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

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 $$\{0\leq t_0, all increments $$\{ 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

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.

• What you are trying to show is that Brownian motion (with drift) is the only continuous Lévy process. As far as I know, there is no elementary proof of this fact and you have to use the Lévy-Itô decomposition. See here Jun 3, 2020 at 12:12

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 $$\sigma_t^2-\sigma_s^2 \ge 0$$.

• It might be better if you can provide a summary of the list of ideas used in Theorem 2... Sep 30, 2020 at 6:42
• @Idonknow See updated post. Sep 30, 2020 at 7:43