Thread: I'm a pure mathematician. HELP!

Quote: Originally Posted by BGM We know that Pr{X1 = 0} = Pr{X1 ≤ a} = 1/(n+1) ∀a∈[0, 2) Sorry, I do not understand the notation used at the end of this line so I can not respond to what you have said directly. Perhaps an example will help illustrate why I made the first query. The second query doesn't matter so much as it follows from the first and the fact that you used 2x and y whereas I used x and y/2....

As a conclusion, ∀a∈[0, 2n] Pr{X1 = x|X1 ≤ a} = 1/(⌊a/2⌋+1) for x = 0, 2, ..., 2⌊a/2⌋ one more correction here, Pr{X1 + X2 = y|X1 ≤ a} = (min{y/2, ⌊a/2⌋} - max{0, (y-2n)/2} + 1)/(n+1)(⌊a/2⌋+1) for y = 0, 2, ..., 2n, 2n+2, ..., 2(n+⌊a/2⌋) as Pr{X2 = y - 2x} = 1/(n+1) for y - 2x = 0, 2, ..., 2n i.e. x = (y -...

Quote: Originally Posted by gccd Am I correct in thinking the sample space is S = {0,2,...,2n}? X1: S -> R is defined by th identity function X1(s) = s. Hence, f1(x) = P(X1 = x) = P({s € S: X1(s) = x}) = P({s€S: s = x) = 0 if x does not belong to {0,2,...,2n}; 1/(n+1) if x € S (the discrete uniform distribution?). Since X2: S -> R is defined by X2(s) = s we have f1(x) = f2(x). Let Y = X1 + X2. Y: SxS -> R is defined by...