How to prove that $X_s=\int^s_0 f(u)dW_u$ is independant from $X_t-X_s$

Question

I am asked to prove that $X_s$ and $X_t-X_s$ are independant for $s<t$ then $$X_t=\int^t_0f(u)dW_u$$ for a deterministic function $f$ and brownian motion $W_t$. For the proof I am giving a hint to compute $E[\exp(a_1X_s+a_2(X_t-X_s))]$

How can prove independence using the hint?

So far I have done following:

I find $X_s \sim N(0,\int^s_0f^2(u) du)$ and $X_t - X_s\sim N(0,\int^t_sf^2(u) du)$

$$E[\exp(a_1X_s+a_2(X_t-X_s))] =\exp(\frac{1}{2}a^2_1\int^s_0f^2(u) du+\frac{1}{2}a^2_2\int^t_sf^2(u) du)$$

How Can I prove independence from here?

Answer 1

I'll only answer the question in your title. One way to see it is by discretising the integrals.

From your title, let's define $$I(a,b):=\int_a^b f(u)dW_u$$ Discretising: $$I(a,b)=\lim_{n\to\infty} \sum_{k=0}^{n-1}f(a + i\frac{b-a}{n})(W(a + (i+1)\frac{b-a}{n}) - W(a + i\frac{b-a}{n}))$$ in $L^2$.

Now simply note that any increment $\Delta W$ in $[t, s]$ is independent of any increment in $[0,s]$. Therefore $I(s,t)$ must be independent of $I(0,s)$.

Answer 2

What follows is not rigorous, but hopefully has the main idea. First, you probably want to justify $$X_t-X_s \sim N\left(0, \int_s^t f(u)^2 du \right), $$ which can be done by approximating $f$ by a simple function $g = a_1 1_{(s,t_1)} + a_21_{(t_1,t_2)} + \ldots + a_n1_{(t_{n-1},t)}$ and then using $$\int_s^t g(u)dW_u = a_1(W_{t_1} - W_s) + a_2 (W_{t_2}-W_{t_1}) + \ldots a_n (W_{t}-W_{t_{n-1}}) \sim N(0, a_1^2t_1 + a_2^2 (t_2-t_1)+\ldots a_n^2(t_n-t_{n-1})) = N\left(0, \int_s^t g(u)^2 du \right)$$ for $W_{t_i}-W_{t_{i-1}} \sim N(0,t_i-t_{i-1})$ all independent. The trick then is to show that $X_s$ and $X_t-X_s$ are uncorrelated in order to help us take the expectation $\mathbb{E}[\exp(a_1X_s + a_2 (X_t-X_s))]$. Now, $$\mathbb{E}[(X_t-X_s)^2] = \mathbb{E}[X_t^2] - 2\mathbb{E}[X_tX_s] + \mathbb{E}[X_s^2],$$ hence, $$\mathbb{E}[X_sX_t] = \frac{1}{2}\left(\mathbb{E}[X_t^2] + \mathbb{E}[X_s^2] - \mathbb{E}[(X_t-X_s)^2] \right) = \frac{1}{2} \left( \int_0^t f(u)du + \int_0^s f(u)du - \int_s^t f(u) du \right) = \int_0^sf(u)du = \mathbb{E}[X_s^2]$$ and so $\mathbb{E}[X_s(X_t-X_s)]=0,$ hence $X_s$ and $X_t-X_s$ are uncorrelated.

Now, since two uncorrelated normally distributed random variables $Y_1 \sim N(0,\sigma_1^2)$ and $Y_2 \sim N(0,\sigma_2^2)$ satisfy $\mathbb{E}[\exp(a_1Y_1 + a_2 Y_2)] = \exp(\frac{a_1}{2} Y_1 + \frac{a_2}{2} Y_2) = \mathbb{E}[\exp(a_1Y_1)]\mathbb{E}[\exp(a_2Y_2)]$, we have

$$\mathbb{E}[a_1X_s + a_2(X_t-X_s)] = \mathbb{E}[\exp(a_1X_s)] \mathbb{E}[\exp(a_2(X_t-X_s))]$$

for all $a_1$, $a_2$. Taking $n$ derivatives with respect to $a_1$ and $m$ derivatives with respect to $a_2$, and then setting $a_1=0=a_2$, we get

$$\mathbb{E}[X_s^n(X_t-X_s)^m] = \mathbb{E}[X_s^n]\mathbb{E}[(X_t-X_s)^m].$$

Using this, we now know that for any polynomial functions $p$ and $q$, we have $\mathbb{E}[p(X_s)q(X_t-X_s)] = \mathbb{E}[p(X_s)]\mathbb{E}[q(X_t-X_s)].$ Since polynomial functions are weakly dense, the probability distribution function $\rho_{X_s,X_t-X_s}$ of $X_s$ and $X_t-X_s$ is the product of the probability density functions $\rho_{X_s}$ of $X_s$ and $\rho_{X_t-X_s}$ of $X_t-X_s$. Hence, $X_s$ and $X_t-X_s$ are independent.

Answer 3

We have that

$$ X_{t} - X_{s} = \int_{s}^{t}f(u)dW_{u} $$ Thus, if $f$ was a simple function, then $X_{t} - X_{s}$ would be a linear combination of $W_{k}$'s where $k \in [s,t] $ which is independent of $W_{s}$ by definition of the Wiener process. A fortiori, $X_{t} - X_{s}$ would be independent of $X_{s}$ which is a linear combination of $W_{k}$'s where $k \in [0,s] $. Now use that fact that $f$ can be written as a limit of simple functions.