#16) It is possible to calculate all of the distribution/density
functions, but I think it is easier to argue based more on general principles.
X1: This is a mixture of two uniform r.v. Each of the individual r.v. has
a continuous dist. func., but a discontinuous density func. The
dist./density functions of X1 is just a mixture (linear combination) of the
dist/density functions of the uniform r.v., so it will also have a cont.
dist. func. and a discontinous dens. function. So X1 corresponds to case 2.
X3: The act of censoring the normal distribution at 1 has the effect of
concentrating all of the probability X >= 1 at the single point 1. So
there is a single point with positive probability of occurring. This means
that the distribution function will have a jump discontinuity at 1. So X3
corresponds to case 3.
X2: By elimination, X2 corresponds to case 1. Also, the distribution of a
sum of random variables should generally be a degree smoother than those of
original distributions. In this case, the two uniform distributions have
dist. functions which are continuous, but not differentiable. The
distribution of the sum should then be continuous and have one continuous
derivative (which is the density function).
>Reply-To: "Costas Constantinou" <constantinos@snafu.de>
>From: "Costas Constantinou" <constantinos@snafu.de>
>To: <studygroup4B@lists.casact.org>
>Subject: Re: SPRING 98 Exam
>Date: Thu, 22 Oct 1998 12:00:06 +0100
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>I am working out the 5/98 exam and have difficulty getting the right
answers to >the problems 16, 24 and 28. Can someone help?
>
>Thanks and good luck with your exams. Costas
> P.S. Dave, your solutions (last e-mail) helped me too. Thanks.