TITLE: The amount of time using exponential distribution?
QUESTION [0 upvotes]: I have no idea how to approach this kind of questions...
There is train A which arrives according to an Exponential distribution with parameter $\lambda$, and train B which arrives according to an Exponential distribution with parameter $\mu$.

Solve for the distribution of T, the amount of time a person will wait before either A or B arrives.
John wanted to take train A to go to the party, but he realized that he would arrive too early. So he decides that he will wait for the second arrival of train A. Find the distribution of
Z = $X_1 + X_2$, the amount of time that John will wait to take the train.

REPLY [0 votes]: The time waiting before either A or B arrives is the minimum of the two waiting times for each train. If $X_{1}$ and $X_{2}$ are exponentially distributed, then $T = \min\{X_{1},X_{2}\}$ is also exponentially distributed. 
If you get stuck on showing this, check out this link How to Prove that the minimum of two exponential random variables is another
The distribution of the sum of exponentials can be found using convolutions
$$f_{Z}(z) = \int_{0}^{z}f_{X_{1}}(s)f_{X_{2}}(z-s)ds$$
Then plug in the density function for the exponential and evaluate the integral. (I am not sure whether you are defining parameter $\lambda$ as $f(x) = \lambda e^{-\lambda x}$ or $f(x) = \frac{1}{\lambda}e^{-x/\lambda}$
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