Can two normal random variables compare?

Question

$\displaystyle\int_{0}^{1} e^{as}dB_{s}\leq\int_{0}^{2} e^{as}dB_{s}$, is this true?

In my opinion,

$\displaystyle\int_{0}^{1} e^{as}dB_{s}=\int_{0}^{1} e^{as}dB_{s}+\int_{1}^{2} e^{as}dB_{s}$ and $\int_{1}^{2} e^{as}dB_{s}\sim N(0,\frac{e^{2as}-e^{as}}{as})$.

where $B_{t}$ is a Brownian motion. Thanks for all your help?

Answer 1

$\displaystyle P\bigg(\int_0^1e^{as}dB_s\le\int_0^2e^{as}dB_s\bigg)=P\bigg(\int_1^2e^{as}dB_s\ge 0\bigg)=1/2$, since $\int_1^2e^{as}dB_s\sim N(0,\sigma^2)$ and $\sigma^2>0$.