Find $r>0$ such that $\mathsf{E}\mathrm{e}^{r\xi} = 1$

Let $\xi:\Omega\to\mathbb{R}$ be a random variable such that $\mathsf{E}\xi < 0$. I would like to find $r>0$ such that $g(r):=\mathsf{E}\mathrm{e}^{r\xi} = 1$.

Maybe you can help me to derive some necessary of sufficient conditions for such $r$ to exist? Say, if $\xi = -1$ then there is no such positive $r$. I tried to investigate the behavior of $g(r)$ since $g(0) = 1$, but I was not successful in it.

Edited: maybe one of you can help with an example when $\mathsf{P}(\xi>0)>0$ but there is no positive root of $g(r) = 1$?

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If $\xi \leq 0$ a.e. then there is no such $r \gt 0$. In fact, we have $0 \leq e^{r\xi} \leq 1$, so $Ee^{r\xi} = 1$ implies that $e^{r\xi} = 1$ a.e. Therefore $\xi = 0$ a.e. contradicting $E\xi \lt 0$. – t.b. Jun 17 '11 at 10:53
@Theo: that's right if $\xi\leq 0$ a.e. thanks, waiting for more general case. – Ilya Jun 17 '11 at 10:56
Concerning the edit, let $\xi+1$ have a Student t distribution with more than 1 df. Its expectation is $-1$. Yet, for all $r \ne 0$, $\mathbb{E}[e^{r\xi}]$ diverges, whence there are no roots at all of $g(r)-1$ apart from $r=0$. – whuber Jun 17 '11 at 15:49
Can you please make the title more specific? Seeing "Solve an equation" on the front page is not very informative. – Rahul Jun 17 '11 at 21:43

Let us forget integrability questions for a moment and assume that $\mathrm{e}^{r|\xi|}$ is integrable for every positive $r$. Then the function $g$ is well defined and continuous everywhere.

First step: as you noted, $g(0)=1$ and $g'(0)=E(\xi)$, hence $g'(0)<0$ and $g(r)<1$ for $r>0$, $r$ close to $0$.

Second step: assume that $P(\xi>0)\ne0$ (otherwise, as shown by Theo, the result is false), and choose $x>0$ such that $P(\xi\ge x)=p>0$. Then $\mathrm{e}^{r\xi}\ge\mathrm{e}^{rx}\mathbf{1}_{\{\xi\ge x\}}$ for every nonnegative $r$, hence $g(r)\ge\mathrm{e}^{rx}p$ for every nonnegative $r$, and one sees that $g(r)\to+\infty$ when $r\to+\infty$.

Conclusion: since the function $g$ is continuous, the intermediate value theorem implies that there exists a positive $r$ such that $g(r)=1$.

When $\mathrm{e}^{r|\xi|}$ is not integrable for at least some positive $r$, three different things may happen: $g(r)$ may be infinite except at $r=0$, or finite for every $0\le r<r_0$ and infinite for $r\ge r_0$, for a given finite $r_0$, or finite for every $0\le r\le r_0$ and infinite for $r>r_0$.

The conclusion that there exists a positive $r$ such that $g(r)=1$ would still hold if for example, there exists a positive $r_1$ such that $g(r_1)$ is finite and $g(r_1)>1$. It would fail, for example, if $g(r)$ is finite for every $0\le r\le r_0$ and infinite for $r>r_0$, with $g(r_0)<1$.

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how can it be that $g(r),r\geq r_0$ will become infinite without $\lim\limits_{r\to r_0}g(r) = \infty$? – Ilya Jul 11 '11 at 12:01
Let me try to rescue the question and assume that you ask for an example where $g(r)$ is infinite for every $r>r_0$ while $g(r_0)$ is finite. Here is one: $r_0>0$ and $P(\xi\ge x)\sim x^{-2}\mathrm{e}^{-r_0x}$ when $x\to+\infty$. (By the way, any thoughts on chat.stackexchange.com/rooms/741 ?) – Did Jul 11 '11 at 14:19
Thank you, with regards to the chat - I'll answer. Btw, I don't think I saw the invitation. – Ilya Jul 11 '11 at 14:46
The invitation was automatically communicated to you as a comment addressed to you on math.stackexchange.com/q/49597/6179. – Did Jul 11 '11 at 14:52
I've responded you in chat. – Ilya Jul 11 '11 at 15:12