Is $<r_tdt,r_tdt>$ when $r_t$ stochastic?

Question

Let $D_t=e^{-\int_{0}^{t}r(s)ds}$ with $r_t$ being a stochastic process.

Set $X_t = -\int_{0}^{t}r(s)ds$

By Ito :

$$ \frac{dD_t}{D_t} = dX_t+\frac{1}{2} <dX,dX>_t $$ $$ =-r_tdt+\frac{1}{2} <r_tdt,r_tdt> $$

My question is : What is $<r_tdt,r_tdt>$ (or using another notation : $(r_tdt)^2$ ) if we want to go on with the computation? Is it equal to zero when $r_t$ is stochastic by applying the fact that $<dt,dt>=0$ regardless of $r_t$ being stochastic?

Thank you

Answer 1

Yes, it is zero.

Generally, we have \begin{align} {\rm d}X_t&=Y_t{\rm d}t&\Longrightarrow&&{\rm d}\left<X\right>_t&=0,\\ {\rm d}X_t&=Y_t{\rm d}W_t&\Longrightarrow&&{\rm d}\left<X\right>_t&=Y_t^2{\rm d}t, \end{align} where $X_t$ and $Y_t$ are general stochastic processes, while $W_t$ is the Wiener process (standard Brownian motion).

Answer 2

It's Zero indeed as hypernova stated, however it's not zero for any stochastic process, Y_t needs to be progressive (meaning Y_t needs to be measurable for the Borel tribe B([0,t]) x F_t for each t ), the proof of this statment is based on the stochastic dominated convergence theorem for which the integal needs to be defined and bounded for each t... Using your notations I suspect this is to be applied to finance, r_t being a rate it's generally admitted that we are within the mentionned Framework so it's Zero for Financial purposes...