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QUESTION

Let $W_t$ and $\tilde{W}_t$ be two independent Brownian motions
and $p$ a constant between $-1$ and 1. Is the process
$X_t=pW_t+\tilde{W}_t\sqrt{1-p^2}$ continuous? What is its
distribution? Is $X_t$ a Brownian motion?


ANSWER

$$X_t=PW_t+\tilde{W}_t(1-\rho^2)^\frac{1}{2}$$ The process is
continuous since $W_t,\ \tilde{W}_t$ and $\rho$ are.

Mean:
\begin{eqnarray*}
\left&lt;X_t\right&gt;&amp;=&amp;\left&lt;PW_t+\tilde{W}_t(1-p^2)^\frac{1}{2}\right&gt;\\
&amp;=&amp;\rho\left&lt;W_t\right&gt;+(1-p^2)^\frac{1}{2}\left&lt;\tilde{W}_t\right&gt;\\
&amp;=&amp;0
\end{eqnarray*}
since $W_t\sim N(0,t)$ and $\tilde{W}_t\sim N(0,t)$ by definition.

Variance:
\begin{eqnarray*}
\left&lt;(X_t-0)^2\right&gt;&amp;=&amp;\left&lt;(PW_t+\tilde{W}_t(1-\rho^2)^\frac{1}{2})^2\right&gt;\\
&amp;=&amp;\left&lt;\rho^2W_t^2+\tilde{W}_t^2(1-\rho^2)+2PW_t\tilde{W}_t(1-p^2)^\frac{1}{2}\right&gt;\\
&amp;=&amp;\rho^2\left&lt;W_t^2\right&gt;+(1-\rho^2)\left&lt;\tilde{W}_t^2\right&gt;+2\rho\sqrt{1-\rho}\left&lt;W_t\tilde{W}_t\right&gt;
\end{eqnarray*}

Now if $W_t$ and $\tilde{W}_t$ are independent they are
uncorrelated, thus $\left&lt;W_t\tilde{W}_t\right&gt;=0$.

Hence
\begin{eqnarray*}
\left&lt;(X_t-0)^2\right&gt;&amp;=&amp;\rho^2\sigma^2_t+(1-\rho^2)\tilde{\sigma}_t^2\\
&amp;=&amp;\rho^2t+(1-\rho^2)t\\ &amp;=&amp;t
\end{eqnarray*}

Where $\sigma_t^2$ is the variance of $W_t^2$ (mean is zero) and
$\tilde{\sigma}^2_t$ is the variance of $\tilde{W}_t^2$ (mean is
zero).

Thus since $W_T$ and $\tilde{W}_t$ are normal, $X_t\in N(0,t)$

Is it Brownian? Check conditions:

\begin{description}

\item[(i)]
$X_t$ is continuous and $X_0=0$: It is continuous and by
definition $W_0$ and $\tilde{W}_0$ are zero so condition is
satisfied.

\item[(ii)]
$X_t\in N(0,t)$

\item[(iii)]
$X_{s+t}-X_t\in N(0,t)$ and independent of time $&lt;s$:

$$X_{s+t}-X_s\in \rho\underbrace{\underbrace{(W_{t+s}-W_s)}_{\in
N(0,t)}+\sqrt{1-\rho^2}\underbrace{(\tilde{W}_{t+s}-\tilde{W}_s)}_{\in
N(0,t)}}_{\textrm{by definition if $W_t,\ \tilde{W}_t$ are
brownian}}$$

Thus from the above calculations we have

$$X_{s+t}-X_s\in N(0,t)$$

The $W_S$ and $\tilde{W}_s$ are independent of $S$ by definition
too, so $X_{s+t}-X_s$ must be independent of $S$. So (iii) is
satisfied.

\end{description}

Thus $X_t$ is Brownian.




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