# $\lim_{t\rightarrow 0}P\left(X_t >0\right)=\frac 1 2$ for continuous semimartingales?

I am trying to prove the following Lemma, which seems intuitive, but I still have doubts:

Lemma

Given a Brownian motion $\{W_t,\mathcal F_t:0\le t \le1\}$, two bounded processes, $\mu$ and $\sigma$, with $\sigma$ continuous and $\sigma_0\neq 0$, such that the integral

$$X_t=\int_0^t \sigma_t dW_t + \int_0^t \mu_t dt$$

exists, then $$\lim_{t\rightarrow 0}P\left(X_t >0\right)=\frac 1 2.$$

Proof Attempt (wrong)

Define $g_n := \sqrt{n} X_{1/n}$ and clearly it holds $P(X_{1/n}>0)=P(g_n>0)$, for all $n\in\mathbb N$. By the time change formula for stochastic integrals (Karatzas & Shreve, Th. 3.4.8) and regular integration by substitution we have:

$$g_n = \int_0^1 \sigma_{s/n}dB_s + \frac 1 {\sqrt{n}}\int_0^1 \mu_{s/n}ds \quad\text{ a.s. }$$

where $B_s:=\sqrt{n} W_{s/n}$ is a Brownian motion. Furthermore, by bounded convergence, we have $g_n\stackrel{L^2}\longrightarrow \sigma_0 B_1$: (as noted by @sinusx, this does not work as $B_1$ depends on $n$)

$$\mathbb E\left(g_n-\sigma_0 B_1\right)^2\le\mathbb E\left(\int_0^1 (\sigma_{s/n}-\sigma_0)^2ds+\frac 1 {n}\;\left(\int_0^1\mu_{s/n}ds\right)^2\right)\rightarrow 0$$

It remains to show that $P(g_n>0)=\mathbb E(1_{g_n>0})$ converges to $\mathbb E(1_{\sigma_0 B_1>0})$ $=$ $P(\sigma_0 B_1>0)=\frac 1 2$. By bounded convergence, it suffices to show almost sure convergence for $1_{g_n>0}$:

$L^2$ convergence implies almost sure convergence for $g_n$. As the step function is contiuous everywhere except at $0$, we conclude with $P(\lim_{n\rightarrow \infty}g_n=0)=P(B_1=0)=0$.

(BTW, this is the main difference between $g_n$ and $X_{1/n}$: $P(X_{1/\infty}=0)=1\neq P(\sigma_0 B_1=0)=0$.)

Questions

Is the Lemma true? Is the proof correct? Is there an easier proof or does it follow from some other result? Can you point me to similar results?

(Meta: Is question suitable for MO?)

Edit: Simpler proof, suggested in @Sinusx's answer

Define $f_t := \sigma_0\frac{W_t}{\sqrt{t}}$ and $g_t := \frac{X_t}{\sqrt{t}}$. It clearly holds

$$g_t=f_t + \frac{1}{\sqrt{t} }\int _ 0 ^t (\sigma _s - \sigma _0 ) dW _s+ \frac{1}{\sqrt{t} } \int _0 ^t \mu _s ds,$$

implying $g_t - f_t\longrightarrow 0$ in $L^2$ by bounded convergence

$$\mathbb E(g_t-f_t)^2 \le\mathbb E\left(\int_0^1 (\sigma_{st}-\sigma_0)^2ds+t\;(\sup_{s\le t}\mu_s)^2\right)\rightarrow 0, \text{ as } t\rightarrow 0 \tag{1}$$

and thus convergence in probability. Together with $f_t\sim \mathcal N(0,\sigma_0^2)$ (by the scaling property of the Brownian motion) we can apply this condition for convergence in distribution to show $g_t\stackrel{\text{(d)}}\longrightarrow \mathcal N(0,\sigma_0^2)$ as $t\rightarrow 0$. The lemma now follows from $P(X_t>0)=P(g_t>0)\rightarrow \frac{1}{2}$.

PS: However, I think (1) still needs a dominator for the $\sigma$ part.

I think that the lemma is true if you add some additional continuity assumptions on $\sigma$. In the proof you are using the right scaling, however you may want to replace convergence in $L^2$ with convergence in distribution, as $g _ \infty$ is not defined and does not exist. Note that $\frac{1}{\sqrt{t}} W _{t} \overset{(d)}{=}W_1$, so $\frac{1}{\sqrt{t}} W _{t}$ converges in distribution to $N(0,1)$, $t \to 0$, but $\limsup \frac{1}{\sqrt{t}} W _{t} = \infty$, $\liminf \frac{1}{\sqrt{t}} W _{t} = - \infty$ by the law of iterated logarithm.

For example, let in addition to your assumptions $\sigma$ be mean square continuous, that is, $E |\sigma _s| ^2 < \infty$ for all $s$ and and $E|\sigma _t - \sigma _0|^2 \to 0$, $t \to 0$. We have $$\frac{X _t}{\sqrt{t} } =\sigma _0 \frac{W _t}{\sqrt{t} } + \frac{1}{\sqrt{t} }\int _ 0 ^t (\sigma _s - \sigma _0 ) dW _s+ \frac{1}{\sqrt{t} } \int _0 ^t \mu _s ds,$$ You can prove then that the second and third summands on the right hand side go to $0$ in $L^2$ as $t \to 0$, but the first does not and you can get the statement of the lemma.

• While I agree that adding some $L^2$ integrability will make the lemma and in essence the proof correct, I fail to see why the stochastic integral may not exist. In general it is not in $L^2$ and might even not define a martingale, bu exists as local martingale... Mar 6, 2015 at 19:08
• Thanks you for your answer. Some clarifications: 1) I intended to use $g_\infty$ for $\sigma_0 B_1$ which is misleading, so I changed it in the question. 2) I also intended the integral to exists, so I changed it. 3) I think the only thing missing is $\mathbb E\int_0^t\sigma^2(s)ds <\infty$ (as @StephanSturm suggests). Is your suggestion of square continuity more general than that? 4) Will the bounded convergence argument in the last step also work if $L^2$ convergence is replaced by convergence in distribution?
– JSG
Mar 6, 2015 at 19:13
• Ans 5) I think I do not need to look at the limit of $\frac{W_t}{\sqrt{t}}$ because I replaced it for any $t>0$ by another Brownian motion (whose existence is guaranted by the time change theorem) not just with the same distribution, but almost surely equal.
– JSG
Mar 6, 2015 at 19:20
• Well, as it is now, your proof is wrong. Your $B_1$ depends on $n$, so it is not clear what is meant by '$g_n \to \sigma _0 B_1$' Mar 6, 2015 at 20:25
• @Stephan Sturm You are right, the integral always exists if $\sigma$ is continuous. Mar 6, 2015 at 20:27

The proof is not correct, as without additional integrability condition you will not be able to conclude that $g_n \in L^2$ for $n$ large enough, and therefore the $L^2$ convergence argument fails. As for a concrete counterexample, something like $\sigma_t = e^{e^{W_t}}$ (and $\mu_t =0$) should do the trick.

• So I basically need $\mathbb E\int_0^t \sigma_{s}^2ds<\infty$?
– JSG
Mar 6, 2015 at 19:23
• Yes. Note also that $L^2$ convergence implies a.s. convergence only along a subsequence, but this is sufficient for your argument. Mar 6, 2015 at 19:28
• what about @Sinusx's objection?
– JSG
Mar 6, 2015 at 20:45