Concavity of a function of 2 variables

[NotEuler]

Hi all,

I'm trying to show that a particular function of two variables is concave for all positive values of x and y. I'm pretty sure it is, but I haven't been able to prove it.

The function is (1-e^t(y-x))xy)/(x+e^t(y-x)y)

t is a positive parameter, and as mentioned, x and y are always positive.

I've tried looking at second derivatives and the Hessian, but haven't really got anywhere. I wonder if this is even possible to prove generally, or just for specific values of x and y?

Actually, for my pyrposes it would be sufficient to show that for any given fixed value of y, the function is a concave function of x, and vice versa. I'm not sure if this is any easier...

Any input and suggestions would be much appreciated!

 

[mfb]

Actually, for my pyrposes it would be sufficient to show that for any given fixed value of y, the function is a concave function of x, and vice versa. I'm not sure if this is any easier...

That looks easier, but I would expect the more general approach works as well.

There are mismatching brackets in the numerator that make the expression ambiguous.

 

[NotEuler]

There are mismatching brackets in the numerator that make the expression ambiguous.

Ah yes, sorry! A sign typo too, unfortunately. I now tried to figure out how to copy an expression from mathematica in latex... let's see if this works, hopefully will help to avoid any typos in the future.The correct expression is:

$\frac{\left(1-e^{t (-x+y)}\right) x y}{x-e^{t (-x+y)} y}$

Thanks for your reply, and sorry for the mistake!

 

[mfb]

WolframAlpha gets a positive second x derivative for y=2, x=1, t > -1.

 

[NotEuler]

WolframAlpha gets a positive second x derivative for y=2, x=1, t > -1.

It looks to me like the 2nd derivative is positive only for values t<-ln(2)=-0.69. Would you agree? I only need to show concavity for t>0.

If you're correct, then it obviously isnt't concave, if I'm right, then it could be.

 

[mfb]

Example query - if the url does not work: d^2/dx^2 ( (1-e^(t*(2-x)))*x*2/(x-e^(t*(2-x))*1) ) with t=1 and x=1

t>0, x>0, y>0, but the second derivative is positive.

 

[NotEuler]

I think

( (1-e^(t*(2-x)))*x*2/(x-e^(t*(2-x))*1) )

should be
( (1-e^(t*(2-x)))*x*2/(x-e^(t*(2-x))*2) ) ?

Seems that the 2nd derivative is then negative.

 

[mfb]

Oops, forgot to change that when I changed y=2 to y=1.

$$xy\frac{1-e^{t(y-x)}}{x-ye^{t(y-x)}}$$
Let's introduce c=e^ty > 0 and z=xt > 0. Also ignore the first factor of y because it does not change the sign of the second derivative.
$$\frac{z}{t}\frac{1-ce^{-z}}{\frac{z}{t}-yce^{-z}} = z\frac{1-ce^{-z}}{z-ycte^{-z}}$$
Define yct=d
$$z\frac{e^z-c}{ze^z-d} = 1-\frac{zc-d}{ze^z-d}$$
With positive constants c and d and z>0. That looks much easier.

 

[NotEuler]

Thanks, that looks very helpful! Unfortunately the 2nd derivative still looks very complicated, and I got stuck once again. (By the way, there's a - sign missing in the denominator of the first expression, which carries over into all the others).

Here's another suggestion, which makes it even simpler: Introduce z=x/y and a=e^ty. Then (ignoring again the first factor y) the function simplifies to
$\frac{z \left(1-a^{1-z}\right)}{z-a^{1-z}}$
or
$\frac{z \left(a^z-a\right)}{z a^z-a}$

So now we're only left with a parameter a>1, and z>0. It should be sufficient to show that the 2nd derivative for z is negative, but still this looks hard. I wonder if there's some other way to go about this..

One thing that becomes clear from this form is that the function is not defined at z=1 (equivalent to x=y in the original function). So 0<z<1 and 1<z<∞ would probably have to be investigated separately.

 

[mfb]

Ah, I got that sign error from post 1 where the denominator had a +.

You can do the same trick with your expression:
$$\frac{z \left(a^z-a\right)}{z a^z-a} = \frac{z a^z-a+a-za}{z a^z-a} = 1-a\frac{z-1}{z a^z-a}$$
So we have to show the second derivative of $\frac{z-1}{z a^z-a}$ is positive. This is still too messy, but we can do more (using a>1): for z<1, z a^z < a and therefore the whole expression is positive. For z>1, the expression is positive as well. Let's define
$$f(x)=\frac{z a^z-a}{z-1}$$
Now $$\left(\frac{1}{f(x)}\right)'' = \frac{1}{f^3}(2f'^2-f''f)$$
That is more complicated, but the derivatives are easier to calculate. I'll go back to the expression with c and d because it avoids log(a) in the derivatives, the term is still positive for the same reason.

$$f(z)(d-cz)=d-z e^z$$
$$f'(z)(d-cz)^2=cd +ce^zz^2-de^zz-de^x$$
$$f''(z)(d-cz)^3=2c^2d - c^2 e^z z^3 + 2cd e^z (z^2+2z-1) - d^2 e^x x -2d^2 e^x$$
So... is two times the square of the middle expression always larger than the product of the first and last expression? I took (d-cz) terms out because they'll appear in 4th power in both cases and cancel (they are always positive as well, ignoring x=y).
I see some terms that cancel but it still looks messy.

 

[NotEuler]

Thanks - but like you said, still looks a bit messy.
I might be clutching at straws here, but the function can also be written as

$xy \frac{e^{t x}-e^{t y}}{x e^{t x}-y e^{t y}}$

Could the symmetry of this form be used somehow to help solve the problem?

 

[mfb]

14 and 10 terms, significantly easier than the expressions before - did you try it?

The symmetric form invites to set x'=tx, y'=ty, which gives
$$xy\frac{e^x-e^y}{xe^x-ye^y}$$
after ignoring a constant factor of t and using x and y again. And that second derivative looks interesting as well, would need several x>y, y>x estimates.