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I applied a smoothing function to a Brownian equation and obtained a stochastic differential equation by using Ito's lemma. The smoothing function is exp(Bt).

How do I get the expected value and variance of this function? Just looking for the required approach rather than a full fledged solution.

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  • $\begingroup$ Do you mean $\exp\{B(t)\}$ where $B(t)$ is Brownian motion? $\endgroup$
    – Marco
    Feb 17 at 9:31
  • $\begingroup$ Yes, exactly that $\endgroup$
    – atastix
    Feb 17 at 10:49
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For a Normally distributed random variable, $X$, with mean $\mu$, and variance $V$, the following is true:

\begin{equation} \mathbb{E} \{\exp(\theta X)\} = \exp\left(\theta\mu+\frac{1}{2}\theta^2V\right) \end{equation} In your example, conditional upon time $0$, and assuming $B(0)=0$, $B(t)$ is Normally distributed variable with zero mean and variance $t$. Applying this formula with $\mu = 0$, $V=t$ and $\theta=1$, we find:

\begin{equation} \mathbb{E} \{\exp(B(t))\}=\exp \left( \frac{1}{2}t \right) \end{equation}

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  • $\begingroup$ Maybe add somewhere that the question has the case $\theta=1$ for completeness. $\endgroup$
    – Bob Jansen
    Feb 17 at 15:51
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    $\begingroup$ @BobJansen - sure, just added. $\endgroup$
    – Marco
    Feb 17 at 16:30
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As $B(t)$ follows a normal distribution with mean zero and variance $t$, then $\exp\{B(t)\}$ will (by definition) follow a log-normal distribution with parameters $\mu=0$ and $\sigma^2=t$. The moments of the log-normal distribution are known.

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