I have the following process: $$ Z_t=\frac{1}{t}\Big(\frac{X^2_t}{\sigma^2_X}-\frac{Y^2_t}{\sigma^2_Y})+2 \Big(\frac{\mu_X}{\sigma^2_X}X_t-\frac{\mu_Y}{\sigma^2_Y}Y_t\Big)+\Big(\frac{\mu^2_X}{\sigma^2_X}-\frac{\mu^2_Y}{\sigma^2_Y}\Big)t $$ where the processes $X_t$ and $Y_t$ are given by $$ dX_t=\mu_X dt+\sigma_X dB_t $$ $$ dY_t=\mu_Y dt+\sigma_Y dB_t $$ where $B_t$ is a standard Brownian motion and the $\mu$'s and $\sigma$'s are constants. Is there a way to define $Z_0$? If I can show that $Z_t$ converges in probability toward a random variable $\hat Z_0$, can I define $Z_0=\hat Z_0$?
As suggested in the comment, I applied Ito's lemma to the first term and got: $$ d\Big(\frac{1}{t}\Big(\frac{X^2_t}{\sigma^2_X}-\frac{Y^2_t}{\sigma^2_Y}\Big)\Big)=\Bigg(-\frac{1}{t^2}\Big(\frac{X^2_t}{\sigma^2_X}-\frac{Y^2_t}{\sigma^2_Y}\Big)+\frac{2}{t}\Big(\frac{\mu_X}{\sigma^2_X}X_t-\frac{\mu_Y}{\sigma^2_Y}Y_t\Big)\Bigg)dt+\frac{2}{t}\Big( \frac{X_t}{\sigma_X}-\frac{Y_t}{\sigma_Y}\Big)dB_t $$ It seems that neither $dt$ nor $dB_t$ coefficients of the above process go to zero as $t$ approaches zero (since their means do not go to zero).