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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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    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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    @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