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This question was previously posted on https://math.stackexchange.com/questions/3981156/let-x-t-be-a-solution-of-a-sde-does-the-set-x-t-in-p-has-null-meas.

I think this question is easy. However, I have not been able to solve it.

Let $a,\sigma:\mathbb{R}\times \mathbb R\to\mathbb{R}$, smooth functions such that $\sigma>0$. Consider the 1-dimensional SDE, $$dX_t = a(X_t,t) dt + \sigma(X_t,t) dW_t$$ $$X_0 = x_0\in\mathbb{R}. $$ where $W_t$ is the standard Brownian motion.

Fixing $y\in\mathbb R$ and $t>0$, I was interested in showing that$$\mathbb{P}\left(\{\omega \in \Omega;\ X_t = y\}\right)=0.$$ Where $(\Omega,\mathcal F, \mathbb P)$, is the probability space being considered.

Does anyone know if the above equation is true? A reference would be enough for me.

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  • $\begingroup$ What assumptions on $W_t$ are you making? With literally no assumptions, the answer could be "no" by the (not so interesting) counterexample $W_t = 0, \forall t$, $a(X_t, t) = 0$, and $x_0 = y$. $\endgroup$
    – stats_model
    Commented Jan 13, 2021 at 20:58
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    $\begingroup$ More generally though, one easy way to proceed is to begin by "forgetting" about the fact that $X_t$ is defined from an SDE in the first place. Then $X_t$ is just a bona fide random variable, so the usual things you can say about random variables apply to $X_t$ as well. In particular, $\mathbb P(\{\omega \in \Omega, X_t = y\}) = 0$ would certainly be true as long as $X_t$ was a continuous random variable, which would certainly be true, for example, if $W_t$ were Brownian motion. $\endgroup$
    – stats_model
    Commented Jan 13, 2021 at 21:03
  • $\begingroup$ @stats_model $W_t$ is the Brownian motion sorry. Are you able to provide a proof of such a fact, or provide a reference? I really want to know how to prove this. $\endgroup$
    – Matheus Manzatto
    Commented Jan 13, 2021 at 21:39

2 Answers 2

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You say that this question is 'easy' but I would say that it is not so easy, having had a quick look at the relevant literature. Possibly the most elementary and accessible answer to this question is contained in the paper by N. Fournier, J. Printems, Absolute continuity of some one-dimensional processes, Bernoulli, 16 (2010), pp. 343-360. Section 3 of the paper considers SDEs of the form $$X_t=x+\int_0^t \sigma(X_s)\kappa (s,(X_u)_{u\leq s} H_s) \text{d}B_s+\int_0^t b(s,(X_u)_{u\leq s},H_s)\text{d}s$$ for suitable functions $\sigma$, $\kappa$ and $b$ and an auxiliary process $(H_t)_{t\geq 0}$. Theorem 3.1 states that, given suitability nice coefficient functions, the law of $X_t$ has a density on $\{x\in\mathbb{R}: \sigma(x)\neq 0\}$ whenever $t>0$. I think that this theorem should cover your situation if you place suitable restrictions on your coefficient functions e.g. that they have at most linear growth. The paper also has references to alternative approaches such as the use of the Malliavin calculus.

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With my washed-up SDE knowledge from option pricing, I came up with the following. I am not sure if it is correct reasoning, but I am open to discuss it so I can also learn the right answer If I am wrong.

We have:

$$X_t - X_{t-1} = a(X_{t-1}, t-1)(t - t + 1) + \sigma (X_{t-1}, t-1)(W_t - W_{t-1})$$

$$X_t = X_{t-1} + a(X_{t-1}, t-1)(t - t + 1) + \sigma (X_{t-1}, t-1)(W_t - W_{t-1})$$

From the properties of Brownian motion, we know that:

$$(W_t - W_{t-1}) \sim N(0, 1)$$

Also note that because we are at time $t$, $X_{t-1}$ is realized, thus $a(X_{t-1}, t-1)$ and $\sigma (X_{t-1}, t-1)$ is deterministic and the only random component is Brownian motion increment. Thus, distribition of $X_t$ is given by:

$$X_t \sim N(X_{t-1}+a(X_{t-1}, t-1), \sigma (X_{t-1}, t-1))$$

Since it is a normal distribution and normal distribution is a continuous distribution. And for all continuous distributions, we have:

$$P(X=x) = 0 \ \forall x$$

Thus:

$$P(X_t=y) = 0 \ \forall y$$

Looking forward to your feedback

Edit:

This is the explanation of what @stats_model commented.

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  • $\begingroup$ I was not able to understand why $X_t = X_{t-1} + a(X_{t-1},t-1) (t-t+1) + \sigma (X_{t-1},t-1) (W_t - W_{t-1})$. Does it hold in general or is it only in a discrete context? $\endgroup$
    – Matheus Manzatto
    Commented Jan 20, 2021 at 18:09
  • $\begingroup$ @MatheusManzatto This is called discretizing the SDE, and as far as I know it holds in general. It is discretizing a continous process. $\endgroup$
    – CheeseBurger
    Commented Jan 21, 2021 at 8:33
  • $\begingroup$ Discretizing would be an approximation even if the timestep is very small. If the timestep is fixed at 1, this doesn't seem correct. $\endgroup$
    – S. Catterall
    Commented Jan 21, 2021 at 15:07
  • $\begingroup$ @S.CatterallReinstateMonica Result holds in limit because brownian motion increment follows normal distrb. holds in limit. That is take $W_t, W_{t-k}$ as $k \to 0$ increment converges to normal dıstrb. I just gave 1 as example, time gap can be anything. $\endgroup$
    – CheeseBurger
    Commented Jan 21, 2021 at 16:52
  • $\begingroup$ That's true, the time gap could be anything - but it's a fixed constant in what you've written. To rigorously prove the required result you would have to study what happens as the time gap converges to zero. $\endgroup$
    – S. Catterall
    Commented Jan 21, 2021 at 17:45

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