stochastic process variance

Question

when I have a equation that is

$y_t=x_t+ε_t$

and I assume that $x_t$ and $ε_t$ are independent. $$dx_t=σ_tdW_t,\; E(ε_t)=0,\; var(ε_t)=ω^2$$ where $W_t$ is Brownian motion.

What is the variance of $\;\; var(y_{t}^2)?$

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I know $$var(y_t^2)=var((x_t+ε_t)^2)=var(x_t^2+ε_t^2+2x_tε_t)$$

and I unpick the $var(x_t^2+ε_t^2+2x_tε_t)$

it should equal to $$var(x_t^2)+var(ε_t^2)+var(2x_tε_t)+2cov(x_t,ε_t)+2cov(x_tε_t,ε_t)+2cov(x_t,x_tε_t)$$

and because $x_t$ and $ε_t$ are independent so $cov(x_t,ε_t)=0.$

but when I calculate the $cov(x_tε_t,ε_t)$,

$$cov(x_tε_t,ε_t)=E(x_t(ε_t)^2)-E(x_t)E(ε_t^2)$$

and I don't know how to calculate it. should it equal to $0$? because $x_t$ and $(ε_t)^2$ are independent?

plz help me.

Answer 1

You state $x_t$ and $\xi_t$ are independent, which means that $E[f(x_t)g(\xi_t)] = E[f(x_t)]E[g(\xi_t)]$. Taking $f(x)=x$ and $g(\xi)=\xi^2$ we get:

$cov(x_t \xi_t^2) = E(x_t \xi^2_t)-E(x_t)E(\xi^2_t)=E(x_t)E(\xi^2_t)-E(x_t)E(\xi^2_t)=0$

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As a caveat, we normally call $W(t)$ a standard Brownian motion, or Weiner process. It is important to establish that $W(t)$ is standard and hence has variance $t$.

Answer 2

Rather odd downvote. Whatever... :-)

Simplifying somewhat your notations, it seems you are asked to compute the variance of $y^2$, where $$y=x+z$$ for some independent random variables $x$ and $z$ such that $E(z)=0$, $E(z^2)=\omega^2$, and $E(x)=0$ (can you prove that this very last identity holds in your setting?). Introduce $E(x^2)=a^2$ and assume furthermore that $x$ and $z$ have finite fourth moments (do you see why this is necessary?).

Then, by independence of $x$ and $z$, $$E(y^2)=E(x^2)+2E(x)E(z)+E(z^2)=a^2+2\cdot0\cdot0+\omega^2=a^2+\omega^2$$ and $$E(y^4)=E(x^4)+4E(x^3)E(z)+6E(x^2)E(z^2)+4E(x)E(z^3)+E(z^4)$$ reduces to $$E(y^4)=E(x^4)+6a^2\omega^2+E(z^4)$$ hence $$\mathrm{var}(y^2)=E(x^4)+6a^2\omega^2+E(z^4)-(a^2+\omega^2)^2$$ Note that the last expression cannot be further simplified significantly since, without additional information, $E(x^4)$ and $E(z^4)$ are not known.

Of course, a distinct possibility is that you made a typo and that you are in fact asked for the variance of $y$. Then, from a small part of everything above, the answer would simply read $$\mathrm{var}(y)=E(y^2)=a^2+\omega^2$$