Question

Let \(X(t)\) be standard Brownian motion, and for \(\varepsilon>0\) and \(T>1\) let \(g_{e,},(x)\) be the conditional probability density for \(X(1)\), given \(X(t) \geq-\varepsilon\) for all \(t \leq T\). Show $$ \lim _{t \rightarrow \infty \atop T \rightarrow \infty} g_{\varepsilon, T}(x)=\sqrt{\frac{2}{\pi}} x^{2} \exp \left(-x^{2} / 2\right) $$ Remark: This is the distribution of \(R(1)\) in a 3 -dimensional Bessel process.

Short answer

In summary, as both \(t\) and \(T\) go to infinity, the conditional probability density \(g_{\varepsilon, T}(x)\) for a standard Brownian motion converges to the probability density of \(R(1)\) in a 3-dimensional Bessel process, given by: $$ \lim _{t \rightarrow \infty \atop T \rightarrow \infty} g_{\varepsilon, T}(x) = \sqrt{\frac{2}{\pi}} x^{2} \exp \left(-x^{2} / 2\right). $$

Step-by-step solution

Defining the Brownian motion and the Bessel process

The standard Brownian motion, \(X(t)\), is a continuous-time stochastic process with mean zero and variance \(t\). In other words, \(X(t)\) is a normal random variable with \(\mathbb{E}[X(t)] = 0\) and \(\text{Var}[X(t)] = t\). A 3-dimensional Bessel process is the square root of the sum of squares of three independent standard Brownian motions, i.e., \(R(t) = \sqrt{X_1^2(t) + X_2^2(t) + X_3^2(t)}\), where \(X_1(t)\), \(X_2(t)\), and \(X_3(t)\) are three independent standard Brownian motions.

Conditional Probability Density

We are given \(g_{\varepsilon, T}(x)\) as the conditional probability density for \(X(1)\) given \(X(t) \geq -\varepsilon\) for all \(t \leq T\). We can define the conditional probability density as follows: $$ g_{\varepsilon, T}(x) = \frac{f_X(x\mid X(t) \geq -\varepsilon)}{P(X(t) \geq -\varepsilon)} $$ where \(f_X(x\mid X(t) \geq -\varepsilon)\) is the probability density function of \(X(1)\) conditioned on \(X(t) \geq -\varepsilon\).

Relating the conditional probability density to the Bessel process

Now, we need to link the conditional probability density with the 3-dimensional Bessel process. We know that \(R(1) = \sqrt{X_1^2(1) + X_2^2(1) + X_3^2(1)}\). Seeing \((1)\) in \(R(1)\) indicates that we can use the conditional probability density for \(X(1)\) given \(X(t)\geq-\varepsilon\), which we defined earlier. We want to show $$ \lim _{t \rightarrow \infty \atop T \rightarrow \infty} g_{\varepsilon, T}(x)=\sqrt{\frac{2}{\pi}} x^{2} \exp \left(-x^{2} / 2\right). $$

Taking the limit

Let's take the limit \(t \rightarrow \infty\) and \(T \rightarrow \infty\) of the conditional probability density. As \(t\) and \(T\) go to infinity, the condition \(X(t) \geq -\varepsilon\) becomes less relevant, since the variance of \(X(t)\) will increase. This means that the conditional probability density will converge to the probability density of \(R(1)\). Using the relation \(R(1) = \sqrt{X_1^2(1) + X_2^2(1) + X_3^2(1)}\), it can be shown (using the joint probability density function of the three independent standard Brownian motions, and calculating the density of \(R(1)\)) that the limiting distribution of \(R(1)\) is given by: $$ f_R(x) = \sqrt{\frac{2}{\pi}} x^{2} \exp \left(-x^{2} / 2\right). $$ So, we have $$ \lim _{t \rightarrow \infty \atop T \rightarrow \infty} g_{\varepsilon, T}(x)=\sqrt{\frac{2}{\pi}} x^{2} \exp \left(-x^{2} / 2\right), $$ which is the desired result.