Suppose $X$ is a random variable with $\mu=\mathbb E[X]\ne0$ and that $X$ has a finite moment generating function on some open interval containing $0$. Then for what $\theta\ne0$ does the following equation hold? $$\mathbb E\left[e^{\theta X}\right] = 1. $$
The book I am reading (Stochastic Processes by Ross) says "We shall suppose that such a $\theta$ exists (and is usually unique)" without justification. This statement does not seem obvious to me. For example, if $X$ has exponential distribution with mean $\mu$, then $$\mathbb E\left[e^{\theta X}\right] = \frac{\mu^{-1}}{\mu^{-1}-\theta}=1 $$ has only the solution $\theta=0$, while if $X\sim\mathcal N(\mu, \theta^2)$, $$\mathbb E\left[e^{\theta X}\right] = e^{\mu\theta + \frac12\sigma^2\theta^2}=1,$$ has the solution $\theta = -\frac{2\mu}{\sigma^2}$. So under what conditions does this equation have nonzero solution(s)?
Edit: The map $x\mapsto e^{\theta x}$ is convex, so Jensen's inequality yields $$e^{\theta\mu}\leqslant\mathbb E[e^{\theta X}]=1 $$ and hence $\theta\mu<0$. This means $\theta$ and $\mu$ must have opposite signs.
Expanding on @Did's comment, if $\mathbb P(X>0)=1$ then $\mathbb P(\theta X>0)=1$ for $\theta>0$ so $\mathbb P(e^{\theta X}>1)=1$ and $\mathbb E[e^{\theta X}]>1$. By the same logic, $\mathbb E[e^{\theta X}]<1$ for $\theta <0$.
If $\mathbb P(X<0)=1$ then by symmetry we see that $\mathbb E[e^{\theta X}]<1$ when $\theta>0$ and $\mathbb E[e^{\theta X}]>1$ when $\theta<0$.
If $\mathbb P(X>0)$ and $\mathbb P(X<0)$ are both positive, then $\mathbb E[e^{\theta X}]\to\infty$ when $\theta$ converges to a boundary point of its region of convergence. So $\mathbb E[e^{\theta|X|}]$ is finite for some $\theta>0$, we conclude by continuity of $\exp$ that there exists $\theta^\star\ne0$ such that $\mathbb E[e^{\theta^\star X}]=1$.