Let $W_{1},W_{2}$ be independent, exponentially distributed random variables, $\mathbb{E}(W_{i})=\lambda$. Find $\mathbb{E}(W_{1}|\min(W_{1},W_{2})).$
I think I have the correct solution, but it's based solely on intuition.
Let $X=\min(W_{1},W_{2})$. $$\mathbb{E}(W_{1}|X)=\mathbb{E}(W_{1}|X=W_{1},W_{1}<W_{2})\mathbb{P}(W_{1}<W_{2})+\mathbb{E}(W_{1}|X=W_{2},W_{2}<W_{1})\mathbb{P}(W_{2}<W_{1})=\frac{1}{2}(X+\mathbb{E}(W_{1}|W_{1}>X))$$ And now the remaining expectation can be obtained by computing one integral and dividing it by the probability that the random variable $W_{1}$ is greater than $X$. Is this approach correct? Could someone explain, in mathematical terms, why the decomposition in the first line of the above reasoning is valid? (if it is).
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $\frac12$ and $W_1=X+W$ with probability $\frac12$, where $W$ is independent of $X$ and exponential with parameter $\lambda$. Hence, $$ \mathbb E(W_1\mid X)=\frac12X+\frac12(X+\mathbb E(W))=X+\frac1{2\lambda}. $$