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Let $X_1$ and $X_2$ be independent exponentially distributed random variables with parameter $\theta > 0$. I want to compute $\mathbb E[X_1 \wedge X_2 | X_1]$, where $X_1 \wedge X_2 := \min(X_1, X_2)$.

I'm really not sure how to do this. I don't want to use any joint distribution formulas (that's a different exercise in this text). Basically all I know about conditional expectations is that $\mathbb E\left[\mathbb E[X | Y] \mathbb 1_A \right] = \mathbb E[X \mathbb 1_A]$, for any $A \in \sigma(Y)$. I thought about using this property to calculate $\mathbb E\left[(X_1 \wedge X_2) \mathbb 1_{\{X_1 \leq X_2\}}| X_1\right]$ and $\mathbb E\left[(X_1 \wedge X_2) \mathbb 1_{\{X_1 > X_2\}}| X_1\right]$ separately, but it's not clear to me that either of these sets are necessarily in $\sigma(X_1)$. Any hints?

Edit: I want to avoid using conditional probability over expectations while conditioning over zero-probability events. That's a different section of the book I'm reading out of (Achim Klenke's "Probability Theory: A Comprehensive Course").

Edit 2: I eventually found my own solution, which I've posted as an answer below.

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  • $\begingroup$ You cannot answer this without any independence assumption. $\endgroup$ – Kavi Rama Murthy Jan 25 at 5:47
  • $\begingroup$ Assuming independence, you might to try to find $\mathbb E[k \wedge X_2 ]$ for some constant $k \ge 0$ $\endgroup$ – Henry Jan 25 at 8:49
  • $\begingroup$ Ah yes, they are independent. I edited it. $\endgroup$ – D Ford Jan 25 at 16:33
  • $\begingroup$ @Henry I'm not sure how that helps if we assume independence. Is there an independence property you're invoking? $\endgroup$ – D Ford Jan 25 at 16:34
  • $\begingroup$ With independence, knowing $X_1=k$ tells you nothing new about the distribution of $X_2$, so my hint is helpful. Without independence, my suggestion would not help with the original question $\endgroup$ – Henry Jan 25 at 16:41
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$$ E[X_1\wedge X_2|X_1=s]=\int_0^\infty (t\wedge s)\hspace{-.5cm}\underbrace{\theta e^{-\theta t}}_{\text{conditional pdf}\\\text{of $X_2$ given $X_1=s$}}\hspace{-.5cm}\,dt=\int_0^st\theta e^{-t\theta}\,dt+\int_s^\infty s\theta e^{-t\theta}\,dt $$

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  • $\begingroup$ See my above comment: $\mathbb P[X_1 = s] = 0$ for each fixed $s$, and I'd prefer to avoid conditioning on zero-probability events. Plus I'm not sure how to calculate $\mathbb E[X_1 \wedge X_2 | X_1] $ from knowing $\mathbb E[X_1 \wedge X_2 | X_1 = s]$ for any given $s$. $\endgroup$ – D Ford Jan 25 at 20:16
  • $\begingroup$ @DFord The above calculation will give you the correct answer, and then you can prove it is correct by showing it satisfies $E[E[X_1\wedge X_2|X_1]1_A]=E[X_1\wedge X_2\cdot 1_A]$ for all $A\in \sigma(X_1)$. $\endgroup$ – Mike Earnest Jan 25 at 20:20
  • $\begingroup$ Also, $E[X_1\wedge X_2|X_1=s]$ works out to be $f(s)$ for some function $s$. Then $E[X_1\wedge X_2|X_1]=f(X_1)$. @DFord $\endgroup$ – Mike Earnest Jan 25 at 20:21
  • $\begingroup$ If $\mathbb E[X_1 \wedge X_2 | X_1 = s] = f(s)$, wouldn't it follow that $f(X_1) = \mathbb E[X_1 \wedge X_2 | X_1 = X_1] = \mathbb E[X_1 \wedge X_2]$? $\endgroup$ – D Ford Jan 25 at 20:33
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    $\begingroup$ @DFord It doesn't quite work like that. If $F(x)=P(X\le x)$ is the cdf of a random variable $X$, do you conclude $F(X)=P(X\le X)=1$? No, $F(X)$ is just another random variable, which takes values between $0$ and $1$. Same kind of thing here. $\endgroup$ – Mike Earnest Jan 25 at 20:57
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Here you are an other way to see it (to not condition on zero probability events).

The conditional expectation with respect to a continuous random variable $\mathbb{E}[X\mid Y]$ can be defined as (see https://en.wikipedia.org/wiki/Conditional_expectation) \begin{align} g(y)=\lim _{\epsilon \to 0}\mathbb{E} [ X| \{\omega: \|Y(\omega)-y \|\le \epsilon\} ] \end{align} Applying this to your case and letting $H_{\epsilon,y}=\{\omega: |X_1(\omega)-y |\le \epsilon\}$, we get \begin{align} g(y) & =\lim _{\epsilon \to 0}\mathbb{E} [ X_1 \vee X_2 \mid H_{\epsilon,y} ] \\ % & =\lim _{\epsilon \to 0}\frac{\mathbb{E} [ (X_1 \vee X_2) I_{H_{\epsilon,y}} ]}{P(H_{\epsilon,y} )} \\ % & =\lim _{\epsilon \to 0} \frac { % \mathbb{E} [ (X_1 \vee X_2) I_{H_{\epsilon,y}} ] \int_{\mathbb{R}_+^2} (x_1 \vee x_2) I_{\{|x_1-y |\le \epsilon\}} \theta e^{-\theta x_1} \theta e^{-\theta x_2} {\rm d }x } {\int_{y-\epsilon}^{y+\epsilon} \theta e^{-\theta z}{\rm d }z} \\ % & =\lim _{\epsilon \to 0} \frac { % \mathbb{E} [ (X_1 \vee X_2) I_{H_{\epsilon,y}} ] \int_{y-\epsilon}^{y+\epsilon} \int_{\mathbb{R}_+} (x_1 \vee x_2) \theta e^{-\theta x_1} \theta e^{-\theta x_2} {\rm d }x_2{\rm d }x_1 } { e^{-\theta(y-\epsilon)} -e^{-\theta(y+\epsilon)} } \\ % & =\lim _{\epsilon \to 0} \frac { \int_{y-\epsilon}^{y+\epsilon} \theta e^{-\theta x_1} % \left( \int_{0}^{x_1} x_2\theta e^{-\theta x_2} {\rm d }x_2 + x_1 \int_{x_1}^{\infty} \theta e^{-\theta x_2} {\rm d }x_2 \right) % {\rm d }x_1 } { e^{-\theta(y-\epsilon)} -e^{-\theta(y+\epsilon)} } \\ % & =\lim _{\epsilon \to 0} \frac { \int_{y-\epsilon}^{y+\epsilon} e^{-2\theta x_1} % \left( e^{\theta x_1} - \theta x_1 -1 + \theta x_1 \right) % {\rm d }x_1 } { e^{-\theta(y-\epsilon)} -e^{-\theta(y+\epsilon)} } \\ % & =\lim _{\epsilon \to 0} \frac{1}{\theta } \frac { \int_{y-\epsilon}^{y+\epsilon} \theta e^{-\theta x_1} {\rm d }x_1 - \int_{y-\epsilon}^{y+\epsilon} \theta e^{-2\theta x_1} {\rm d }x_1 } { e^{-\theta(y-\epsilon)} -e^{-\theta(y+\epsilon)} } \\ % & = \frac{1}{\theta }- \frac{1}{2\theta } \lim _{\epsilon \to 0} \frac { e^{-2\theta(y-\epsilon)} -e^{-2\theta(y+\epsilon)} } { e^{-\theta(y-\epsilon)} -e^{-\theta(y+\epsilon)} } \\ % % & = % \frac{1}{\theta }- % \frac{e^{-\theta y}}{2\theta } % \lim _{\epsilon \to 0} % \frac % { % e^{2\theta \epsilon} - 1 - (e^{-2\theta \epsilon} -1) % } % { % e^{ \theta \epsilon} -1 - (e^{- \theta \epsilon}-1) % } \\ & = \frac{1}{\theta }- \frac{e^{-\theta y}}{2\theta } \times \frac { 4 \theta } { 2 \theta } \\ & = \frac{1-e^{-\theta y}}{\theta } \end{align}

Hope it helps.

