The following question seems simple but I am not sure how to handle it correctly because of the integral with respect to a measure. I would be very thankful for any reply.Cheers!
Knowing that $$f(\psi(t)/t,1/t)=\int_0^\infty exp(\theta \psi(t)/t-\theta^2/2t)F(d\theta )=a$$ where a and $\theta $ are positive real constants,F is a positive $\sigma-finite$ measure and $\psi$ an initinity often continuously differentiable function of time, how can we deduce that $\psi(t)/t$ is monotone decreasing?
Let $g(t):=\psi(t)/t$ and $h(t):=-1/(2t)$, so that $$\int_0^\infty e^{\theta g(t)+\theta^2 h(t)}F(d\theta)=a$$ for all $t>0$. Assume that the support set of measure $F$ contains some point in the interval $(0,\infty)$.
Then indeed $g(t)$ is decreasing in $t>0$. Indeed, otherwise there are real $s$ and $t$ such that $0<s<t$ and $g(s)\le g(t)$. Since the function $h$ is increasing on $(0,\infty)$, we have $h(s)<h(t)$ and hence $e^{\theta g(s)+\theta^2 h(s)}<e^{\theta g(t)+\theta^2 h(t)}$ for all $\theta>0$, so that $$ a=\int_0^\infty e^{\theta g(s)+\theta^2 h(s)}F(d\theta) <\int_0^\infty e^{\theta g(t)+\theta^2 h(t)}F(d\theta)=a, $$ which is the desired contradiction.
