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$X,Y$ are independent exponential RV with parameter $\lambda,\mu$. How to calculate

$$ E[\min(X,Y) \mid X>Y+c] $$

Michael Hardy
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hxhxhx88
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    Two related questions (in the last two days!) : http://math.stackexchange.com/questions/261456/independence-between-maximum-and-minimum-of-exponential/261513 http://math.stackexchange.com/questions/259591/using-the-memoryless-property-to-explain-the-expected-value-of-the-maximum-of-ii/259698 – leonbloy Dec 18 '12 at 16:15
  • Do you intend the parameters to be the rates, so that the density is $\lambda e^{-\lambda x}$ on $(0,\infty)$, or the expectations, so that the density is $(1/\lambda)e^{-x/\lambda}$ on $(0,\infty)$? Both conventions are sometimes used. – Michael Hardy Dec 18 '12 at 19:35
  • In TeX you should write \min, not \text{min}. Then something like f\min g has proper spacing before and after $\min$, thus $f\min g$. Also, when in a "displayed" rather than "inline" setting, something like \min_{x\in A} looks like this: $\displaystyle\min_{x\in A}$. – Michael Hardy Dec 18 '12 at 19:38
  • @MichaelHardy, the parameters is the rates and thank you for advise in latex~ – hxhxhx88 Dec 19 '12 at 02:36

2 Answers2

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The result is independent on $c\geqslant0$.

To see this, recall that, for every measurable function $u$, $$ \mathbb E(u(Y);X\gt Y+c)=\int_0^{+\infty} u(y)\mathbb P(X\gt y+c)\mathrm d\mathbb P_Y(y)=\mathbb E(u(Y)\mathrm e^{-\lambda(Y+c)}). $$ Using this for $u:y\mapsto y$ and for $u:y\mapsto1$ and canceling the common factor $\mathrm e^{-\lambda c}$ yields $$ \mathbb E(Y\mid X\gt Y+c)=\frac{\mathbb E(Y\mathrm e^{-\lambda Y})}{\mathbb E(\mathrm e^{-\lambda Y})} $$ The denominator is $$ \mathbb E(\mathrm e^{-\lambda Y})=\int_0^{+\infty}\mathrm e^{-\lambda y}\mu\mathrm e^{-\mu y}\mathrm dy=\frac{\mu}{\mu+\lambda}. $$ The numerator is minus the derivative of the denominator with respect to $\lambda$ hence the ratio is $$ \mathbb E(Y\mid X\gt Y+c)=\frac{\mu/(\mu+\lambda)^2}{\mu/(\mu+\lambda)}=\frac1{\mu+\lambda}. $$

Did
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  • I wonder what the semicolon in $E[A;B]$ means? And why the result is independent on $c\geq 0$? If $c<0$, $\min(X,Y)$ is not necessarily $Y$, right? – hxhxhx88 Dec 19 '12 at 02:43
  • There is no $E(A;B)$ in the picture. If $X$ is a random variable and $A$ an event, $E(X;A)$ is the expectation of $X$ restricted to $A$, that is, $E(X\mathbf 1_A)$. // The result is independent on the value of $c$ as long as $c\geqslant0$. The post says nothing about the case $c\lt0$. – Did Dec 19 '12 at 08:40
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Assuming $c \ge 0$ and so $Y \lt X$, this is equivalent to $$E[Y\mid X\gt Y+c]$$ $$= \frac{\int_{y=0}^{\infty} y \, \mu \exp(-\mu y)\, \int_{x=y+c}^\infty \lambda \exp(- \lambda x)\, dx \, dy}{\int_{y=0}^{\infty} \mu \exp(-\mu y)\, \int_{x=y+c}^\infty \lambda \exp(- \lambda x)\, dx \, dy}$$

Michael Hardy
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Henry
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