# Maximum of exponential rv's and its Laplace-Stieltjes transform

Suppose we have two indendent random variables $B_1$ and $B_2$ who are exponentially distributed with mean 10, ie. $$P(B_i \leq t) = 1-e^{-\frac{1}{10}t}$$ for $i = 1,2$. I want to compute the Laplace-Stieltjes transform of $B = \max(B_1,B_2)$.

This is an exercise from a reader I have (not homework). The answer should be $$\tilde{B}(s) = \frac{72}{(s+12)(s+6)}$$ but I get a different answer. Here's what I did:

$P(B \leq t) = P(\max(B_1,B_2) \leq t) = P(B_1 \leq t, B_2 \leq t) = P(B_1 \leq t)P(B_2 \leq t)$ by independence, so $P(B \leq t) = (1 -e^{-\frac{1}{10}t})^2$

The Laplace-Stieltjes tranform of a rv $X$ is defined as $\tilde{X}(s) = \int_{0}^{\infty}e^{-st}f_{X}(t)dt$ and since $$f_{B}(t)=\frac{d}{dt}(P(B\leq t)) = \frac{1}{5}(1-e^{-\frac{1}{10}t})e^{-\frac{1}{10}t}$$ we have $$\tilde{B}(s) = \int_{0}^{\infty}e^{-st}\frac{1}{5}(1-e^{-\frac{1}{10}t})e^{-\frac{1}{10}t} dt$$ After some calculations I found $\tilde{B}(s)$ to be equal to $$\frac{1}{10s+1} - \frac{1}{5s+1} = \frac{1}{(10s+1)(5s+1)} \neq \frac{72}{(s+12)(s+6)}$$ What did I do wrong?

Edit: the exercise is stated as follows:

Customers arrive to a shop according to a Poisson process at rate 2 per hour. Arriving customers wait in order of arrival if another customer is already in service. When the service of a customer starts, two tasks are generated that are processed simultaneously. The time it takes to perform one task is exponentially distributed with mean 10 minutes, and the service time of the customer is over when both tasks have been done.;

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Maybe it is too early in the morning, but the calculation seems right. There is a typo in the first item, last displayed line, but the "answer" (middle) is what I got. My feeling is that if the correct answer is the one with $72$, the problem must be different. –  André Nicolas Jun 25 '11 at 13:56
I don't see a question in the quote of the exercise. What's the question? So far, rate of arrival of the customers hasn't been used. –  cardinal Jun 25 '11 at 14:53

The formula for $\tilde B(s)$ given by your reader is correct if the time unit is the hour (while your computations use the minute). For example, the random variables $B_1$ and $B_2$ should be such that $P(B_i\ge t)=\mathrm{e}^{-6t}$ for every nonnegative $t$, and the rest follows.