Is the average of independent Brownian Motions still a Brownian Motion?

Question

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?

Answer 1

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.

Answer 2

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

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

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 $X_t$.

Answer 3

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 $t\geq 0$.