Proof of Cramér-Lundberg inequality I'm trying to prove the Cramér-Lundberg inequality, which deals with the probability of ruin for an insurance company given a certain initial capital. Specifically, if $Y_1, Y_2, \ldots$ are the differences between the premiums and payments of an insurance company at time $n$, and $X_n = Y_1 + \cdots + Y_n$ is the total gain of the insurance company at time $n$, and $k_0$ is the initial capital, then the probability of eventual ruin $p(k_0)$ satisfies the Cramér-Lundberg inequality:
$$
p(k_0) := \mathbb P\left[ \inf\left\{ X_n + k_0 : n \in \mathbb N_0 \right\} < 0 \right] \leq \exp\left(\theta^* k_0\right)
$$
where $\theta^* < 0$ satisfies $\log\left( \mathbb E\left[ \exp(\theta^* Y_1 )\right]\right) = 0$.
My reference text proposes proving this in the following steps. Suppose $Y_1, Y_2, \ldots$ are i.i.d. integrable random variables that are not almost surely constant. Let $X_n = Y_1 + \cdots + Y_n$, and suppose there is $\delta > 0$ so that $\mathbb E\left[\exp\left(\theta Y_1 \right)\right] < \infty$ for all $\theta \in (-\delta, \delta)$. Define $\psi : (-\delta, \delta) \to \mathbb R$ by $$\psi(\theta) := \log \left(\mathbb E\left[\exp\left(\theta Y_1 \right)\right]\right)$$
and define the process $Z^\theta = \left(Z^\theta_n\right)_{n \geq 1}$ by $Z_n^\theta := \exp\left(\theta X_n - n\psi(\theta)\right)$. We are suggested to show the following:

*

*$Z^\theta$ is a martingale for all $\theta \in (-\delta, \delta)$.

*$\psi$ is strictly convex.

*$\mathbb E\left[\sqrt{Z_n^\theta}\right] \xrightarrow{n \to \infty} 0$ for $\theta \neq 0$.

*$Z_n^\theta \xrightarrow{n \to \infty} 0$ almost surely.

*If $\mathbb E[Y_1] > 0$ and if $\psi(\theta) = 0$ has a nonzero solution $\theta^*$, then $\theta^* < 0$.

*Prove that if such a $\theta^* < 0$ exists, and if $\mathbb E[Y_1] > 0$, then $p(k_0) \leq \exp\left(\theta^* k_0\right)$.

I've been able to show 1, 4 (from 3), and 5 (from 2). I'm close for 2 but having some trouble: we need to show $\psi(\lambda\theta + (1-\lambda)\phi) < \lambda \psi(\theta) + (1-\lambda)\psi(\phi)$ whenever $\theta \neq \phi$ and $\lambda \in (0,1)$. Clearly we have $$\psi(\lambda\theta + (1-\lambda)\phi) = \log\mathbb E\left[ \exp\left(\lambda\theta Y_1 + (1-\lambda)\phi Y_1 \right)\right]$$ Meanwhile by Jensen's inequality and concavity of $x \mapsto x^\lambda$ for $0 < \lambda < 1$,
\begin{align*}
\lambda \psi(\theta) + (1-\lambda)\psi(\phi) &= \log \left( \mathbb E\left[\exp (\theta Y_1) \right]^\lambda\right) + \log\left(\mathbb E\left[ \exp(\phi Y_1)\right]^{1-\lambda}\right) \\
&\geq \log\left(\mathbb E\left[\exp\left(\lambda \theta Y_1\right)\right]\right) + \log\left(\mathbb E\left[\exp\left((1-\lambda)\phi Y_1\right)\right]\right) \\
&= \log\left(\mathbb E\left[\exp\left(\lambda \theta Y_1\right)\right]\mathbb E\left[\exp\left((1-\lambda)\phi Y_1\right)\right]\right).
\end{align*}
If I could show $\mathbb E\left[\exp\left(\lambda \theta Y_1\right)\right]\mathbb E\left[\exp\left((1-\lambda)\phi Y_1\right)\right] \geq \mathbb E\left[\exp\left(\lambda \theta Y_1\right)\exp\left((1-\lambda)\phi Y_1\right)\right]$,that would solve this problem, but this is very far from obvious to me (especially since the integrands aren't independent).
Then 3 and 6 I'm really stuck on. Any help on any of these three would be greatly appreciated. Note I would prefer not to use martingale convergence theorems because these results have yet to appear in my textbook; I can only work with square integrable martingales and stopping times.
 A: These are my ideas:

*

*For point 2, I get by Hölder's inequality for the pair $(1/(1-\lambda), 1/\lambda)$ that
\begin{align*} \psi(\lambda\theta+(1-\lambda)\phi)
&=\log\left(\mathbb{E}\left[\exp(\lambda\theta Y_1)\exp((1-\lambda)\phi Y_1)\right]\right) \newline
&\le\log\left({\mathbb{E}\left[\exp(\theta Y_1)\right]}^{\lambda}
{\mathbb{E}\left[\exp(\phi Y_1)\right]}^{1-\lambda}\right)\newline
&= \lambda \psi(\theta)+(1-\lambda)\psi(\phi)\;.
\end{align*}

