The conditional expectation $E(f(X) \mid \min(X,t))$ if $X \sim Exp(\lambda)$

Question

Let $X \sim Exp(\lambda)$, $t \ge 0$ and $Y_t := \min(X,t)$. Let further be $f$ some measurable function with $\mathbb{E}(\lvert f(X) \rvert) < \infty$. Does it hold that:

$$\mathbb{E}(f(X) \mid Y_t) = f(Y_t)I_{[0,t)}(Y_t) + \mathbb{E}_{\{Y_t = t\}}(f(X))I_{\{t\}}(Y_t) \quad?$$

I recognise that the intent here is to split up according to cases:

1. The left summand covers the case that $X<t$, but I do not understand why there is $f(Y_t)$ instead of $\mathbb{E}(f(Y_t))$.
2. The right summand covers the case that $X \ge t$.

So all in all I get the feeling that this identity is true, however, I can not quite see how to argue that the left summand is correct.

For context; here is the definition of conditional expectation from my lecture:

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and let $X$ be a real RV with $E(\lvert X \rvert) < \infty$. $\mathbb{E}(X \mid \mathcal{F})$ is uniquely defined as a RV via the conditions

1. $\mathbb{E}(X \mid \mathcal{F})$ is $\mathcal{F}$ measurable,

2. $\mathbb{E}(X \mid \mathcal{F}) \in L^1(\mathbb{P})$

3. $\int_A \mathbb{E}(X \mid \mathcal{F}) d \mathbb{P} = \int_A X d \mathbb{P}$ for all $A \in \mathcal{F}$.

Answer 1

Note that $f(X)=f(X)\mathbf{1}_{\{X<t\}}+f(X)\mathbf{1}_{\{X\geqslant t\}}$ and

$$ \operatorname{E}[f(X)\mathbf{1}_{\{X<t\}}|Y_t]=f(X)\mathbf{1}_{\{X<t\}} $$

as $f(X)\mathbf{1}_{\{X<t\}}=f(Y_t)\mathbf{1}_{\{Y_t<t\}}$ is $\sigma (Y_t)$ measurable. By the other hand, for every Borel set $A\subset \mathbb{R}$ set $A_{\geqslant t}:=A\cap [t,\infty )$ and note that

$$ \{Y_t\in A_{\geqslant t}\}=\{Y_t\in A\}\cap \{Y_t=t\}=\{Y_t\in A\}\cap \{X\geqslant t\} $$

Then for any chosen Borel set $A$ we find that

$$ \begin{align*} \int_{\{Y_t\in A\}}\operatorname{E}[f(X)\mathbf{1}_{\{X\geqslant t\}}|Y_t]\,d P&=\int_{\{Y_t\in A\}}\mathbf{1}_{\{X\geqslant t\}}\operatorname{E}[f(X)|Y_t]\,d P\\ &=\int_{\{Y_t\in A_{\geqslant t}\}}\operatorname{E}[f(X)|Y_t]\,d P\\ &=\int_{\{Y_t\in A_{\geqslant t}\}}f(X)\,d P \end{align*} $$

as the function $\mathbf{1}_{\{X\geqslant t\}}=\mathbf{1}_{\{Y_t=t\}}$ is $\sigma (Y_t)$-measurable and $\{Y_t\in A_{\geqslant t}\}\in \sigma (Y_t)$. Finally note that $$ \{Y_t\in A_{\geqslant t}\}=\begin{cases} \{X\geqslant t\},&\text{ when }t\in A\\ \emptyset ,&\text{ otherwise } \end{cases} $$ Therefore $$ \operatorname{E}[f(X)\mathbf{1}_{\{X\geqslant t\}}|Y_t]=\mathbf{1}_{\{Y_t=t\}}\operatorname{E}[f(X)|X\geqslant t] $$ ∎

Answer 2

Maybe this will help:

Let $s<t$. Then $\mathbb{E}(f(X)\mid Y_t=s) = f(s)$ (do you understand why?). If $s=t$ then $\mathbb{E}(f(X)\mid Y_t=s) = \mathbb{E}(f(X)\mid X>t) = \mathbb{E}f(X)1[X>t]$. If $s>t$ then conditioning by $Y>t$ does not make sense since $Y\overset{a.s.}{\leq}t$. All together,

$$\mathbb{E}(f(X)\mid Y_t=s) = f(s)1[s<t] + \mathbb{E}f(X)1[X>t]1[s=t].$$

Now, look at your notes and you will find somewhere that $$ \mathbb{E}(f(X)\mid Y) = g(Y) $$ for appropriate function $g$. But $g(s) = f(s)1[s<t] + \mathbb{E}f(X)1[X>t]$ Hence, $\mathbb{E}(f(X)\mid Y) = f(Y)1[Y_t<t] + \mathbb{E}f(X)1[Y=t]$