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Let $\{W_t\}_{t \in \mathbb{R}_+}$ be a Wiener process. I wish to calculate $E \left[ W(t)e^{\lambda W(t)} \right], \lambda \in \mathbb{R}$.

Solution attempt

First define a function given by $f(t,x) = xe^{\lambda x}$. Then, \begin{align} \frac{\partial f(t,x)}{\partial x} &= e^{\lambda x} + \lambda x e^{\lambda x}, \\ \frac{ \partial^2 f(t,x)}{\partial x^2} &= 2\lambda e^{\lambda x} + \lambda^2 xe^{\lambda x}.\end{align}

So by Ito's formula, the differential of $f(t,W(t))$ is given by \begin{align} df = \left[ \lambda e^{\lambda W} + \frac{\lambda^2}{2}We^{\lambda W} \right]dt + \left[ e^{\lambda W} + \lambda We^{\lambda W} \right] dW\end{align}

Or, rather, $$f_s = f_0 + \int_0^s \left[ \lambda e^{\lambda W} + \frac{\lambda^2}{2}We^{\lambda W} \right]dt + \int_0^s \left[ e^{\lambda W} + \lambda We^{\lambda W} \right] dW.$$ Then we take expectations, so that the stochastic integral disappears, then we define $m(t) = E f_s = E f(s,W(s))$, and then we take the derivatives. We then get $$m'(t) = E \lambda e^{\lambda W} + \frac{\lambda^2}{2}m(t).$$ The first term on the left hand side equals $\lambda e^{t\frac{\lambda^2}{2}}$.

At this point, as I am not good with differential equations, I am kind of stuck. And, either way, I would like to have my solution so far checked out as well, as I am unsure whether the stochastic integral really does disappear when I take expectations. What condition needs to be satisfied for this to hold, and how do I check whether it does?

Jaood
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1 Answers1

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Since $W_t\sim\mathcal{N}(0,t)$ $$\mathbb{E}\left[\lambda e^{\lambda W_t}\right]=\lambda\mathbb{E}\left[e^{\lambda W_t}\right]=\lambda e^{\frac 12 \lambda^2 t}$$ thus $$m'(t) -\frac{\lambda^2}{2}m(t)=\lambda e^{\frac 12 \lambda^2 t}$$ This ODE is a First-order equation with variable coefficients, thus $$\mu(t)=e^{\int -\frac 12 \lambda^2dt}=e^{-\frac 12 \lambda^2 t}$$ and $$m(t)=\frac{1}{e^{-\frac 12 \lambda^2 t}}\left(\int e^{-\frac 12 \lambda^2 t}\lambda e^{\frac 12 \lambda^2 t}dt +c\right)$$ Where $c\in \mathbb{R}$. As a result $$m(t)=e^{\frac 12 \lambda^2 t}(\lambda t+c)$$ In other words $$m(t)=\lambda e^{\frac 12 \lambda^2 t} t+c e^{\frac 12 \lambda^2 t}$$ On the other hand $m(0)=0$, so $c=0$ and $$m(t)=\lambda e^{\frac 12 \lambda^2 t} t$$

Edit for Jin5

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Behrouz Maleki
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  • All right. I really should look up common DEs and make a note of their solutions. Why does the stochastic integral disappear when taking expectations though? This does not seem to be explained in the answer given in the duplicate thread, and so I am wondering if it's obvious? I've found a lemma which guarantees it, but it assumes that the expectation of the integrand squared is Riemann-integrable, and when I calculate the expectation of the integrand squared, I seemingly get something more complicated than my original problem (e.g., the term $E W^2 e^{\lambda W}$). – Jaood Dec 24 '16 at 15:00
  • http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf. – Behrouz Maleki Dec 24 '16 at 15:04
  • Also you can see it http://math.stackexchange.com/questions/232932/it%C5%8D-integral-has-expectation-zero – Behrouz Maleki Dec 24 '16 at 15:07
  • The links you give don't seem to give me more information than I already expressed in the comment above. – Jaood Dec 24 '16 at 22:59
  • In the Oksendal 's book, you can see why Ito's integral has expectation zero. – Behrouz Maleki Dec 25 '16 at 05:59
  • This does not hold in general, so no, I am not going to go through an entire book in search of a result that does not exist. – Jaood Dec 25 '16 at 13:28
  • please read section 3.2 Some properties of the It^o integral :Theorem 3.2.1. – Behrouz Maleki Dec 25 '16 at 13:31
  • That theorem imposes restrictions on the integrand. It does not hold for arbitrary integrals. Can you show that those restrictions are met in this instance? – Jaood Dec 25 '16 at 13:36
  • $$\int _{0}^{t}\mathbb{E}[W_s^2e^{2\lambda W_s}] ds<\infty$$ – Behrouz Maleki Dec 25 '16 at 13:45
  • I solved this question in other link . Look at it: http://math.stackexchange.com/questions/2068309/what-is-mathbbews-mathrmews-where-ws-is-a-standard-brownian-mo – Behrouz Maleki Dec 25 '16 at 14:10
  • Please look at other answers and compare these. – Behrouz Maleki Dec 25 '16 at 14:12
  • Even in your other answer, you do not explain why the integrand satisfies the assumptions made in Oksendal Theorem 3.2.1. You seem to be having just as trouble with this question as me.... – Jaood Dec 25 '16 at 21:56