I think you'll find this thread interesting. It gives you a possible choice of what should be and there are hints for a proof; this should allow you to understand how must be chosen.
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I'm not quite sure what part (a) is asking.
For part (b), I'm thinking that this has something to do with Chebyshev's inequality, but the examples for Chebyshev's inequality in my textbook and notes don't involve the use of 'max {...}'. So I'm pretty much stuck. But intuitively, I think it will be 0 when n tends to infinity, but I don't really know how to prove it.
I think you'll find this thread interesting. It gives you a possible choice of what should be and there are hints for a proof; this should allow you to understand how must be chosen.
What you did is fine. Now you are asked to choose to be any sequence such that has a limit when . There are many possible choices.
The most obvious would be or (or with an arbitrary ). You indeed get or (or ), which is the distribution function of an exponential distribution with some parameter.
What happens for other choices? Suppose for instance or any sequence with . Then you can see that the limit would be 0, which is not a distribution function.
If to the contrary (still with ), for instance , then the limit is 1, which is the distribution function of the Dirac probability measure at 0, i.e. the distribution of a r.v. that is constant, equal to 0. This would be a correct answer, but it is called a "degenerate" limit, which is not very interesting. The best choice is when is on the order of .
Of course, any sequence such that would give an exponential as the limit, not only . But you're only asked for one, so let's choose simple.