*For point 3 we know that Jensen's inequality is strict, whenever the random variable in question is almost surely not constant, which is part of the assumptions. Hence,  using concavity of the square root we have
\begin{align*}
\mathbb{E}\left[\exp\left(\frac{\theta}{2}Y_1\right)\right]=\mathbb{E}\left[\sqrt{\exp\left(\theta Y_1\right)}\right]< \sqrt{\mathbb{E}\left[\exp\left(\theta Y_1\right)\right]}\;.
\end{align*}
Hence by the independence of ${(Y_i)}_{i\in\mathbb{N}}$
\begin{align*}
\mathbb{E}\left[\sqrt{Z_n^{\theta}}\right]
&= \mathbb{E}\left[\exp\left(\frac{\theta}{2} X_n - \frac{n}2 \psi(\theta)\right)\right] \newline
&={\left(\mathbb{E}\left[\exp\left(\frac{\theta}{2} Y_1-\frac12 \psi(\theta)\right)\right]\right)}^n\newline
&={\underbrace{\left(\frac{\mathbb{E}\left[\exp\left(\frac{\theta}{2} Y_1\right)\right]}{\sqrt{\mathbb{E}\left[\exp\left(\theta Y_1\right)\right]}}\right)}_{<1}}^n\rightarrow 0\;.
\end{align*}

*In your question you write $\mathbb{N}_0$ which I assume are the natural numbers including $0$, but I am not sure what $X_0$ is, so I assume the process starts at $X_1$ and write $\mathbb{N}$. For $m\in \mathbb{N}$ define
\begin{align*}
\tau^{(m)}&:=\inf\{n\in\mathbb{N}:X_n< -k_0\}\wedge m\;,
\end{align*}
By construction we have $1=\mathbb{E}\left[Z_1^{\theta^*}\right]$. By point 1 we know that ${\{Z_n^{\theta^*}\}}_{n\in \mathbb{N}}$ is a martingale. Hence, by the optional sampling theorem we get
\begin{align*}
1=\mathbb{E}\left[Z_1^{\theta^*}\right]
&=\mathbb{E}\left[Z^{\theta^*}_{\tau^{(m)}}\right]\\
&\ge\mathbb{E}\left[Z^{\theta^*}_{\tau^{(m)}}\mathbf{1}_{\tau^{(m)}<m}\right]\\
&\ge\exp\left(-\theta^* k_0\right)\mathbb{E}\left[\mathbf{1}_{\tau^{(m)}<m}\right]\\
&=\exp\left(-\theta^* k_0\right)
\mathbb{P}\underbrace{\left[\inf\{n\in\mathbb{N}:X_n+k_0< 0 \}<m\right]}_{A_m}\;,
\end{align*}
where in the second inequality we used that $\theta^*<0$.
The only dependence on $m$ is in $A_m$, and since ${\{A_m\}}_{m\in\mathbb{N}}$ is an increasing set of events you may take the limit and multiply with $\exp\left(\theta^*k_0\right)$ to get
\begin{align*}
\exp\left(\theta^* k_0\right)
\le\mathbb{P}\left[\inf\{n\in\mathbb{N}:X_n+k_0< 0 \}<\infty\right]
=\mathbb{P}\left[\inf\{X_n+k_0: n\in\mathbb{N} \}<0\right]
\end{align*}
which is the desired inequality.

A: Here is an idea for the convergence results 3. and 4. 
I will start with 4. : Write $Z_n^\theta=e^{n\big(\theta\frac{X_n}{n}-\psi(\theta)\big)}$ and let $A^\theta_n=\theta\frac{X_n}{n}-\psi(\theta)\;,n\in\mathbb{N}$. By the strong law of large numbers, it follows that for all $\theta\in(-\delta,\delta)$ $$A^\theta_n\underset{n\to\infty}\longrightarrow a(\theta):=\mathbb{E}[\theta \;Y_1]-\psi(\theta)\;\;\text{a.s.}$$  Now, observe that for all $\theta\neq0$
$$a(\theta)= \mathbb{E}[\theta\;Y_1]-\log \mathbb{E}[exp\{\theta\;Y_1\}]< 0$$ by strict convexity of the exponential (or Jensen). Thus, on an event $\tilde{\Omega}$ of probability $1$, there exists an $n_0\in\mathbb{N}$ and $M<0$, such that for all $n\geq n_0$, 
$$A^\theta_n\leq M\;\;(on\;\tilde{\Omega})\;$$
Hence for $\theta\neq 0$ $$ |Z^\theta_n|\leq e^{nM}\longrightarrow 0\;\;a.s.$$ (For $\theta=1$, $Z_n^1=1$ for all $n$)
Finally, for 3. it suffices to use the continuity of the square root to conclude that $\sqrt{Z_n^\theta}$ converges to $0$ almost surely (provided that $\theta\neq 0$) and the Dominated convergence theorem to pass the limit inside the expectation.
