Ito Integral of exponential function

Question

I start studying stochastic calculus by myself. There are many concepts confusing to me. Could someone explain this integral below? $$ \int_0^\infty e^{-t}dB_t $$ Does anyone know how to derive the distribution for this integral? And how to work on the quadratic variation? Thank you very much.

Answer 1

Presumably $B_t$ is a Brownian motion. For a finite upper bound we know that $M_T=\int_0^Te^{-t}\,dB_t$ is a martingale and that its quadratic variation is $\langle M\rangle_T=\int_0^Te^{-2t}\,dt=\frac{1-e^{-2T}}{2}\,.$ The inverse of this function is $$ T_t=-\frac{\log(1-2t)}{2}\,,\quad t\in[0,\textstyle\frac{1}{2})\,. $$ The time changed martingale $[0,\textstyle\frac{1}{2})\ni t\mapsto M_{T_t}$ has quadratic variation $\langle M\rangle_{T_t}=t$ and is therefore a Brownian motion. Let's call this $W_t\,.$ Because $\lim\limits_{t\to\frac{1}{2}}T_t=\infty$ we see that $$ M_\infty=W_{\frac{1}{2}}\,. $$ In particular $$ \int_0^\infty e^{-t}\,dB_t $$ esists as the almost sure limit of $\int_0^Te^{-t}\,dB_t$ and has normal distribution with mean zero and variance $\frac{1}{2}\,.$