How to prove that this process is always positive?

Question

I would like to ask is there any way to prove that following process $$ \mathrm dY_t=\left(a+\frac{b}{Y_t}\right)\mathrm dt +\mathrm dW_t, \ \ Y_0=y_0>0, $$ where $a\neq 0$ and $b\geq 1/2$, is always positive, i.e. $P(0<Y_t)=1$ a.s. Thank you !

Answer 1

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it.

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If we define

$$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$

then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ and $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply Itô's formula:

$$f_n(Y_t) - f_n(y_0) = \int_0^t f_n'(Y_s) \, dY_s + 0.$$

Taking expectation on both sides and using that $\mathbb{E}\left( \int_0^t f_n'(Y_s) \, dW_s \right)=0$, we obtain

$$\begin{align*} \mathbb{E}f_n(Y_t)-f_n(y_0) &= \mathbb{E} \left( \int_0^t f_n'(Y_s) \left(a + \frac{b}{Y_s} \right) \, ds \right) \\ &= n \mathbb{E} \left( \int_0^t 1_{[0,1/n)}(Y_s) \left(a+ \frac{b}{Y_s} \right) \, ds \right). \end{align*}$$

For $n$ sufficiently large, we have $f_n(y_0)=1$. As $\mathbb{E}f_n(Y_t) \leq 1$ (since $\|f_n\| \leq 1$), the previous identity gives

$$ n \mathbb{E} \left( \int_0^t 1_{[0,1/n)}(Y_s) \left(a+ \frac{b}{Y_s} \right) \, ds \right) \leq 0.$$

For any $Y_s \in [0,1/n)$, the inequality $a+ \frac{b}{Y_s} \geq a+bn$ holds true and therefore

$$n (a+bn) \int_0^t \mathbb{P}(Y_s \in [0,1/n)) \, ds \leq 0.$$

As $b>0$, we know that $(a+bn)>0$ for $n$ sufficiently large. Consequently, we get

$$\int_0^t \mathbb{P}(Y_s \in [0,1/n)) \, ds \leq 0.$$

This implies $\mathbb{P}(Y_s \in [0,1/n))=0$ for (Lebesgue)almost all $s \geq 0$. Since $(Y_t)_{t \geq 0}$ has continuous sample paths, we find $\mathbb{P}(Y_s \in [0,1/n))=0$ for all $s \geq 0$; in particular $\mathbb{P}(Y_s>0)=1$.

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As already mentioned above, this argumentation is not correct because $f_n$ is not twice differentiable and so we cannot simply apply Itô's formula. However, this calculation can be made rigourous by choosing a smooth function which does the job, e.g.

$$f_n(x) := \exp \left(- \frac{1}{n} \frac{1}{x^2} \right).$$

The calculations when applying Itô's formula get more technical (in particular we have to discuss away the term $\int_0^t f_n''(Y_s) \, ds)$), but they are not hard and very similar to the ones above.