1(x)>0$, $f_2(x)>0$, and $g_1(x)\geq0$, $g_2(x)\geq0$.

Basically, I am wondering if the lower-bound "equivalent" of the following upper bounds from Massart and Laurent (see Lemma 1 on page 1325) exist:

$$P(X_n-n\geq 2\sqrt{xn}+2x)\leq\exp(-x)$$ $$P(n-X_n\geq 2\sqrt{xn})\leq\exp(-x)$$

I tried Cramer-Chernoff Theorem (i.e. Theorem 1 in here), but the lower bound isn't sharp enough...

  • $\begingroup$ The left hand side of the second Massart-Laurent inequality above is zero if $x\ge n/4$, hence can't be bounded from below by an eponential function. $\endgroup$ Apr 6, 2013 at 20:31

3 Answers 3


This is not quite the form you requested. But if your goal was to control chi-squared left tails, I have found these lower bounds helpful.

For $X_1 \sim \chi^2_1$:

Let $Z \sim N(0,1)$, so that $Z/\sqrt{2} \sim N(0, 1/2)$. For $x>0$, we have $\text{erf}(x) = P\left(|Z/\sqrt{2}| \leq x\right) = P(X_1 \leq 2x)$.

Apply this bound or this refinement: $\text{erf}(x) \leq \sqrt{1-\exp(-4x^2/\pi)}$.

Then $P(X_1 \geq y) \geq 1- \sqrt{1-\exp(-2y/\pi)}$ for $y>0$.

For $X_2 \sim \chi^2_2$:

Directly from the chi-squared CDF for $n=2$, we see that $P(X_2 \geq x) = e^{-x/2}$.

Also, since the chi-squared distribution shifts right as the df increases, $P(X_n \geq x) \geq e^{-x/2}$ for $n\geq 2$.

For other $n$:

See this answer to another question.


You should quantify the range of $x$ that you are interested in. For example, for $1<<x<<\sqrt{n}$ you should be in the moderate deviations regime, and you could give lower bounds by performing the appropriate $n$-dependent change of measure and computing sharp asymptotics.


I believe you can find your answer in this link:

  • 1
    $\begingroup$ I had somehow the impression that the question is about possible lower bounds... $\endgroup$ Sep 23, 2013 at 20:01
  • $\begingroup$ @AndrasBatkai is correct. I obtained the upper bounds that I put in the statement of my question from the paper cited in that response on stats.SE that you put in your answer. I may have even gotten the reference from that answer (which prompted my question), so I should probably link the two... $\endgroup$
    – Bullmoose
    Sep 23, 2013 at 23:07

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