I have the following problem:
$X_1,X_2,…$ are positive identically distributed random variables with the distribution function $F(x) :=P(X_n \leq x)$ and we assume that $F(0)<1$ for all $n$. Let $$S_0 :=0 \qquad \qquad \quad S_n := \sum_{i=1}^{n} X_i$$
and $N(t) := \sup\{n \mid S_n \leq t\}$. Let $m(t):=E[N(t)]$
Show that:
- $m(t) = \sum_{n=1}^{\infty}F_n(t)$ where $F_n(t) := P(S_n \leq t)$.
- For $t \geq 0$, $m(t) < +\infty$ ($F(0) < 1$ is important).
Regards, Raxel.
Best Answer
Let $\omega \in \Omega$ such that $N(t)(\omega)=n$. Then $S_k(\omega) \leq t$ for all $k \leq n$ and $S_k(\omega) > t$ for all $k>n$, i.e. $$N(t)(\omega) = \sum_{k=1}^n 1_{[S_k \leq t]}(\omega) = \sum_{k=1}^{\infty} 1_{[S_k \leq t]}(\omega)$$ Thus, by monotone convergence: $$m(t) = \mathbb{E}(N(t))= \sum_{k=1}^{\infty} \mathbb{P}[S_k \leq t] \tag{1}$$
By the monotonicity of $(S_n)_n$ we have $$\mathbb{P}[S_n \leq t] = \mathbb{P}[e^{-S_n} \geq e^{-t}] \leq \frac{1}{e^{-t}} \cdot \mathbb{E}(e^{-S_n}) = e^t \cdot (\mathbb{E}e^{-X_1})^n \tag{2}$$ where we used in the last step that the random variables $(X_n)_n$ are independent and identically distributed. Since $F(0)= \mathbb{P}[X_1=0]<1$, there exists $\delta>0$ such that $\mathbb{P}[X_1>\delta]>0$. Hence $$\begin{align} \mathbb{E}e^{-X_1} &= \mathbb{E} \big( e^{-X_1} \cdot 1_{[X_1 \leq \delta]} + e^{-X_1} \cdot 1_{[X_1 > \delta]} \big) \leq \mathbb{P}[X_1 \leq \delta]+\underbrace{e^{-\delta}}_{<1} \cdot \mathbb{P}[X_1 > \delta] \\ &< \mathbb{P}[X_1 \leq \delta]+\mathbb{P}[X_1>\delta] =1 \end{align} \tag{3}$$ Putting it all together, we arrive at $$m(t) \stackrel{(1)}{=} \sum_{k=1}^{\infty} \mathbb{P}[S_k \leq t] \stackrel{(2)}{\leq} e^t \cdot \sum_{k=1}^{\infty} (\mathbb{E}e^{-X_1})^k \stackrel{(3)}{<} \infty$$