MHB Find the expectation and covariance of a stochastic process

i_a_n
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The problem is:Let $W(t)$, $t ≥ 0$, be a standard Wiener process. Define a new stochastic process $Z(t)$ as $Z(t)=e^{W(t)-(1/2)\cdot t}$, $t≥ 0$. Show that $\mathbb{E}[Z(t)] = 1$ and use this result to compute the covariance function of $Z(t)$. I wonder how to compute and start with the expectation cause it is not any case with a formula to use. Thanks in advance!
 
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From the definition of the Wiener process we have that $W(t) \sim N(0,t)$. Calculating the expected value gives
$$\mathbb{E}[Z(t)] = e^{\frac{-1}{2}t} \mathbb{E}[e^{W(t)}]$$
To complete the proof use the moment generating function of the normal distribution.

I guess with the covariance function you mean $\mbox{cov}(Z(t),Z(s))$ for $s,t \geq 0$?
 
ianchenmu said:
The problem is:Let $W(t)$, $t ≥ 0$, be a standard Wiener process. Define a new stochastic process $Z(t)$ as $Z(t)=e^{W(t)-(1/2)\cdot t}$, $t≥ 0$. Show that $\mathbb{E}[Z(t)] = 1$ and use this result to compute the covariance function of $Z(t)$. I wonder how to compute and start with the expectation cause it is not any case with a formula to use. Thanks in advance!

I suggest You reading this excellent written by Steven R. Dunbar ...

http://www.math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/StochasticCalculus/GeometricBrownianMotion/geometricbrownian.pdf

... which describes the properties of the 'Geometric Brownian Motion'. This process is described by the formula ...

$\displaystyle Z(t) = Z_{0}\ e^{\mu\ t + \sigma\ W(t)}\ (1)$

... where W(t) is a standard Brownian Motion. The mean and the variance are...

$\displaystyle E \{ Z(t)\} = Z_{0}\ e^{(\mu + \frac{\sigma^{2}}{2})\ t}\ (2)$

$\displaystyle Var \{Z(t) \} = Z_{0}^{2}\ (e^{\sigma^{2}\ t} - 1)\ e^{(2\ \mu\ + \sigma^{2})\ t}\ (3)$

Kind regards

$\chi$ $\sigma$
 
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