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Neeraj
Neeraj
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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))$$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 $X_t$.

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:

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

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 $X_t$.

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

$W_t$ ~ $N(0, s^2_1)$$N(0, t)$,
$B_t$ ~ $N(0, s^2_2)$$N(0, t)$
$Cov(W_t,B_t)=0$

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

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(s^2_1+s^2_2)$$$$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(var(W_t)+var(B_t))=.5t$$

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

$1/2(B_t+W_t)$ ~ $N(0, 1/4(s^2_1+s^2_2))$$N(0, .5t))$

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

$W_t$ ~ $N(0, s^2_1)$,
$B_t$ ~ $N(0, s^2_2)$
$Cov(W_t,B_t)=0$

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

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(s^2_1+s^2_2)$$ because $W_t$ and $B_t$ are independent. So:

$1/2(B_t+W_t)$ ~ $N(0, 1/4(s^2_1+s^2_2))$

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:

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

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Bob Jansen ♦
Bob Jansen ♦
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$W_t$ and $B_t$ are tow independent Brownian motion, where :

$W_t$ ~ $N(0, s^2_1)$,
$B_t$ ~ $N(0, s^2_2)$
$Cov(W_t,B_t)=0$

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

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(s^2_1+s^2_2)$$ because $W_t and B_t$$W_t$ and $B_t$ are independent. So:

$1/2(B_t+W_t)$ ~ $N(0, 1/4(s^2_1+s^2_2))$

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

$W_t$ ~ $N(0, s^2_1)$,
$B_t$ ~ $N(0, s^2_2)$
$Cov(W_t,B_t)=0$

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

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(s^2_1+s^2_2)$$ because $W_t and B_t$ are independent. So:

$1/2(B_t+W_t)$ ~ $N(0, 1/4(s^2_1+s^2_2))$

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

$W_t$ ~ $N(0, s^2_1)$,
$B_t$ ~ $N(0, s^2_2)$
$Cov(W_t,B_t)=0$

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

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$ $$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(s^2_1+s^2_2)$$ because $W_t$ and $B_t$ are independent. So:

$1/2(B_t+W_t)$ ~ $N(0, 1/4(s^2_1+s^2_2))$

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Neeraj
Neeraj
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