# Using the definition of a Chi-square Distribution, prove that $P(\chi_n^2>1)$ is increasing in $n$.

I tried to prove using the definitions and characteristics of the Chi-square distrbution.

Let $X_1,X_2,X_3,...,X_n$ be iid random variables and let $S_n=\sum_{k=1}^{n}X_k$, then $$(a) \space S_n\sim \chi_{n}^{2} \Longleftrightarrow X_1\sim \chi_{1}^{2},\space\space \text {and}\\ (b) \space X_1\sim \mathcal {N}(0,1) \Longrightarrow \sum_{k=1}^{n}X_k^2\sim \chi_n^2\\ (c) \space S_n\sim \chi_{n}^{2}, \space S_k\sim \chi_{k}^{2}, \Longrightarrow T_n=S_n+S_k\sim \chi_{n+k}^{2}$$.

If $X$ has a Chi-square distribution with $n \space \text {d.f.}$ ,i.e, $X\sim\chi_n^2$, then it's pdf is given by $$f(x)= {{\frac{x{^{\frac{n}{2}-1}\space e{^\frac{-x}{2}}}}{2{^\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}\space for \space\space x\geq0\\0 \space\space\space\space\space for \space \space\space\space x<0}}$$

Mean of a Chi-square distribution is the $d.f = n$, variance = $2. df.=2n$ and mode is $df-2=n-2$.

Setting $$f'(x)=\mathbb e{^\frac{-x}{2}}.x{^{\frac{n}{2}-2}}(\frac{x+2-n}{2})=0\space (\because e{^\frac{-x}{2}}\geq0,\space x{^{\frac{n}{2}-2}}>0, \space\therefore f'(x)>0)\\ \Longrightarrow x=n-2 \\ \text {and} f''(x)|_{x=n-2}<0$$

Looking at the characteristics obviously I can say that $f(x)$ is increasing and mean and variance of this distribution increase as $d.f$ increases. (Does mode decrease?)

$\therefore P(\chi_n^2>1)\Longrightarrow P(X_1^2+X_2^2+X_3^2+...+X_n^2>1)\\ \Longrightarrow P(\frac{x{^{\frac{n}{2}-1}\space e{^\frac{-x}{2}}}}{2{^\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}>1)\\\Longrightarrow \int_{1}^{\infty}\frac{x{^{\frac{n}{2}-1}\space e{^\frac{-x}{2}}}}{2{^\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}dx\\\Longrightarrow \int_{0}^{\infty}\frac{x{^{\frac{n}{2}-1}\space e{^\frac{-x}{2}}}}{2{^\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}dx \space - \space \int_{0}^{1}\frac{x{^{\frac{n}{2}-1}\space e{^\frac{-x}{2}}}}{2{^\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}dx$, where each $X_i \sim \mathcal {N}(0,1)$

I am not sure how to proceed further to prove that $P(\chi_n^2>1)$ is increasing in $n$ using the definition of a Chi-square Distribution or particularly I am stuck at the part which says increasing in n. I cannot prove that. Any help or explanation is highly appreciated.

• I think it is easiest to use $\sum_{k=1}^n X_k^2\sim \chi_n^2$, at least if this counts as "using the definition". Commented Apr 1, 2018 at 12:39
• @PhoemueX, would you please explain this? I am trying but stuck at the part which says increasing in n. Cannot use $P(\chi_n^2 >1)$, to show $n$ is increasing.
– vbm
Commented Apr 1, 2018 at 13:28

Let it be that $X,Y$ are nonnegative random variables.

Then: $$\{X>1\}\cup\{Y>1\}\subseteq\{X+Y>1\}\tag1$$ so that:$$P(X>1)\leq P(X>1)+P(X\leq1\wedge Y>1)\leq P(X+Y>1)\tag2$$If $P(X\leq1\wedge Y>1)>0\tag3$ then we have also the strict inequality: $$P(X>1)<P(X+Y>1)\tag4$$

Applying this on $X=\chi^2_{n-1}$ and $Y=X_n^2$ we find:$$P(\chi_{n-1}^2>1)<P(\chi_{n}^2>1)\tag5$$

• Just to clear my confusion I ask you this question. Definition of Chi-square assumes the r.vs are iid but is here (in this solution) $X$ and $Y$ independent? And is it $P(X>1)<P(X+Y+1)$? or $P(X>1)<P(X+Y>1)$?. Also following your steps I think if $P(X>1)<P(X+Y>1)$, then $P(\chi_{n-1}^2>1)<P(\chi_{2n-1}^2>1)$. Am I wrong?
– vbm
Commented Apr 1, 2018 at 14:02
• $(1)$ and $(2)$ are valid for nonnegative $X,Y$ and dependency (or not) does not matter. $(3)$ is evidently true in your case. In $(4)$ there was a typo ($+$ was wrong and must be $>$, sorry, it is repaired now). Note that $Y=X_n^2$ where $X_n\sim\text{Norm}(0,1)$ as defined in your question. So not $Y=\chi_n^2$ as you interpreted. Commented Apr 1, 2018 at 15:00
• it is clear and (+1) for this solution. Thank you so much.
– vbm
Commented Apr 1, 2018 at 15:18
• You are welcome. Commented Apr 1, 2018 at 15:21