Hello,
I have two increasing concave functions f and g where f is more concave than g.
Let x be a random variable.
Does f more concave than g imply that Cov(x,f(x))>=Cov(x,g(x))?
I would say yes, but I am not sure...
Many thanks for your help!
Hello,
I have two increasing concave functions f and g where f is more concave than g.
Let x be a random variable.
Does f more concave than g imply that Cov(x,f(x))>=Cov(x,g(x))?
I would say yes, but I am not sure...
Many thanks for your help!
Thanks a lot so far. You are right, I was not very precise. f more concave than g means that f''(x)<g''(x)<0 for every x. (I assume that f and g are two times continuously differentiable on their domain). Moreover, f(x) equals -g'(x). Do you mean with your example of a linear function that we care more about the slope of the functions that is if f'(x)>g'(x) or in other words if -g''(x)>g'(x)?
I have two increasing concave functions f and g.
properties:
g is three times continously differentiable.
f(x)=-g'(x)
-g'''(x)< g''(x)< 0 ----> f''(x)<g''(x)
-g''(x)>g'(x)> 0 ----> f'(x)>g'(x)
x is a random variable.
Is the following statement true (for any distribution of x):
Cov(x,f(x))>=Cov(x,g(x)) ?