Conditional probability involving two random times, where only the distribution of one of them is used.

Question

Consider two random times $\tau_{1}$ and $\tau_{2}$ defined on a common probability space $\left(\Omega, \mathcal{G}, \mathbb{Q}\right)$ with $\mathbb{Q}(\tau_{k}=0)=0$ and $\mathbb{Q}(\tau_{k}>t)>0$ for $k=1,2$ and all $t >0$. We also include a reference filtration, $\mathbb{F}=\left\{ \mathcal{F}_{t}\right\}_{t\geq0}$, say, on the probability space mentioned above.

For the random time $\tau_{1}$, its cumulative distribution function is denoted by $F_{\tau_{1}}(t) \left(=\mathbb{Q}(\tau_{1}\leq t) \right)$.

I would like to formally prove (using measure-theoretic arguments) the following equality:

$$ \mathbb{E}_{\mathbb{Q}} \left[ \mathbb{Q}\left\{ \tau_{1} < \tau_{2} , \tau_{1} \leq T \mid \mathcal{F}_{t}\right\} \right] = \mathbb{E}_{\mathbb{Q}} \left[ \int_{0}^{T} \mathbb{Q}\left\{ u < \tau_{2} \mid \mathcal{F}_{t}\right\} dF_{\tau_{1}}(u.) \right]$$

Remark: Intuitively this equality seems to make sense, as you "freeze" the random time $\tau_{2}$, and "integrate over the values of $\tau_{1}$". I would like to formalise this using theorems/lemmas, etc. of measure/probability-theory.

Thanks in advance.

Answer 1

After you edited the question you can use the tower law to write the LHS more simply as \begin{align} \mathbb Q\big\{\tau_1<\tau_2,\tau_1\le T\big\}\,. \end{align} Assuming that the joint distribution function $F_{\tau_1,\tau_2}$ has a density $f_{\tau_1,\tau_2}$ we can write $$ f_{\tau_1,\tau_2}(u,v)=\frac{f_{\tau_1,\tau_2}(u,v)}{f_{\tau_1}(u)}f_{\tau_1}(u) =:f_{\tau_2\color{red}{|u}}(v)f_{\tau_1}(u)\,. $$ Where $f_{\tau_2\color{red}{|u}}(v)$ is the conditional density of $\tau_2$ given $\tau_1=u\,.$ Then, \begin{align} &\mathbb Q\big\{\tau_1<\tau_2,\tau_1\le T\big\}=\int_0^T\int_u^Tf_{\tau_2\color{red}{|u}}(v)f_{\tau_1}(u)\,dv\,du=\int_0^T\left\{\int_u^Tf_{\tau_2\color{red}{|u}}(v)\,dv\right\}\,f_{\tau_1}(u)\,du\\ &=\int_0^T\mathbb Q\big\{u\le\tau_2\,\big|\,\tau_1=u\big\}\,dF_{\tau_1}(u)\,. \end{align} Fixing a typo in the upper integration limit of your RHS we can pull $\mathbb E$ into the integral and use the tower law which gives $$ \int_0^T\mathbb Q\big\{u\le \tau_2\big\}\,\,dF_{\tau_1}(u)\,. $$ This is wrong unless $\tau_1$ and $\tau_2$ are independent.