I'm pretty sure, that "exponential with parameter " means that cumulative distribution function is
(with support )
Hence probability density function is
And the expected value is
Because series is convergent, we know that
Hence
I have a problem proving the following theorem...
Let be independent random variables of exponential distributions with parameters . Let . Provided prove that for every (except for zero-probability set) exists exactly one index such that . What is more and are independent, and .
To solve the problem we shall use the following theorem
Let S, T be random variables of exponential distributions with parameters and . The random variable has exponential distribution with parameter . Probability . What is more, events , are indpendent for all
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The theorem implies that for random variables random variable has exponential distribution with parameter . For this distribution converges to exponential distribution with parameter , which is also distribution of T (because converges to T almost surely).
Let where is the number for which . If such (unique!) j does not exist we put
First of all we will calculate . We can easily see that .
has exponential distribution with parameter . Thus by the theorem. What is more, we can see that and are independent (we use the theorem putting , ).
Now, it is easy to prove that J is finite for almost all . Indeed, we see that: