Stochastic Process Help, stuck on question

So i've worked through most my questions and then am stuck on the one below, any advice or help would be very much appreciated as the notes and tutorials I have been using are quite limited and hard to transfer the methods over as the question is confusing.

At a discus throwing competition two athletes (1 and 2) are competing against each

other. Each is given just one attempt to throw a discus. Let us assume that the

achieved distances of a throw (**L1 and L2**) are exponentially distributed random variables

and the performance of each athlete is not affected in any way by the performance

of his competitor. The average throwing distances for athletes 1 and 2 are L1 and L2

respectively. Find

(a) Explicit expressions for the probabilities P(Li = l) := pi(l) for each athlete to

throw a discus at distance l

(b) The probability P(**L1 - L2** = L) := P(L) that the result of the athlete 1 is

dierent from the result of the athlete 2 by L

(c) The probability that athlete 1 will beat athlete 2

(d) The probability that the world record 'lo' will not be beaten at this competition

(e) The probability that a winner of the event will beat the world record

"In the problem above I called processes **L1 and L2**. They are typed as 'boldfaced' (the third line, in brackets). In the fifth line I introduced mean values. Although I have used the same letters please pay attention, these L1 and L2 ARE MEAN, i.e. they are just some numbers. Not processes! These are numbers which you must get if you calculate mean distance from the distribution function, in other words they are the expectations of the processes defined above."

^^ extra notes provided by lecturer when asked for more information at office.

Any help would be greatly appreciated. Thanks

Re: Stochastic Process Help, stuck on question

Hey tommylee.

I'll get you started for a) by noting that if one person throw doesn't affect another then they are independent. If this is the case then P(A = a and B = b) = P(A = a) * P(B = b) and so you have PDF distributions for any throwers distance (L1 and L2).