Solution to SDE conditional on high maxima of driving Brownian motion

Question

Let $W$ be a standard one dimensional Brownian motion, and let $X$ be the solution to the SDE

$$dX_t = X_t \, dW_t \;, \quad X_0 = 1 \;.$$

For every $\varepsilon > 0$, let $A_\varepsilon$ denote the event

$$\{\underset{0 \leq t \leq 1}{\text{max}} W_t \geq \frac{1}{\varepsilon}\} \;, $$

and let $\mathbb P^\varepsilon$ be the probability measure given by

$$\mathbb P^\varepsilon (E) = \frac{\mathbb P(E \cap A_\varepsilon)}{\mathbb P(A_\varepsilon)} \;, $$

for all measurable events $E$.

We denote by $\mathbb E_{\mathbb P^\varepsilon}$ the expectation under $\mathbb P^\varepsilon$.

Question: Is it true that

$$\lim_{\varepsilon \to 0} \, \mathbb E_{\mathbb P^\varepsilon} [|X_1^\varepsilon - e|] = 0?$$

Remarks:

The above limit does exist in probability, that is, for every $\delta > 0$,

$$\lim_{\varepsilon \to 0} \mathbb P^\varepsilon [|X_1^\varepsilon - e| > \delta] = 0.$$

This can be seen by taking logarithms, and applying an earlier result.

Probablistic control of the logarithm of $X$ gives probablistic control of $X$ itself, hence the result.

We have also $L^1$ control of the logarithm, as can be seen by applying the result here.

However, the difficulty is that $L^1$ control of the logarithm does not give $L^1$ control of $X$ itself.

Further, if the limit in question holds, then it can be easily extended to the entire path before time $1$. That is, for all $0 < t < 1$,

$$\lim_{\varepsilon \to 0} \, \mathbb E_{\mathbb P^\varepsilon} [|X_t^\varepsilon - e^t|] = 0.$$

Again, the limit above does hold in probability, but $L^1$ is uncertain.

Answer 1

Partial answer

First, an heuristic argument. When we condition by events with low probability, the main is given by behaviour the less improbable situation. Here we condition by $S_1 := \max_{0 \le s \le 1} W_s$ at least equal to the huge number $1/\epsilon$. The most probable situation when this event holds is that the maximum is close to $1/\epsilon$, is achieved close to time $1$ and the sample path $W$ goes almost in straight line on the time interval $[0,1]$.

We have $X_t = \exp(W_t-t/2)$ for any $t \ge 0$. So $X_1^\epsilon = \exp(\epsilon(W_1-1/2))$. Under $\mathbb{P}^\epsilon$, $W_1$ is close to $1/\epsilon$, so $X_1^\epsilon$ is close to $e^1$ with probability close to $1$. Yet, very rare events where $W_1$ is still larger may affect significantly the expectation of $W_1$ under $\mathbb{P}^\epsilon$, so computations or fine estimations are necessary.

Set $S_t = \max_{0 \le s \le t} W_s$ for $t \ge 0$ and $\tau_a = \inf\{t \ge 0 : S_t>a\}$ for $a \ge 0$.

The distribution of $(W_t,S_t)$ can be computed using the reflexion principle. The random variable $(W_t,S_t)$ takes values in $\{(a,b) \in \mathbb{R}^2 : b \ge \max(0,a)\}$.

For every real numbers $a$ and $b$ such that $b \ge \max(0,a)$, $$\mathbb{P}[W_t<a~;~S_t>b] = \mathbb{P}[W_t<a~;~\tau_b < t] = \mathbb{P}[W_t>2b-a~;~\tau_b < t] = \mathbb{P}[W_t>2b-a],$$ since $S_t>2b-a>b$ on the event $[W_t>2b-a]$. One deduces the joint density of $(W_t,S_t)$ by computing $$-\frac{\partial^2}{\partial a\partial b}(\mathbb{P}[W_t<a~;~S_t>b]).$$ Hence the distribution of $W_1$ and the expectation of $X_1$ under $\mathbb{P}^\epsilon$ can be computed...

Answer 2

This is not a complete answer, but a continuation of the ideas in Christophe Leuridan's post.

Write

$$M_t := \underset{0 \leq s \leq t}{\text{max}} \, W_s.$$

Using the suggestion in the aforementioned post, we deduce the joint density $f_{(W_1, M_1)}$ of $(W_1, M_1)$ to be

$$f_{(W_1, M_1)} (x, y) = \sqrt{\frac{2}{\pi}} (2y - x) \,\text{exp} \left ( -\frac{1}{2}(2y - x)^2 \right ) \mathbf 1_{y > 0, x \leq y}.$$

We also know that $M_1$ has a half normal distribution, thus its probability density function is given by

$$f_{M_1} (y) = \sqrt{\frac{2}{\pi}} e^{-\frac{1}{2}y^2}.$$

Thus the conditional density $f_{W_1| M_1}$ of $W_1$ given $M_1$ is given by

$$f_{W_1| M_1} (x|y) = \frac{f_{(W_1, M_1)} (x, y)}{f_{M_1} (y)}$$

$$= (2y - x) \, \text{exp}\left (-\frac{1}{2}(3y^2 + x^2 - 4xy)\right) \, \mathbf 1_{y > 0, x \leq y}.$$

Hence we may compute

$$\lim_{\varepsilon \to 0+} \mathbb E_{\mathbb P^\varepsilon} (|X_1^\varepsilon - e|)$$

$$ = \lim_{\varepsilon \to 0+} \mathbb E_{\mathbb P^\varepsilon} (| e^{\varepsilon (W_1 - \frac{1}{2})} - e|)$$

$$= \lim_{\varepsilon \to 0+} \frac{\int_{\frac{1}{\varepsilon}}^\infty \int_{-\infty}^y |e^{\varepsilon (x - \frac{1}{2})} - e| \, f_{W_1| M_1} (x|y) \, dx \, dy}{\mathbb P(M_1 \geq \frac{1}{\varepsilon})}.$$

$$= \lim_{\varepsilon \to 0+} \frac{\int_{\frac{1}{\varepsilon}}^\infty \int_{-\infty}^y |e^{\varepsilon (x - \frac{1}{2})} - e| \, (2y - x) \, \text{exp}\left (-\frac{1}{2}(3y^2 + x^2 - 4xy)\right ) \, dx \, dy}{2 \Phi(\frac{1}{\varepsilon})}.$$

Where $1 - \Phi$ is the CDF of the standard normal distribution.

As of now, it is unclear to me how to evaluate the above limit.