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how do I compute the expected value and variance of a stochastic process where I only know the SDE? In particular for the process $dB_t = ( 1/B_t - B_t/(T-t)) dt + dW_t$ where $W$ is a standard Brownian motion. Do I need to get $B_t$ first? Is there any other way? thanks!

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For every $n$, $\mathrm d(B^n)=nB^{n-1}\mathrm dB+\frac12n(n-1)B^{n-2}\mathrm d\langle B,B\rangle$ and $\mathrm d\langle B,B\rangle=\mathrm d\langle W,W\rangle=\mathrm dt$ hence $$ \mathrm d(B^n)=nB^{n-1}(B^{-1}-c(t)B)\mathrm dt+nB^{n-1}\mathrm dW+\tfrac12n(n-1)B^{n-2}\mathrm dt, $$ where $c(t)=(T-t)^{-1}$. Each function $M_n(t)=E[(B_t)^n]$ solves the differential equation $$ M'_n=n(M_{n-2}-cM_n)+\tfrac12n(n-1)M_{n-2}=\tfrac12n(n+1)M_{n-2}-ncM_n. $$ Thus, $M_1$ and $M_2$ can be deduced from the initial conditions $M_k=B_0^k$ and from the system $$ M'_1=M_{-1}-cM_1,\qquad M'_{-1}=cM_{-1},\qquad M'_2=3-2cM_2, $$ which yields something similar to $$ M_1=\frac{t}{B_0}+B_0\left(1-\frac{t}T\right),\qquad M_2=3t\left(1-\frac{t}T\right)+B_0^2\left(1-\frac{t}T\right)^2. $$

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