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Suppose $X_{1}$ and $X_{2}$ are two random variables, what does $E\left ( X_{1}|min\left \{ X_{1},X_{2} \right \} \right )$ mean. I am not getting the intuition behind it. While describing conditional probability terms on the right side of vertical bar are things that we take for granted have already occurred and therefore it should be an observed value on the right side not the random variable itself but $min\left (X_{1},X_{2} \right )$ is a random variable here. Can someone please explain this to me and probably use an example to show how the value of $E\left ( X_{1}|min\left \{ X_{1},X_{2} \right \} \right )$ is calculated

redenzione11
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  • @Glen_b This notation is standard: it's in the form $E[X\mid Y]$ and represents a random variable that is [measurable with respect to the sigma field generated by $Y$.](https://stats.stackexchange.com/questions/74332) I realize that likely doesn't help the OP, but it ought to make it clear that the expression is well-defined and understandable. – whuber Oct 04 '17 at 13:41
  • So from what I understand now, since $E(X_{1}|X_{2}=x $ is a scalar quantity but $E(X_{1}|X_{2})$ is a function over all the values that $ X_{2}$ can take. TO tackle the probable in hand , let's take three cases –  redenzione11 Oct 04 '17 at 21:35
  • Suppose $X_{1}$ and $X_{2}$ are two independent and identically distributed discrete random variables. Both are uniformly distributed taking the following values {1,2,3,...,N} and we have to calculate $E(X_{1}|min\left\{X_{1},X_{2}\right\})$. So we will start by taking three cases, $X_{1}>X_{2}$, $X_{1} –  redenzione11 Oct 05 '17 at 00:20
  • For $X_{1} –  redenzione11 Oct 05 '17 at 00:29

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