Rigorous explanation for a conditional expectation

Question

I am looking to answer in fundamental rigour the following, but I'm unsure how to explain a crucial step:

Let $N^{(1)}(t), N^{(2)}(t)$ be independent Poisson processes with respective rates $\lambda_1,\lambda_2$. Let $T=\inf\{t: N^{(1)}(t)>0\}$ and $X=N^{(2)}(T)$. What is $\mathbb{E}[X|\sigma(T)]?$

My idea is this.

See that $T\sim \text{Exp}(\lambda_1)$, $\sigma(T)=\sigma(\{T\geqslant t\}:t\geqslant 0)$, and that $X\mathbb{1}_{\{T=t\}}\sim \text{Po}(\lambda_2t)$, where $\mathbb{1}_\cdot$ is an indicator function.

Then

\begin{align} \mathbb{E}[X\mathbb{1}_{\{T\geqslant t\}}]& = \sum_{x=0}^\infty x \mathbb{P}(X\mathbb{1}_{\{T\geqslant t\}}=x) \\ & =\sum_{x=0}^\infty x \int_t^\infty \mathbb{P}(X\mathbb{1}_{\{T= s\}}=x) \lambda_1 e^{-\lambda_1 s}ds \end{align}

and after an application of Fubini and using $$ \sum_{x=0}^\infty x\mathbb{P}(X\mathbb{1}_{\{T= s\}}=x)= \mathbb{E}[X\mathbb{1}_{\{T=s\}}]=\lambda_2 s $$ the result follows straightforwardly to show that $\lambda_2 T$ is the answer almost surely.

However I am not sure if it is correct to say $$ \mathbb{P}(X\mathbb{1}_{\{T\geqslant t\}}=x)= \int_t^\infty \mathbb{P}(X\mathbb{1}_{\{T= s\}}=x) \lambda_1 e^{-\lambda_1 s}ds $$ or how I could explain it rigorously. I see the issue that $\{T=t\}$ is null so technically it doesn't work.

Any input would be great!

Answer 1

Write \begin{align} N^{(i)}(t)&=\sum_{k=1}^\infty 1_{\{\tau^{(i)}_k\le t\}}\,,\\ N^{(2)}(T)&=N^{(2)}(\tau^{(1)}_1)=\sum_{k=1}^\infty 1_{\{\tau^{(2)}_k\le \tau^{(1)}_1\}}\, \end{align} where $\tau^{(i)}$ are the jump times of $N^{(i)}\,.$

Then use the following standard

Lemma. When $U,V$ are two independent random variables and $f:\mathbb R\times\mathbb R\to\mathbb R_+$ a Borel function then $$ \mathbb E[f(U,V)|\sigma(V)]=\mathbb E[f(U,v)]\Big|_{v=V}\,. $$ This lemma is shown by assuming first that $f$ is of the form $f(u,v)=g(u)h(v)$ (for which it is trivial) and then using a monotone class theorem to show it for general $f\ge 0\,.$

In your case $k$ is fixed but arbitrary and $U=\tau^{(2)}_k$ and $V=T=\tau^{(1)}_1$ and $$f(U,V)=1_{\{\tau^{(2)}_k\le \tau^{(1)}_1\}}\,.$$ This gives $$ \mathbb E\Big[1_{\{\tau^{(2)}_k\le T\}}\,\Big|\,\sigma(T)\Big]=\mathbb P\Big\{\tau^{(2)}_k\le t\Big\}\Big|_{t=T}\,. $$ Therefore, $$ \mathbb E\Big[N^{(2)}(T)\,\Big|\,\sigma(T)\Big]=\mathbb E\Big[\sum_{k=1}^\infty 1_{\{\tau^{(2)}\le t\}}\,\Big]\Big|_{t=T}=\mathbb E\Big[N^{(2)}(t)\Big]\,\Big|_{t=T}=\lambda_2T\,. $$