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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.

  • 0
    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
  • 0
    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
  • 0
    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
  • 2
    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
  • 0
    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

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