If $t\mapsto X_t$ is continuous almost everywhere and $(X_t)$ has independent increments, then $X_t - X_s$ follows a normal distribution?

Question

The following statements can be found at Glasserman's Monte Carlo Methods in Financial Engineering.

Given a stochastic process $(X_t)_{t\in [0,T]},$ if

$(i)$ the mapping $t\mapsto X_t$ is continuous almost everywhere on $[0,T],$ and

$(ii)$ $X_t$ has independent increments (ie for any $\{0\leq t_0<t_1<\dots <t_n\leq T\}$, all increments $\{ W(t_n) - W(t_{n-1}), W(t_{n-1}) - W(t_{n-2}),...,W(t_1)-W(t_0)\}$ are independent.) and $X_t - X_s$ has mean $0$ and variance $t - s$ for all $s<t,$

then $X_t - X_s$ follows a normal distribution.

When I tried to show that $X_t- X_s$ follows a normal distribution, I tried to show that its MGF is the same as a normal distribution, that is, if $Y = X_t - X_s$, then $$\mathbb{E} [e^{uY}] = e^{\frac{1}{2} u^2(t-s)^2}.$$ However, I am not able to do it. Any hint is appreciated.

Answer 1

The problem is solved in here (see Theorem 2 and proof) in full detail using Itô's lemma.

On OP's demand, below I outline the main steps of the proof. First note that using the independences of the increments, we have that the characteristic function of $X_t-X_s$ is

$$ \mathbb E[e^{ia(X_t-X_s)}] = \phi(t,a)/\phi(s,a),\;\forall a \in \mathbb R, $$ where $\phi(t,a):=\mathbb E[e^{ia(X_t-X_0)}]$ for all $(t,a) \in \mathbb R_+ \times \mathbb R$. The proof then proceeds as follows.

• Step 1: Show that (see Lemma 3) there exists a unique continuous function $\psi:\mathbb R_+ \times \mathbb R \rightarrow \mathbb R$ such that for all $(t,a) \in \mathbb R_+ \times \mathbb R$, it holds that

  • $\psi(0,a)=0$,
  • $\mathbb E[e^{ia(X_t-X_0)}] = e^{\psi(t,a)}$,
  • $e^{iX_t - \psi(t,a)}$ is a martingale (w.r.t the natural filtration of the process $X$).

• Step 2: Show that there exists a continuous function $\hat{b}:\mathbb R_+ \to \mathbb R$ such that for all $(t,a) \in \mathbb R_+ \times \mathbb R$, it holds that

  • $\hat{X}_t := X_t - \hat{b}_t$ is a martingale,
  • $ia(X_t-X_0)-\psi(t,a)-a\hat{X}_t/2$ is a square-integrable supermartingale.

• Step 3: Show that (see Lemma 5) for every $(t,a) \in \mathbb R_+ \times \mathbb R$, it holds that $\psi(t,a) := iab_t - \sigma_t^2 a^2/2$, where $b_t := \mathbb E[X_t-X_0]$ and $\sigma_t := \mathbb E[\hat{X}_t^2]$.

Putting things together, you've got that for $0 \le s \le t$, $$ \mathbb E[e^{ia(X_t-X_s)}] = e^{\psi(t,a)} = e^{ia(b_t-b_s)-(\sigma_t^2-\sigma_s^2)a^2/2}, $$ which is the characteristic function of a normal distribution with mean $b_t$ and variance $\sigma_t^2-\sigma_s^2 \ge 0$.