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I've got a little problem: if $X_{t}$ and $Y_{t}$ are two indipendent Brownian motions, is then $$Z_{t}:=X_{t}+Y_{t}$$ a Brownian motion too? I've got some troubles only with showing that $Z_t$ is Gaussian. Thank you for your help.

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First we show that that if $X\colon \Omega \rightarrow \mathbb{R}^n$ is normal $\mathcal{N}(m, C)$, $Y\colon \Omega \rightarrow \mathbb{R}^n$ is normal $\mathcal{N}(m^{\prime}, C^{\prime})$ and they are independent then $X+Y$ is normal $\mathcal{N}(m_1+m_2, C_1+C_2)$. Here $m$, $m^{\prime} \in \mathbb{R}^n$ and $C=[c_{jk}]$, $C^{\prime}=[c_{jk}^{\prime}]$ are non-negative definite $n \times n$ matrices.

Let $\phi_X$ and $\phi_Y$ be the corresponding characteristic functions. Since $X$ and $Y$ are independent we have that $$ \phi_{X+Y}= \phi_X\phi_Y.$$ Furthermore,

$$\phi_{X+Y}(x_1, \ldots, x_n)= \phi_X(x_1, \ldots, x_n)\phi_Y(x_1, \ldots, x_n)= \exp\left( -\frac{1}{2}\sum_{j,k}x_jc_{jk}x_k + i\sum_{j}x_jm_j \right)\exp\left( -\frac{1}{2}\sum_{j,k}x_jc_{jk}^{\prime}x_k + i\sum_{j}x_jm_j^{\prime} \right)= \exp\left( -\frac{1}{2}\sum_{j,k}x_j(c_{jk}+c_{jk}^{\prime})x_k + i\sum_{j}x_j(m_j+m_j^{\prime}) \right). $$ Note that this is the characteristic function of $\mathcal{N}(m+m^{\prime}, C + C^{\prime})$ and since the characteristic function determines the distribution uniquely the distribution of $X+Y$ is $\mathcal{N}(m_1+m_2, C_1+C_2)$.

Now, I believe you will be able to finish the answer.

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    $\begingroup$ Yes, but thus $Z_{t}\sim N(0,2t)$ but in order to show that's a BM $Z_{t}$ needs to be $N(0,t)$, isn't it? $\endgroup$ – Giuseppe Guarnuto Oct 27 '14 at 11:47
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    $\begingroup$ Correct, so you can see that instead you should consider $Z_t=\frac{X_t+Y_t}{\sqrt{2}}$. $\endgroup$ – m_gnacik Oct 27 '14 at 12:49
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    $\begingroup$ The question is a bit misleading because the poster was really trying to prove for a standard Brownian Motion. Adding two independent standard Brownian Motion won't added up to another standard Brownian Motion. $\endgroup$ – SmallChess Oct 28 '14 at 5:05

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