True or false? Positiveness of the parameter of the moment generating function We have a random variable with expected value $\mathbb{E}\left [X \right] < 0$, paired with a parameter $\theta \neq 0$. The expectation $\mathbb{E} \left[ e^{\theta X} \right] = 1$. Is it true that $$ \theta > 0 ? $$ My answer is that the affirmation is clearly FALSE, since the hypotheses are wrong: the first moment shouldn't be less than zero, because if we differentiate the mgf we get exactly zero. I'd say, even better, that given those specific hypotheses, nothing can be concluded about the value of $\theta$, it might simply be $\theta \in \mathbb{R}$. Am I wrong? Hope that anyone can help :)

The moment generating function $f:t\mapsto E[e^{t X}]$ is convex, so its graph lies above all its tangent lines, including the one tangent at $(0,f(0))$. In particular $f(\theta)\ge f(0)+f'(0)\theta$. The mgf also has the property that its derivatives, evaluated at $0$, are the moments of $X$, so that $E[X^k] = f^{(k)}(0)$: that's why its called the _moment_ generating function. But we know $f(0)=1$ and are told that $f(\theta)=1$ and that $E[X]<0$. Put these together and we have $1\ge 1+ f'(0)\theta = 1+E[X]\theta$ and so $0\ge E[X]\theta.$ This implies $\theta\ge0$.

For example, suppose $X\sim N(-1,1)$, for which $E[X]<0$ and whose mgf is $$f(t)=E[\exp(tX)]=\exp(t^2/2-t)=\exp((t-1)^2/2-1/2).$$ There are two values of $\theta$ such that $f(\theta)=1$, namely $\theta=1\pm1$, that is, $0$ or $2$. Of these, the value of $\theta$ for which $\theta\
e0$ is $\theta=2$, which is positive.