$N(t)/t\to 1/\mu$

Question

Let $X_i$‘s are iid with $0<X_i<\infty$ almost surely.

Define $T_n=X_1+\cdots+X_n$

E$(X_1)>0$

N$(t)=\sup[n\ge 0: T_n\le t]$

How to show using Strong law of Large Numbers that $N(t)/t\to1/\mu$ almost surely?

As per the SLLN requirement, how to convert N(t) into summation of iid terms?

Answer 1

Step 1

$$N_t\rightarrow \infty \ \ \ as \ \ t\rightarrow\infty \ \ \ a.e. \ P$$

Step 2

$$\frac{T_{N_t}}{N_t}\rightarrow\mu \ \ as \ \ t\rightarrow \infty \ \ \ a.e. \ P$$

Step 3

Since $\frac{T_{N_t}}{N_t}\le \frac{t}{N_t}$ taking $\lim\ inf$ we get $$\lim_{t\rightarrow\infty} inf \ \ \frac{T_{N_t}}{N_t}\leq \lim_{t\rightarrow\infty} \ inf \ \ \frac {t}{N_t}$$ $$\implies \mu\le \lim_{t\rightarrow\infty} \ inf \ \ \frac {t}{N_t} $$

We also have $\frac{T_{N_t+1}}{N_t}\ge \frac{t}{N_t}$ and taking $\lim \ sup$ we have $$\lim_{t\rightarrow \infty} \ sup \ \frac {T_{N_t+1}}{N_t}\geq \lim_{t\rightarrow \infty} \ sup \ \frac{t}{N_t}$$ $$\implies \mu \geq \lim_{t\rightarrow \infty} \ sup \ \frac{t}{N_t} $$

Thus you have

$$\lim _{t\rightarrow \infty} \ \frac{t}{N_t}=\mu $$

Conclusion follows

Answer 2

The strong law tells us that $\frac{T_n}{n}\to\mu$ almost surely where $\mu=E[X_1]$. Also it is clear by definition that $T_{N(t)}\leq t<T_{N(t+1)}$. Hence:

$\frac{T_{N(t)}}{N(t)}\leq\frac{t}{N(t)}<\frac{T_{N(t+1)}}{N(t)}=\frac{T_{N(t+1)}}{N(t+1)}\frac{N(t+1)}{N(t)}$

Now let's look at the set of all points $\omega$ such that $T_n(\omega)<\infty$ for all $n$. This set has probability $1$ and in this set we have that $N(t)$ is increasing to $\infty$. So $N(t)\to\infty$ almost surely and hence $\frac{N(t+1)}{N(t)}\to 1$ almost surely. Hence by the squeeze theorem $\frac{t}{N(t)}\to\mu$ almost surely. And of course this implies that $\frac{N(t)}{t}\to\frac{1}{\mu}$ almost surely.

Note that I assumed all the time that $\mu$ is finite. Though you can also prove that if $\mu=\infty$ then $\frac{N(t)}{t}\to 0$ almost surely.