I didn't quite understand the solution(part (i)) in the examiner's report. Additionally, the solution in the revision notes has me wondering as to how the second part of the double integral was evaluated. Thank you for reading
Oh, I see The examiners are saying that Z1 needs to be positive - and that's going to happen with a probability of 0.5. By setting Z2=Z1+X we can examine the increment X and place requirements on it such that Z2<0. Firstly we need X to be negative, and secondly it needs to be so negative that it wipes out all of the positive progress made by Z1. These two requirements can be represented by having X<0 and abs(X)>abs(Z1). These are independent and both occur with a probability of 0.5. Does that do it? Maybe you could indicate how far you've got with the integral method and then we can fill in the gaps.
Wow that was a great explanation. I'm grateful. In the integral approach I'm baffled by this integral: \(\large{ \int_0^{\infty} \phi(w) \Phi(-w) dw }\)
First thing to notice is that \(\phi(w)\) is the derivative of \(\Phi(w)\) with respect to \(w\). Secondly, \(\frac{d}{dw}\Phi^2(w)=2\phi(w)\Phi(w)\), which is almost what you want to integrate. Now this result only helps if the variables driving the \(\phi\) and \(\Phi\) are the same - but they're different in the integral you're trying to solving. The last observation we need to make is that \(\phi(-w)=\phi(w)\), which means that if we combine these results together the integral should be solvable. Let me know if that doesn't do it
There's certainly nothing to feel stupid about - it only looks straightforward once you've seen how it's done
