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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.

zlbi
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  • the title is not about the joint distribution of $X$ and $X\wedge t$, what is the relation of $\mathbb{E}[X|X\wedge t]$ with the content of the question? – Masacroso Mar 12 '23 at 02:03
  • I tried to use the joint distribution to find the conditional expectation but failed. Since if the joint distribution is discrete or absolutely continuous, then a version of the conditional mean is $\int_{\mathbb{R}} xf(x|Y)dx$. Unfortunately, the distribution is neither discrete nor absolutely continuous. – zlbi Mar 12 '23 at 02:20

3 Answers3

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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 $\mathbf{E}[X \mid X \vee t]$.

Sangchul Lee
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  • beautiful way to see the question!!! – Masacroso Mar 12 '23 at 06:06
  • Thank you for your answer. I have two questions. One is can equation (1) be used as an equivalent definition for $\mathbb{E}[X|Y]$?; the other is how to find $h$, it seems a bit difficult to come up with its form just from the functional equation. – zlbi Mar 12 '23 at 08:55
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    @zlbi, (1) Yes, the equality is equivalent to the definition for $\mathbf{E}[X\mid Y]$. Note that, if we restrict the class of test functions to indicator functions, we retrieve the usual definition of $\mathbf{E}[X\mid Y]$. – Sangchul Lee Mar 13 '23 at 13:58
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    (2) Although it is possible to read out $h$ in an honest way using our ability of pattern matching, it is perhaps better to start from the naïve idea that we generally expect $$\mathbf{E}X\mid\mathcal{A}\quad\text{“}=\text{”}\quad\text{[average of $X$ over the "smallest event" in $\mathcal{A}$ containing $\omega$]}$$ to hold. In particular, it is reasonable to guess that $$\mathbf{E}[X\mid f(X)]=h(f(X))$$ holds, where $h$ is given by $$h(y)=\mathbf{E}[X\mid f(X)=y]=\mathbf{E}[X\mid X\in f^{-1}({y})].$$ This guess is indeed what we confirmed to be true in the answers. – Sangchul Lee Mar 13 '23 at 14:00
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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 $c=\frac1{P[X\geqslant t]}\int_{\{X\geqslant t\}}X\,d P$.

Masacroso
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  • Hi, thanks for your answer. What does $dtdP$ mean, also why does $t$ in $B$ or not matter? – zlbi Mar 12 '23 at 08:57
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    sorry, $dtdP$ was a typo, its corrected now. If $t\notin B$ then ${Y\in B\cap [t,\infty )}=\emptyset $ as the support of $Y$ is $[0,t]$, and if $t\in B$ then ${Y\in B\cap [t,\infty )}={Y=t}={X\geqslant t}$ – Masacroso Mar 12 '23 at 09:29
  • thank you for the clarification – zlbi Mar 15 '23 at 00:04
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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: $$E[X\mid Y]=\left\lbrace\begin{array}{lcc} Y&if&Y<t\\ E[X\mid X\geq t]&if&Y=t\\ indifferent&if&Y> t \end{array}\right.$$

Speltzu
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