What are $\mathbb{E}[X|X \wedge t]$ and $\mathbb{E}[X|X \vee t]$?

Question

Here $X$ is a random variable with mean $\frac{1}{\lambda}$ exponential distribution, and $t$ is a fixed nonnegative number. A similar question has been asked here, I hope to get an answer where each step is rigorous and explained in more details.

I have tried to find the joint distribution of $X$ and $X \wedge t$, however it is neither discrete nor absolutely continuous, hence I can't use the elementary method where you find $f(x|y)$ and use $\int_{\mathbb{R}}xf(x|Y)dx$ as a version of the conditional mean.

Answer 1

Let $X$ be any random variable with finite mean. To find $\mathbf{E}[X \mid X \wedge t]$, note that we may write

$$ \mathbf{E}[X \mid X \wedge t] = h(X \wedge t) $$

for some Borel-measurable function $h$, which is a solution of the functional equation of the form

$$ \bbox[color:navy; padding: 10px; background: #F0F8FF;]{ \mathbf{E}[X \psi(X \wedge t)] = \mathbf{E}[h(X \wedge t)\psi(X \wedge t)] } \tag{1} $$

for any bounded Borel-measurable test function $\psi$. However,

\begin{align*} \bbox[color:navy; padding: 3px; background: #F0F8FF;]{\text{LHS of (1)}} &= \mathbf{E}[X \psi(X) \mathbf{1}_{\{X < t\}}] + \mathbf{E}[X \mathbf{1}_{\{X \geq t\}}]\psi(t) \\ &= \mathbf{E}[X \psi(X) \mathbf{1}_{\{X < t\}}] + \mathbf{E}[X \mid X \geq t]\psi(t)\mathbf{P}(X \geq t) \end{align*}

and

\begin{align*} \bbox[color:navy; padding: 3px; background: #F0F8FF;]{\text{RHS of (1)}} &= \mathbf{E}[h(X) \psi(X) \mathbf{1}_{\{X < t\}}] + h(t)\psi(t)\mathbf{P}(X \geq t). \end{align*}

Plugging these into $\text{(1)}$ and comparing both sides, we find a solution $h$ of the form

$$ h(x) = x \mathbf{1}_{\{x < t\}} + \mathbf{E}[X \mid X \geq t] \mathbf{1}_{\{x = t\}} $$

and hence

$$ \bbox[color: #4B0000; padding: 10px; background: #FFF8F0; border: 1px dotted #8B0000;]{ \begin{aligned} \mathbf{E}[X \mid X \wedge t] &= h(X \wedge t) \\ &= X \mathbf{1}_{\{X < t\}} + \mathbf{E}[X \mid X \geq t] \mathbf{1}_{\{X \geq t\}} \end{aligned} } $$

holds almost surely.

Finally, in OP's case, all we have to do is compute $\mathbf{E}[X \mid X > t]$, which can be easily done by invoking the memoryless property of the exponential distribution.

A similar computation can be done to find $\mathbf{E}[X \mid X \vee t]$.

Answer 2

Let $Y:=X\wedge t$ and note that for every Borel set $B\subset \mathbb{R}$ we have that $B=(B\cap [0,t))\cup (B\cap [t,\infty ))$, therefore $$ \{Y\in B\}=\{Y\in B\cap [0,t)\}\cup \{Y\in B\cap [t,\infty )\}=\{X\in B\cap [0,t)\}\cup \{X\geqslant t\} $$ when $t\in B$. Then from the definition of conditional expectation we find that $$ \int_{\{Y\in B\}}\mathbb{E}[X|Y]\,d P=\int_{\{X\in B\cap [0,t)\}}X\,d P+\int_{\{X\geqslant t\}}X\,d P\\ =\int_{B\cap [0,t)}t f_X(t)\,d t+\int_{[t,\infty )}t f_X(t)\,d t\tag1 $$

when $t\in B$. If $t\notin B$ then the above just reduces to the first integral. From (1) its now easy to find the distribution of $\mathbb{E}[X|Y]$. Also from above you can deduce that $$ \mathbb{E}[X|Y](\omega )=X(\omega )\mathbf{1}_{\{X<t\}}(\omega )+c\mathbf{1}_{\{X\geqslant t\}}(\omega )\quad P-\text{a.s. } $$ where $c=\frac1{P[X\geqslant t]}\int_{\{X\geqslant t\}}X\,d P$.

Answer 3

Another way to look at it is to see that the $\sigma $-algebra in $R$ associated with $Y=X\wedge t$ is generated by the intervals that contain the interval $[t,\infty)$, so: $$P[X\in A\mid Y]=\left\lbrace\begin{array}{lcc} I_A(Y)&if&Y<t\\ P[X\in A\mid X\geq t]&if&Y=t\\ indifferent&if&Y> t \end{array}\right.$$ therefore: $$E[X\mid Y]=\left\lbrace\begin{array}{lcc} Y&if&Y<t\\ E[X\mid X\geq t]&if&Y=t\\ indifferent&if&Y> t \end{array}\right.$$