# What is the order of the lower tail of a Chi-Squared distribution?

Let X be a random variable with having a chi-squared distribution with n degrees of freedom and let y be some real number at most n. Is it known how P (X < y) behaves at least in some reasonable range of y? For instance, could one determine exactly the order of the latter probability when y=cn (for some $0\lt c \lt 1$) when n becomes large?

In the end, I am interested in the upper bound for the latter probability, but I am not sure if the usual Chernoff bound gives the correct magnitude.

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I fixed some broken formatting – Yemon Choi Feb 25 '13 at 10:07

You can get the correct exponential order of the decay of the probability from a large deviation principle. If $X_n$ has chi-squared distribution with $n$ degrees of freedom then $$X_n = Z_1^2 + Z_2^2 + \cdots Z_n^2,$$ where the $Z_i$ are independent standard normal random variables. Then, Cramer's theorem gives the exponential order of decay for both the left and right tails. That is, $$\lim_{n\rightarrow\infty} \frac{1}{n} \log P( X_n < c n ) = - I(c), \quad 0 < c < 1,$$ and $$\lim_{n\rightarrow\infty} \frac{1}{n} \log P( X_n > c n ) = - I(c), \quad 1 < c < \infty,$$ where $$I(c) = \frac{c-1-\ln(c)}{2}.$$
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, $$P(X_n < cn) \leq e^{-n I(c)}, \quad \forall n\geq 1, \text{ and } 0 < c < 1.$$