Brute force... I can't see any simpler way. See the attached pdf.
recent thread, in a slightly more complicated way (more parameters, hence heavier computation), and I devised a computation-free way in my second post (after having spent some time on the direct way in the first one): when ,
is the convolution product of the pdf's of a and a , evaluated at the point , thus it is the pdf of a evaluated at the point (cf. explanation in the quoted post if necessary):
If you want for positive (in fact, for otherwise it diverges), this method doesn't work directly. But you can use analytic continuation: is analytic on (using the integral formula,...), and it coincides with the above function when , therefore the coincidence extends to , i.e. for . This gives:
for any .
Thank you to Moo and Laurent. Your contributions to the Maths web forum are invaluable. It is amazing how you can code and solve the sums at the same time!
I realised that from the solutions there was a factor in the distributions. I used this to solve Q 2B.
Can you guys help also with your thoughts/solutions with Q 1B?