Standard Wiener process continuity

Question

This will be one big question. Basically, I had a lecturer who, to put it mildly, was not able to explain anything well. I am preparing for the exam now. One sample exam question is the following:

Given the transition density function of a standard Wiener process:

$$f_{1\mid1} (x_2,t_2\mid x_1,t_1) = \frac{1}{\sqrt{2\pi (t_2 - t_1)} } \exp\left( -\frac{(x_2-x_1)^2}{2(t_2-t_1)}\right)$$

One has to show that the above conforms to the Lindberg condition:

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x_2-x_1|>\epsilon} f_{1\mid1} (x_2,t+\Delta t \mid x_1,t) dx_2 =0 \ \ \ \forall \epsilon >0 \ \text{uniformly in} \ x_1,t,\Delta t$$

I was actually admitted for the course even though I have no formal background in maths. So I shall highlight the areas below which are problematic for me:

• I know from some analysis reading that I have made, that $\epsilon$ is some small positive value, by convention. So I am really not sure why the question highlights "$\forall \epsilon > 0$". As this is true by convention.
• I have never seen an integral with such a bound before. I, of course, understand what it means. We want to integrate only when the absolute difference is larger than some small positive value. But how that would translate into actually evaluating the integral, I have no clue. (Also remembering that $\epsilon$ is arbitrary).
• I do not understand what the "uniformly in ..." part means.
• To show that the expression evaluates to zero when we take the limit for change in $t$ to go to zero, do we need that the expression in the integral evaluates to $0$? I cannot see what other case there is that would ensure that as $1/\Delta t$ as $\Delta t \rightarrow 0$ becomes infinity.

So with all of my constrained knowledge:

$$f_{1\mid1} (x_2,t_1 + \Delta t_1\mid x_1,t_1) = \frac{1}{\sqrt{2\pi (\Delta t_1)} } \exp\left( -\frac{(x_2-x_1)^2}{(\Delta t_1)^2}\right)$$

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x_2-x_1|>\epsilon} \frac{1}{\sqrt{2\pi (\Delta t_1)} } \exp\left( -\frac{(x_2-x_1)^2}{(\Delta t_1)^2}\right) dx_2$$

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t^{3/2}\sqrt{2\pi}} \int_{|x_2-x_1|>\epsilon} \exp\left( -\frac{(x_2-x_1)^2}{(\Delta t_1)^2}\right) dx_2$$

And that's it.. I should probably mention that I have some elementary understanding of what a Markov process is, what transition density is and that this requirement is for the continuity of the process. Also that if we integrate w.r.t. to $x_2$ in this case then we essentially get rid of that variable.

Answer 1

Your formula for the transition density needs to be corrected; if should be $$ \frac{1}{\sqrt{2\pi (t_2 - t_1)} } \exp\left( -\frac{(x_2-x_1)^2}{2(t_2-t_1)}\right). $$

Your integral can be interpreted as a probability, and this aids in evaluating the limit. Let $X(\Delta t_1)$ be a random variable that is normally distributed with mean $x_1$ and variance $\Delta t_1$. Your integral is then $$ \Bbb P[|X(\Delta t_1)-x_1|>\epsilon]. $$ To demonstrate the limit assertion, we don't need to evaluate this probability explicitly, but merely to find a sufficiently accurate upper bound. One's first impulse is to use Chebyshev's inequality $$ \Bbb P[|X(\Delta t_1)-x_1|>\epsilon]\le{var(X(\Delta t_1))\over \epsilon^2}={\Delta t_1\over \epsilon^2}, $$ but this is not good enough, because when divided by $\Delta t_1$ it does not converge to $0$.

So let re-write the event in question and then use Markov's inequality: $$ \eqalign{ \Bbb P[|X(\Delta t_1)-x_1|>\epsilon] &=\Bbb P[|X(\Delta t_1)-x_1|^4>\epsilon^4]\cr &\le{\Bbb E[(X(\Delta t_1)-x_1)^4]\over \epsilon^4}\cr &={3(\Delta t_1)^2\over\epsilon^4}.\cr } $$ (In the last step here I have used the fact that the fourth central moment of a normal random variable is 3 times the square of its variance.) We have shown the upper bound $$ {1\over \Delta t_1}\Bbb P[|X(\Delta t_1)-x_1|>\epsilon] \le {3\Delta t_1\over\epsilon^4}. $$ This upper bound tends to $0$ as $\Delta t_1\to 0$, and does not depend on $x_1$ or $t$ (whence the "uniformity" in the convergence). The bound does depend on $\epsilon$ in a way that gets worse as $\epsilon$ gets small, so the convergence, or at least the upper bpund, is not uniform with respect to $\epsilon$.

[Incidentally, I haven't seen this condition called "Lindeberg condition" before; rather "Dynkin's condition".]