Let $X=(X_{1},...,X_{d})$ be a random vector with values in $\mathbb{R}^{d}$, defined on some finite probability space $(\Omega,F,\mathbb{P})$. Denote by $F:\mathbb{R}^{d}\rightarrow\mathbb{R}$ the function defined by $F(\theta)=E[e^{{\theta}.X}]$.
Show that if F achieves a minimum at $\theta^{*} \in \mathbb{R}^{d}$, we have $E[Xe^{{\theta}.X}]=0$
My idea is the following:
I take a partial derivative of the function: $F(\theta)=E[e^{{\theta}.X}]$ for $\theta$, therefore I find:
$F(\theta)=E[Xe^{{\theta}.X}]$
I said that the only value that make null the function is when $X=0$ and since $F(X^{°}) \leq F(X)$, I have a minimum.
Is that correct?
Partial differentiation gives $EX_ie^{\theta . X}=0$ for each $i$. $Xe^{\theta . X}$ is a vector valued random variable and its expectation is the vector whose components are the expectations of the components of $Xe^{\theta . X}$. Hence $EXe^{\theta . X}$ is the vecor $(0,0,...,0)$.