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I have the following problem related to a statistics question:

Prove that the function defined for $x\ge 1, y\ge 1$, $$f(x,y)=\frac{\Gamma\left(\frac{x+y}{2}\right)(x/y)^{x/2}}{\Gamma(x/2)\Gamma(y/2)}\int_1^\infty w^{(x/2)-1}\left(1+\frac{xw}{y}\right)^{-(x+y)/2} dw$$

is increasing in $x$ for each $y\ge 1$ and decreasing in $y$ for each $x\ge 1$. (Here $\Gamma$ is the gamma function.)

Trying to prove by using derivatives seems difficult.

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    What is the statistical interpretation of this quantity? It looks like some quantity related to a beta distribution. Something like $U = W/(1+W)$ should be beta distributed. I haven't actually worked it out yet. Maybe that will provide clues.2011-03-13
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    Just a thought: I am pretty sure that if you do a change of variable $w\mapsto xw/y$ you can simplify your function $f(x,y)$ to the [regularized incomplete beta function](http://en.wikipedia.org/wiki/Beta_function) $I_{y/x}(x,y)$.2011-03-13
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    Oops, the correct version: $$ f(2x,2y) = I(\frac{y}{x+y}; y,x) $$ where $I(z;a,b)$ is the regularized incomplete Beta function.2011-03-13
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    Note that the property that $I(z;a,b) + I(1-z;b,a) = 1$ says that the statement that $f$ increases in $x$ and that $f$ decreases in $y$ are equivalent.2011-03-13
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    BTW, for $x = 2$, you have $f(2,y) = \left( \frac{y}{y+2}\right)^{y/2}$, and for $y = 2$, you have $f(x,2) = 1 - \left(1 - \frac{2}{x+2}\right)^{x/2}$ both of which you can check directly having the property that you want.2011-03-13

1 Answers 1

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Let $W \sim F(x, y)$ where $F(x,y)$ stands for an $F$ distribution with degrees of freedom $x$ and $y$. Then, the quantity

$$ \mathbb{P}(W \geq 1 ) = f(x,y)=\frac{\Gamma\left(\frac{x+y}{2}\right)(x/y)^{x/2}}{\Gamma(x/2)\Gamma(y/2)}\int_1^\infty w^{(x/2)-1}\left(1+\frac{xw}{y}\right)^{-(x+y)/2} \mathrm{d}w \> . $$

From this, I think you can find the answer in the reference below.

B. K. Ghosh, Some monotonicity theorems for $\chi^2$, $F$ and $t$ distributions with applications, J. Royal Stat. Soc. B, vol. 35, no. 3 (1973), pp. 480-492.

Incidentally, note that $W = \frac{y}{x} \frac{U_{xy}}{1-U_{xy}}$ where $U_{xy} \sim \mathrm{Beta}(x/2, y/2)$ and so $\mathbb{P}(W \geq 1) = \mathbb{P}(U_{xy} \geq (1+y/x)^{-1})$.

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    @cardinal: I am curious as to how you find this paper. Did you just google "monotoncity F-distribution"?2011-03-13
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    @TCL, no. I came across it years ago. I think I first saw it referenced in F. Graybill, *Theory and Application of the Linear Model*, 1976. When I saw the form of your integral, it reminded me of a beta integral and from there it reminded me of some quantity related to an $F$ distribution and that made me recall the paper.2011-03-13
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    @TCL, where did you come across this problem?2011-03-13
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    @cardinal: This is actually my conjecture, not a problem from somewhere. It arises from my previous question [A question about F-distribution]2011-03-13
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    @cardinal: This is actually my conjecture, not a problem from somewhere. It arises from my previous question [A question about F-distribution]. If this is true, then in testing about hypothesis about equality of variances when level of test is $<0.5$, the rejection region can be simplified. I think Ghosh's paper answers my question, but I have to check closely. Thank you for this reference.2011-03-13
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    @TCL, any update on your examination of the Ghosh paper? I will try to look at it again, though I may not get to it until the weekend.2011-03-17