If $W$ and $B$ are independent Brownian Motions (BM thereafter), then the average of $W$ and $B$ is $X_t=\frac{1}{2}(W_t+B_t)$.

Where do I begin to show that indeed it is still a BM?

Also, if both are martingales, then $X_t$ must be a martingale also. How would I prove this considering it has the two random variables?

  • 2
    $\begingroup$ $X_t = \frac{1}{\sqrt{2}}(W_t+B_t)$ is a BM. $\endgroup$
    – Gordon
    Oct 14 '15 at 15:54

It is nearly a Bronwian motion. Just the variance is not correct:

The question is more tricky than it seems. A Brownian motion has the distribution properties stated below, so does a linear combination of BMs. But after all it is a martingale in a certain filtration (set of information) which has to be defined. $B_t$ is a BM in its own filtration, so is $W_t$. The mean is a Brownian motion in its own filtration and in the filtration generated by $B_t+W_t$.

For the distribution consider that $E[B_t] = E[W_t]=0$ and $Var[B_t] = Var[W_t]=t$. Then $E[X_t]=0$ and $Var[X_t] = 1/2 t$ provided that $(B_t)_{t\ge0}$ and $(W_t)_{t\ge0}$ are independent. The Gaussian law is a known fact.

EDIT: $ VAR[1/2(W_t+B_t)] = 1/4 (VAR[W_t] + VAR[B_t]) = 1/4 (t+t)= 1/2 t$. So the variance is wrong and it is no BM.

  • $\begingroup$ $Var[X_t]$ must be $0.5t$ $\endgroup$
    – Neeraj
    Oct 14 '15 at 6:26
  • $\begingroup$ Correct, and then it is not BM. $\endgroup$
    – Ric
    Oct 14 '15 at 7:00
  • $\begingroup$ It is correct for a Standard Brownian Motion. The incrementation of a Brownian Motion is normally distributed: $W_t-W_s \sim N(0,t-s)$, for $0\leq s \leq t$ $\endgroup$
    – QFi
    Sep 30 '17 at 3:39

Let's $W_t$ and $B_t$ are tow independent Brownian motion, where :

$W_t$ ~ $N(0, t)$,
$B_t$ ~ $N(0, t)$

We know that sum of two Gaussian random variable is also Gaussian.

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(var(W_t)+var(B_t))=.5t$$

because $W_t$ and $B_t$ are independent. So:

$X_t$=$1/2(B_t+W_t)$ ~ $N(0, .5t)$

EDIT : $X_t$ has continuous path and $X_t=0$ for $t=0$ but $Var(X_t) \neq t$( a necessary condition for Brownian Motion). Hence $X_t$ is not Brownian Motion.

@Gordon mention rightly $\sqrt{1/2}(Wt+Bt)$ is a BM but not $X_t$.

  • $\begingroup$ the variance of BM at time $t$ is $t$ not $s_1^2$. $\endgroup$
    – Ric
    Oct 13 '15 at 18:25
  • $\begingroup$ You are right @Richard. It must be $W_t$ ~ $N(0,t)$, $B_t$ ~ $N(0,t)$. So $$E(1/2(W_t+B_t))=0$$ and $$Var(1/2(W_t +B_t))=1/4(t+t)=0.5t$$. $\endgroup$
    – Neeraj
    Oct 14 '15 at 6:23
  • $\begingroup$ yes , correct .. $\endgroup$
    – Ric
    Oct 14 '15 at 7:04
  • $\begingroup$ Also having normal marginal distributions is not enough to be a Brownian motion $\endgroup$
    – AFK
    Oct 14 '15 at 8:51
  • $\begingroup$ @AFK I also agree with you.. $\endgroup$
    – Neeraj
    Oct 15 '15 at 7:13

The OP states that $W(t)$ and $B(t)$ are two independent Brownian motions, which is slightly different from Standard Brownian Motion/Wiener Process, even if they have little in common (both are Markov and Martingale processes). The Wiener process is the standard Brownian motion while a general Brownian motion is of a form: $B(t)=\alpha\,W(t)+\beta$.

The definition of the Brownian motion from Stochastic Calculus for Finance II (Shreve, 2004) is:

Let $(Ω, F, P)$ be a probability space. For each $ω ∈ Ω$, suppose there is a continuous function $W(t)$ of $t ≥ 0$ that satisfies $$W(0) = 0\tag1$$ and that depends on $ω$. Then $W(t)$, $t ≥ 0$, is a Brownian motion if for all $0=t_0 <t_1 <···<t_m$ the increments $W(t_1) = W(t_1)−W(t_0),W(t_2)−W(t_1),...,W(t_m)−W(t_{m−1})$ are independent and each of these increments is normally distributed with $$\mathop{\mathbb{E}}[W(t_{i+1}) − W(t_i)] = 0, \tag2$$ $$Var[W(t_{i+1})−W(t_i)]=t_{i+1}−t_i \tag3$$

So, if $X(t)=\dfrac{W(t)+B(t)}{2}$ is Brownian motion, it must verify the properties $(1)$, $(2)$, and $(3)$. Let's see:

Property $(1)$ $$\begin{align} X(0)&=\dfrac{W(0)+B(0)}{2}\\ &=0 \end{align}$$

Property $(2)$

For $0\leq s\leq t$: $$\begin{align} \mathop{\mathbb{E}}[X(t)-X(s)]&=\mathop{\mathbb{E}}\left[\dfrac{W(t)+B(t)}{2}-\dfrac{W(s)+B(s)}{2}\right]\\ &=\dfrac{\mathop{\mathbb{E}}[W(t)-W(s)]+\mathop{\mathbb{E}}[B(t)-B(s)]}{2}\\ &=0 \end{align}$$

Property $(3)$

For $0\leq s\leq t$: $$\begin{align} Var[X(t)-X(s)]&=\mathop{\mathbb{E}}\left[\dfrac{W(t)+B(t)}{2}-\dfrac{W(s)+B(s)}{2}\right]\\ &=Var\left[\dfrac{W(t)-W(s)}{2}\right]+Var\left[\dfrac{B(t)-B(s)}{2}\right]\\ &=\dfrac{1}{4}\,Var[W(t)-W(s)]+\dfrac{1}{4}\,Var[B(t)-B(s)]\\ &=\dfrac{1}{2}\,(t-s)\\ &\neq t-s \end{align}$$

Since $Var[X(t)-X(s)]\neq t-s$, for $0\leq s \leq t$, we conclude that $X(t)$ is not a Brownian motion for $t\geq 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.