Misunderstanding conditional expectation Suppose $X$ ~ Exp($\lambda$) and $Y$ ~ Exp($\mu$). I want to find $\mathbb{E}(\rm{min} \{X,Y\})$. I already know that the minimum is an exponential with rate $\frac{1}{\lambda + \mu}$, but I can't figure out what is going wrong when I try to use conditional expectation because I don't get the correct answer which is $\frac{1}{\lambda + \mu}$.
Attempt:  $\mathbb{E}(\rm{min}\ \{X,Y\}) = \mathbb{E}(\rm{min}\ \{X,Y\}|X>Y)\mathbb{P}(X>Y)+\mathbb{E}(\rm{min}\ \{X,Y\}|X>Y)\mathbb{P}(Y>X)$. At this point, I want to say that in the first case the expectation is just $\frac{1}{\mu}$ since we know that $X>Y$ and in the second case the expectation is just $\frac{1}{\lambda}$ since $Y>X$, but this isn't correct because the final answer comes out to $\frac{2}{\lambda + \mu}$. I talked to someone and they said that it's because $\mathbb{E}(\rm{min}\ \{X,Y\}|X>Y)$ is not just $\mathbb{E}(Y)$. However, to me I can't reconcile why this is wrong based on an example on Wikipedia. The example is basically that if you get a light bulb that comes from two different factory and light bulbs from factory 1 lasts on average $T_1$ amount of time and light bulb from the other last for $T_2$ amount of time, then the expected lifetime of a light bulb given that it came from the first factory is just $T_1$. How is this different from saying that $\mathbb{E}(\rm{min}\ \{X,Y\}|X>Y)=\mathbb{E}(Y)$ since we know the minimum in this case is just $Y$?
 A: The mistake in your train of thought is that you essentially jumped a step! Indeed your intuitive statement that "we know that the minimum is just Y" holds true, but this leads to the following:
$$\mathbb E[ \min\{X,Y\}\vert X>Y] = \mathbb E[Y \vert X>Y]$$
The problem now is that $\mathbb E[Y \vert X>Y]$ is not equal to $\mathbb E[Y]$. This equality would hold e.q. if $Y$ was independent of $\mathbf{1}_{\{X>Y\}}$, but this is not the case! 
A: To claim that $$\mathbf{E}(Y\vert X>Y) = \mathbf{E}(Y)$$ means that it does not matter that $Y$ is less than $X$ when you want to compute $\mathbf{E}(Y)\ .$ If you know beforehand that $Y$ will always be below $X$, then it is true that $\mathbf{E}(Y\vert X>Y) = \mathbf{E}(Y)\ ;$ knowing that $X$ is an upper bound for $Y$ does not effect $Y$ in any way. 
You have not mentioned whether $X$ and $Y$ are independent random variables, but in the following computation I will assume this to be the case. 
Tail probabilities of the random variable $\min(X,Y)$ are equal to tail probabilities of an exponential distributed random variable, as the following computation shows. 
$$\mathbf{P}(\min(X,Y)>m) = \mathbf{P}(X>m \text{ and } Y>m) = \mathbf{P}(X>m) \cdot \mathbf{P}(Y>m) = e^{-\lambda m}\cdot e^{-\mu m} = e^{-(\lambda+\mu)m} \ .$$ This computation shows that the random variable $\min(X,Y)$ is exponential distributed with intensity $(\lambda + \mu)\ ;$ hence its expected value is $$\mathbf{E}(\min(X,Y)) = \frac{1}{\lambda+\mu}\ \ .$$
