Show that if $E(X) < 0$ and $\theta \neq 0$ is such that $E(e^{\theta X}) = 1$, then $\theta > 0$

Question

Show that if $E(X) < 0$ and $\theta \neq 0$ is such that $E(e^{\theta X}) = 1$, then $\theta > 0$

I'm not sure about how to show $\theta > 0$. This is my try:

We have

$$E\left(e^{\theta X}\right) = E\left(\sum_{n=0}^{\infty} \frac{(\theta x)^{n}}{n!} \right) = 1$$

$$\Longleftrightarrow 1 + E\left(\sum_{n = 1}^{\infty} \frac{(\theta X)^{n}}{n!}\right) = 1,$$

and by Linearity,

$$ 1 + E\left(\frac{\theta X}{1!}\right) + E\left(\frac{\theta^{2}X^{2}}{2!}\right) + \cdots = 1$$

$$\Longleftrightarrow \frac{\theta}{1!} E(X) + \frac{\theta^{2}}{2!} E(X^{2}) + \cdots = 0.$$

But since $E(X) < 0$, $\ldots \ldots$. I don't know how to finish from here. I don't know if this is the right approach, either.

Answer 1

By Jensen's inequality $\exp(\theta E(X))\leq E(\exp(\theta X))=1.$ As $E(X)<1$ and $\theta \neq 0$ we must have $\theta>0.$