How to solve this problem? (exponential distributions) Studying for a re-take exam in a first term probability course and, among other things, I'd like to understand the solutions to the maths problems I didn't understand on my first attempt.
I couldn't seem to find help for this particular question on Google nor on this site, so I was hoping someone here could help.
Alma and Oscar make a phone call every day. It is known that the lengths of Alma's phone calls are Exponential(1)-distributed and that the lengths of Oscar's phone calls are Exponential(2)-distributed. Furthermore, the lengths of the two phone calls are independent.
Let Z denote the length of the longest of the two calls and determine the mean of Z.
I don't understand how to determine which of the calls is longest. Or perhaps I've fundamentally misunderstood the question. If a kind soul could at least point me in the right direction, I'd very much appreciate that. ^^
Apologies if I made any mistakes in this post. It's my first time asking a question here.
 A: Welcome to Math StackExchange!
Lets denote $X \sim \text{Exp}(1)$ the random variable that models the lenghts of Alma's calls. Similarly $Y \sim \text{Exp}(2)$. Briefly, this means that given a number $t \in [0, \infty)$, you can figure the probability that their calls last $t$ (lets say seconds). For example if someone asked you what is the probability that Oscar lasts $5$ seconds in call or less then you would have to calculate:
$$
\mathbb{P}( X \leq 5) = 1-e^{-5}
$$
This is because the cumulative distribution function of exponential distributions is of the form, if $K \sim \text{Exp}(\lambda)$ then
$$ 
\mathbb{P} ( K \leq t) = 1-e^{-\lambda x}
$$
Now we know the basics of the exponential distribution we can try to figure out the distribution of the maximum of two of such random variables. So let $M$ denote the random variable that models the maximum of $X$ and $Y$. That is:
$$
M = \max(X,Y)
$$
We are looking for the average of $M$, but first lets figure out $M$'s C.D.F:
$$
\mathbb{P}(M \leq t) = \mathbb{P}(X \leq t, Y \leq t) = \mathbb{P}(X \leq t)\mathbb{P}(Y \leq t) = (1-e^{-x})(1-e^{-2x})
$$
Now we have the C.D.F of the random variable $M$. In order to find the expected value $\mathbb{E}(M)$ you can differentiate $(1-e^{-x})(1-e^{-2x})$ with respect to $x$ and then find the integral
$$
\int_0^\infty x \left[ \frac{d}{dx}(1-e^{-x})(1-e^{-2x}) \right] dx
$$
This is just differentiating $M$'s CDF, which gives you $M$'s probability density function and then integrating it multiplied by $x$.
