Yes , you're right , but I have a question !
Can I get the same solution, or prove your claim if I let X =\frac{1}{a}\,x^{(1-a)/a} and Y =\frac{1}{a}\,y^{(1-a)/a} and find in which case \phi_{a}(X,Y)=X+Y-XY is positive ?
Thanks again
Hi Uart ,
Thanks for your reply , but I check for example the case
x=y \, \,\,\text{and}\,\,\, a=1/3 ,
and I see that \phi(x,y) can be non-negative for this
value of "a" which is not in the interval , you gave 1/2 \leq a \leq 1
And the other problem, as x and y are in...
0\leq x,y\leq1, \,\,a>0,\,\, \phi_{a}(x,y)=\frac{1}{a}\,x^{(1-a)/a}+\frac{1}{a}\,y^{(1-a)/a}-\frac{1}{a^2}\,(x\,y)^{(1-a)/a} .
In which case depending on the value of the parameter "a" ,the function \phi_{a}(x,y) \geq 0 .
Thanks
I understood that the value will be any constant depending on the choice of f which is arbitry chosen here. So I wanted to say that it is some L2 norm. But someone on the list said that It should be L1 norm (not L2 norm !), because the problem here is on a probabity space. So I wonder why, he...
Yes I'm talking about the bivariate bounded probability density function (pdf) f(x,y).
Sorry I can't understand the difference between L1 or L2 norm on the probability space.
About not using the L2 norm , I thought that the pdf could be written as
\int^{b}_{a} \int^{b}_{a} f^{2}(x,y)dxdy...