6

Let $\left( X_{t}\right) _{t\geq0}$ be a subordinator with the Laplace exponent given by

$$ \Phi\left( \lambda\right) =d\lambda+\int_{0}^{\infty}\left( 1-e^{-\lambda x}\right) \nu\left( dx\right) $$

Show that almost surely

$$ \lim_{t\rightarrow\infty}\frac{X_{t}}{t}=d+\int_{0}^{\infty}x\nu\left( dx\right) $$

I first use the Levy-Khintchine formula, where we have

$$ X_{t}=dt+Y_{t}% $$

and $\left\{ Y_{t}\right\} _{t\geq0}$ is a subordinator whose the Laplace exponent is given by

$$ \Phi^{\prime}\left( \lambda\right) =\int_{0}^{\infty}\left( 1-e^{-\lambda x}\right) \nu\left( dx\right) $$

Thus,

$$ \frac{X_{t}}{t}=d+\frac{Y_{t}}{t}% $$

Now, I think it suffices to show that almost surely $\lim_{t\rightarrow\infty }\frac{Y_{t}}{t}=\int_{0}^{\infty}x\nu\left( dx\right) .$

However, I don't know if I can do this and what should I do next.

Lost1
  • 8,085
BigMike
  • 361

1 Answers1

0

This is a simple application of Law of large numbers.

Note that

$$E(Y_1) = \int^\infty_0 x\nu(\text{d}x)$$

Assume that $E(Y_1)<\infty$. Let $n$ be the largest integer smaller than $t$, then

$$\frac{Y_t}{t}=\frac{Y_1+(Y_2-Y_1)+...+(Y_n-Y_{n-1})}{n}\frac{n}{t}+\frac{Y_t-Y_n}{t}$$

By the strong law of large numbers, the first part of the sum converges to $E(Y_1)$. Note that $\frac{Y_t-Y_n}{t}$ is a convering almost surely to $0$.

For $E(Y_1)=\infty$, then obverse that $Y$ admits the following decomposition

$$Y=Y_1+Y_2$$

where the Laplace exponent of $Y_1$ and $Y_2$ are given by

$$\int_{(0,1]}(1-e^{\lambda x})\nu(dx)$$

and

$$\int_{(1,\infty)}(1-e^{\lambda x})\nu(dx)$$

repsectively. note that $Y_2$ is a compound Poisson process with intensity $\lambda = \nu((1,\infty))$. Notice that $EY_1$ is finite, so

$$\frac{Y_t}{t}= \frac{Y^1_t}{t}+\frac{Y_t^2}{t}$$.

Note that $Y_t^1/t$ is convering to $\int_{(0,1]}x\nu(dx)$ by the previous argument.

$$Y_t^2=\frac{N_t}{t}\frac{\sum_{i=1}^{N_t}\zeta_i}{N_t}$$

Well, $N_t/t\rightarrow\nu((1,\infty))$, but $\frac{\sum_{i=1}^{N_t}\zeta_i}{N_t}$ is going to infinity by SLLN because the expection of $\zeta_i$

$$\frac{1}{\nu (1,\infty)}\int_{(1,\infty)}x\nu(dx)=\infty$$

Lost1
  • 8,085