I have a question regarding the probability distribution of the difference of two random variables.
Suppose we have $n$ exponential random variables, and they are all independent and identically distributed with parameter $\lambda$. Then, suppose we define random variable $Y_i$ as the $i$-th smallest $X_k$. For example, $Y_1 = \min(X_1, X_2, \ldots, X_n)$. $Y_2$ is the second smallest $X$.
I understand the following two facts: 1) Exponential random variables are memoryless. That is, if $X$ is an exponential random variable, $P(X \geq t + t_0 | X \geq t_0) = P(X \geq t)$.
and 2) If $X_1, X_2, \ldots, X_k$ are independent, each with parameter $\lambda$, then the random variable $Y_1 = \min(X_1, X_2, \ldots, X_k)$ has the probability density function: $f(y_1) = k \lambda e^{-k \lambda y_1}$
Then, my question is: how do I get the probability density function of $Y_2 - Y_1$? $Y_2 - Y_1$ is the second smallest exponential random variable minus the smallest exponential random variable. I understand that the expected value of $(n-1)$ exponential random variables with the same parameter $\lambda$ is $\frac{1}{(n-1)\lambda}$ (since we can use fact 2 above and set the number of random variables to $k-1$), but I do not see how $E(Y_2-Y_1)$ equals that quantity.
I believe strongly that fact 1 is involved, but I do not see how $Y_2 - Y_1$ is related to $(Y_2 - Y_1 | Y_1 = y_1)$.
Thanks everyone.