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  • $\begingroup$ Yes, that's the answer I got. Though I ended up finding it using image measures and densities instead of explicit limits, although I like your solution. Thanks! $\endgroup$ – D Ford Jan 27 at 20:03
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We're looking for a $\sigma(X_1)$-measurable function $\mathbb E\left[X_1 \wedge X_2 | X_1\right] : \Omega \to \mathbb R$ for which for every $A \in \sigma(X_1)$, $$ \mathbb E\left[\left(X_1 \wedge X_2\right) \mathbb 1_A\right] = \mathbb E\left[\mathbb E\left[ X_1 \wedge X_2 | X_1 \right] \mathbb 1_A \right]. $$ Let $f_\theta(x) = \theta e^{-\theta x}$ be the probability density of $X_1$ and of $X_2$, and let $\lambda_n$ denote the $n$-dimensional Lebesgue measure. Then $f_\theta \, d\lambda_1 = d\left(\mathbb P \circ X_1^{-1}\right)$. Furthermore, by independence of $X_1$ and $X_2$, the joint probability density is $\tilde f(x,y) = \theta^2 e^{-\theta(x+y)}$, and $d\left(\mathbb P \circ \left(X_1 \times X_2 \right)^{-1}\right) = \tilde f \, d\lambda_2$, where $X_1 \times X_2 : \Omega \to \mathbb R^2$ is $\omega \mapsto \left(X_1(\omega), X_2(\omega)\right)$. Let $A \in \sigma(X_1)$. Then, \begin{align*} \mathbb E\left[\left(X_1 \wedge X_2 \right)\mathbb 1_A \right] &= \int_A X_1 \wedge X_2 \, d\mathbb P = \int_{X_1(A) \times [0,\infty)} x \wedge y \, d\left(\mathbb P \circ \left(X_1 \times X_2 \right)^{-1}\right) \\ &= \int_{X_1(A)} \int_0^\infty \left(x \wedge y\right) \theta^2 e^{-\theta(x+y)} \, dy \, dx\\ &= \int_{X_1(A)} \int_0^x \theta^2 ye^{-\theta(x+y)} \, dy \, dx + \int_{X_1(A)} \int_x^\infty \theta^2 x e^{-\theta(x+y)} \, dy \, dx \\ &= \int_{X_1(A)} \left(-\theta x e^{-2\theta x} - e^{-2\theta x} + e^{-\theta x}\right) \, dx + \int_{X_1(A)} \theta x e^{-2\theta x}\,dx \\ &= \int_{X_1(A)} \left(e^{-\theta x} - e^{-2\theta x}\right)\,dx = \int_{X_1(A)} \frac 1 \theta \left(1-e^{-\theta x}\right)\theta e^{-\theta x}\,dx \\ &= \int_{X_1(A)} \frac 1 \theta \left(1-e^{-\theta x}\right) \, d\left(\mathbb P \circ X_1^{-1}\right)(x) = \int_A \frac 1 \theta \left(1-e^{-\theta X_1}\right) \, d\mathbb P \\ &= \mathbb E\left[\frac 1 \theta \left(1-e^{-\theta X_1}\right) \mathbb 1_A \right]. \end{align*} From this it follows that $\boxed{\mathbb E\left[X_1 \wedge X_2 | X_1\right] = \frac 1 \theta \left(1-e^{-\theta X_1}\right).}$

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