Prove Chi Square is Stochastically Increasing

[Mogarrr]

Homework Statement

Prove that the X^2 distribution is stochastically increasing in its degrees of freedom; that is if p>q, then for any a, P(X^2_{p} > a) \geq P(X^2_{q} > a), with strict inequality for some a.

Homework Equations

1.(n-1)S^2/\sigma^2 \sim X^2_{n-1}
2.The Chi squared(p) pdf is
f(x|p)= \frac 1{\Gamma(p/2) 2^{p/2}}x^{(p/2) - 1}e^{-x/2}

The Attempt at a Solution

Since p>q, this implies \forall a, \frac{\sigma^2 a}{p}< \frac{\sigma^2 a}{q}.
Also, X^2_{k} \sim kS^2/\sigma^2.

Therefore \forall a, P(X^2_{p}>a) = P(S^2 > \sigma^2 a/p) \geq P(S^2 > \sigma^2 a/q) = P(X^2_{q}>a).

If a>0, we observe strict inequality, as the support of S^2 is [0,\infty)...

What do you think? If I am going in the wrong direction, please steer me in the right one.

 

[Ray Vickson]

Homework Statement

Prove that the X^2 distribution is stochastically increasing in its degrees of freedom; that is if p>q, then for any a, P(X^2_{p} > a) \geq P(X^2_{q} > a), with strict inequality for some a.

Homework Equations

1.(n-1)S^2/\sigma^2 \sim X^2_{n-1}
2.The Chi squared(p) pdf is
f(x|p)= \frac 1{\Gamma(p/2) 2^{p/2}}x^{(p/2) - 1}e^{-x/2}

The Attempt at a Solution

Since p>q, this implies \forall a, \frac{\sigma^2 a}{p}< \frac{\sigma^2 a}{q}.
Also, X^2_{k} \sim kS^2/\sigma^2.

Therefore \forall a, P(X^2_{p}>a) = P(S^2 > \sigma^2 a/p) \geq P(S^2 > \sigma^2 a/q) = P(X^2_{q}>a).

If a>0, we observe strict inequality, as the support of S^2 is [0,\infty)...

What do you think? If I am going in the wrong direction, please steer me in the right one.

I cannot follow your argument. You need two different $S^2$ random variables---one for $\chi^2_p$ and a different one for $\chi^2_q$. Basically, though, you need to know what $\chi^2$ really is, in simple, intuitive terms: if $Y_r$ has the distribution $\chi^2_r$ then
Y_r = Z_1^2 + Z_2^2 + \cdots + Z_r^2,
where $Z_1, Z_2, \ldots, Z_r$ are iid standard normal random variables.

When looking at stochastic ordering, you are entitled to use a common sample space $\Omega$, since all that matters is how the distribution functions compare (not, for example, whether the two random variables are independent, or not). We can always "construct" a sample space such that the iid N(0,1) random variables $Z_1, Z_2, \ldots, Z_p$ are functions over $\Omega$ (so their values "observed" on a sample point $\omega \in \Omega$) are $Z_1(\omega), Z_2(\omega), \ldots, Z_p(\omega)$. Then
Y_q(\omega) = \sum_{j=1}^q Z_j(\omega)^2 \; \text{and} \;Y_p(\omega) = \sum_{j=1}^p Z_j(\omega)^2
For $q < p$, what does that tell you?

 

[Mogarrr]

Based on your hint, here's what I'm thinking...

Well, then, if Z_{i} \sim N(0,1) (and i.i.d.) , and X^2_{n} = \sum_{i=1}^{n} Z_{i}, then X^2_{p} = pZ^2_1 and X^2_{q} = qZ^2_1.

Thus for all a, P(X^2_{p} &gt; a) = P(pZ^2_1 &gt; a ) = P(Z^2_1 &gt; \frac {a}{p}) \geq P(Z^2_1 &gt; \frac {a}{q}) = P(X^2_{q} &gt; 1), since p&gt;q. If a&gt;0, then there should be strict equality (I think).

Whatdya think?

 

[Ray Vickson]

Based on your hint, here's what I'm thinking...

Well, then, if Z_{i} \sim N(0,1) (and i.i.d.) , and X^2_{n} = \sum_{i=1}^{n} Z_{i}, then X^2_{p} = pZ^2_1 and X^2_{q} = qZ^2_1.

Thus for all a, P(X^2_{p} &gt; a) = P(pZ^2_1 &gt; a ) = P(Z^2_1 &gt; \frac {a}{p}) \geq P(Z^2_1 &gt; \frac {a}{q}) = P(X^2_{q} &gt; 1), since p&gt;q. If a&gt;0, then there should be strict equality (I think).

Whatdya think?

I think you are writing down a bunch of material that makes no sense at all. Chi-squared = a sum of squares of iid N(0,1) random variables, not just a sum. I suggest you sit down, relax, and proceed slowly and carefully. Think about the actual definitions, and think about what I wrote in my previous response.

Alternatively, you can try to proceed directly: $X$ (with cdf $F$) dominates $Y$ (with cdf $G$) if $G(x) \geq F(x)$ for all $x$, and $>$ holds for some $x$. In other words, the stochastically larger random variable has a smaller cdf; that is, for each $x$, it lies below the other one---so $F$ describes a distribution the lies to the "right" of $G$. That means that the density functions $f,g$ must satisfy $\int_0^x [g(t) - f(t)] \geq 0$ for all $x > 0$. You have formulas for $f(t)$ and $g(t)$, so you can try to verify the cumulative ordering. That might be hard to do, so that is why I suggested an alternative approach.

 

[Mogarrr]

Oops. Should have written X^2_{n} = \sum^{n}_{i=1} Z^2_{i}. I was thinking that since the Z_{i}'s are identically distributed, that I could just sum up their squares, X^2_1. Why am I wrong here, if they're identically distributed?

I will try the direct method, since X^2_{p} \sim \Gamma(p/2, 2).

Update: Ok, I believe I've found a solid direct proof using the Chi square, Gamma and Poisson distribution. I'm excited, but busy now, so check tomorrow for my reply! And thanks for the help.

 

[Mogarrr]

Ok, there is a relationship between the Gamma and Poisson distribution. Let X \sim Gamma(\alpha, \beta), then
P(X \leq x) = P(Y \geq \alpha), where Y \sim Poisson(x/\beta).

Now let p &gt; q, p,q \in \mathbb{Z}^{+} a&gt;0, then
P(X^2_{p} &gt; a) &lt; P(X^2_{q} &gt; a)
\Leftrightarrow P(Y &gt; p) &lt; P(Y &gt; q)
\Leftrightarrow 0 &lt; P(Y&gt;q) - P(Y&gt;p)
\Leftrightarrow 0 &lt; \sum^{\infty}_{y=q/2} \frac {e^{-a/2}(a/2)^{y}}{y!} - \sum^{\infty}_{y=p/2} \frac{e^{-a/2}(a/2)^{p} }{y!}
= \sum^{p/2}_{y=q/2} \frac {e^{-a/2} (a/2)^{y} }{y!}.
The above is positive, proving a strict inequality for some a.

If a \notin \mathbb{R}^{+}, then
P(X^2_{p} &gt; a) = 1 = P(X^2_{q} &gt; a),
hence \forall a \in \mathbb{R}, P(X^2_{p} &gt; a) \geq P(X^2_{q} &gt; a), if p&gt;q.

I think this is right.