variance of $W_te^{W_t}$

Question

I wanted to compute $\mathrm{var}[W_te^{W_t}]$. I had no problem computing the mean, but I'm not able to do the same with the mean of the squared variable, basically the trick of putting $\mathrm{E}(W_te^{W_t})=f(t)$ (and solving the differential equation) doesn't work, I got stuck here

$\mathrm{E}[W^2_te^{2W_t}]=\int{e^{2t}dt}+4\int{\mathrm{E}[W_te^{2W_t}]dt}+2\int{\mathrm{E}[W_t^2e^{2W_t}]dt}$

(the problem is the second term on the right, it would become $f(t)/W_t$ if I put $\mathrm{E}[W_t^2e^{2W_t}]=f(t)$)

Any advice?

Answer 1

I would start by using Ito's formula on $f(x)=x^2 e^{2x}$. Then take the expectation, the $dWt$ disappears and you're left with the $dt$ terms.

Answer 2

The "differential" approach you describe seems unnecessarily complicated (and if I has to solve this, I would definitely go for the direct computation advocated by @sos440) but this approach can be made to work. Consider the functions $$F(w)=w^2\mathrm e^{2w},\qquad G(w)=\mathrm e^{2w},\qquad H(w)=w\mathrm e^{2w},$$ then one is interested in $$f(t)=E(W_t^2\mathrm e^{2W_t})=E(F(W_t)).$$ Note that $$F''=2(G+4H+2F),\qquad G''=2(2G),\qquad H''=2(2G+2H), $$ hence, considering $$g(t)=E(G(W_t)),\qquad h(t)=E(H(W_t)),$$ Itô's formula applied three times yields $$ f'=g+4h+2f,\qquad g'=2g,\qquad h'=2g+2h. $$ To solve this system, consider $$ \bar f(t)=\mathrm e^{-2t}f(t),\qquad\bar g(t)=\mathrm e^{-2t}g(t),\qquad\bar h(t)=\mathrm e^{-2t}h(t), $$ then $$ \bar f'=\bar g+4\bar h,\qquad \bar g'=0,\qquad \bar h'=2\bar g. $$ Since $f(0)=\bar f(0)=h(0)=\bar h(0)=0$ and $g(0)=\bar g(0)=1$, one gets successively $\bar g(t)=1$, $\bar h(t)=2t$, and $\bar f(t)=t+4t^2$, that is, finally, $$ E(W_t^2\mathrm e^{2W_t})=(t+4t^2)\mathrm e^{2t}. $$