Monotonic behavior of a function

Question

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.

Answer 1

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 $\mathbb{P}(W \geq 1) = \mathbb{P}(U_{xy} \geq (1+y/x)^{-1})$.