P EXAM QUESTIONS OF THE WEEK
S. Broverman, 2005
Week of December 26/05
The random variable \ is defined on the interval ! # . You are given T \ " " , J Bl\ " B# , J Bl\ " B " , I\ " . % Find J " .
The solution can be found below.
Week of December 26 - Solution
Let T \ " - For B ", J Bl\ "
J B T \"
J B -
B# , so that J B -B# for ! B " .
$ %
J " T \ " T \ " - For " B # , J Bl\ "
" %
, so T \ " " J "
J B- " % $ % -
- .
J BJ " T \" " %
B",
so that for " B # , J B -
$ -B " . %
Then %
$ WB " J B is equal to " -B# for ! B " , W" and WB " -
# " %
- ,
$ -B " $ -# B for " B # . % %
" #
I\ '! WB .B '! " -B# .B '" $ -# B .B % "
$
$ - " % #
* #!
T \ " -
so that
$$#!" #% * J " #! " %
p-
"% #!
* #!
.
.
Note that we have used the expectation rule for a non-negative random variable \ , I\ '! WB .B ; since \ is defined on ! # we have J B " for B # and W# ! for B # . Also note that since \ has a discrete point of probability at B ", the " # integral of WB can be split into '! and '" .