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I am wondering if there is a tight lower bound on the probability of a maximum of $n$ i.i.d. chi-square random variables, each with degree of freedom $d$ exceeding a value close to $d$. Formally, I need to lower bound the following expression:

$$P(\max_i X_i\geq d+\delta)$$

where each $X_i\sim\chi^2_d$ for $i=1,2,\ldots,n$ and $\delta$ is small. Ideally, I would like an expression involving elementary functions of $n$, $d$, and $\delta$. I am interested in the asymptotics and assume that $n$ and $d$ are large.

What I tried

We know that $P(\max_i X_i\geq d+\delta)=1-P(X<d+\delta)^n$ where $X\sim\chi^2_d$. Therefore, I tried to upper bound $P(X<d+\delta)$ using the CLT, which yields the normal approximation to chi-squared distribution, and the lower bounds on the Q-function. However, the resultant overall bound is not tight, and I am hoping something better exists.

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  • $\begingroup$ Would it be possible to give any additional information regarding $d$? Is it large, does it help to work with even $d$, etc? $\endgroup$ – Rookatu Jul 24 '13 at 23:16
  • $\begingroup$ Both the number of degrees of freedom $d$ for each chi-square variate $X_i$ as well as their number $n$ is large. However, I can't make assumptions on whether $d$ is even or not. $\endgroup$ – M.B.M. Jul 25 '13 at 3:08
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I'm not sure if this is helpful, but taking a look at what Wikipedia has to say about the cdf of the Chi-Square distribution, it appears that

\begin{align*} P[X < d+\delta]^n &= \left( \left( \frac{d+\delta}{2}\right)^{d/2} e^{-d/2} \sum_{k=0}^\infty \frac{\left(\frac{d+\delta}{2}\right)^k}{\Gamma(d/2 + k + 1)}\right)^n \\ &= \left( e^{-d/2} \sum_{y=d/2}^\infty \frac{\left(\frac{d+\delta}{2}\right)^y}{\Gamma(y + 1)}\right)^n \\ &= e^{n\delta/2}\left(e^{-(d+\delta)/2} \sum_{y=d/2}^\infty \frac{\left(\frac{d+\delta}{2}\right)^y}{\Gamma(y + 1)}\right)^n, \end{align*} and so the term $$ e^{-(d+\delta)/2}\sum_{y=d/2}^\infty \frac{\left(\frac{d+\delta}{2}\right)^y}{\Gamma(y + 1)} $$ is bound above by $1$ (this is certainly clear for even $d$). It may be that the larger $d$ is, the further the term is from $1$.

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  • $\begingroup$ Hmmm... thank you for your thoughts, but is there an arithmetic error in the third equation -- shouldn's the first exponential be $e^{n\delta/2}$, without the minus sign? Otherwise the exponentials don't add up to $e^{-nd/2}$... $\endgroup$ – M.B.M. Jul 25 '13 at 3:25
  • $\begingroup$ Yes, you are right, sorry about the error. I have edited the post. I'm not so confident that this is helpful, but it was all I could think of at the time. I may ponder over it more if I get the chance. Good luck with this! $\endgroup$ – Rookatu Jul 25 '13 at 3:56

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