| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
This problem came up as a homework problem from a Random Signals and Noise
course I am presently taking. The problem has two parts: (1) find the
probability density function (pdf) of Z given that Z=X+2Y and (2) Find
P[0<Z<=1]. The answer from (1) should give the answer to (2) by integrating
from z=0 to z=1.
Here is the problem statement:
Use convolution to find the probabilty density function of Z given that Z=X+2Y
where f (x) and f (y) are defined as:
X Y
[ 2x 0<x<=1
f (x) = [
X [ 0 elsewhere
"Y is uniformly distributed over the range -1 to 1" which I translated to:
[ 1/2 -1 < y <= 1
f (y) = [
[ 0 elsewhere
oo ("oo" is "infinity")
Using convolution, f (z)= integral [ f (x) f ((z-x)/2) dx ]
Z -oo X Y
My answer for f (z) was:
Z
[ 0 z <= 0
[ (1/2)z**2 0 < z <= 1
f (z) = [ (1/2) 1 < z <= 2
Z [ (1/2)-(1/2)(z-2)**2 2 < z <= 3
[ 0 z > 3
This looked good because the integral over -oo to +oo is 1, a property
of pdf's (see note below for other attempts).
Using the above to do part (2), I get an answer of 1/6 when I integrate f(z)
from 0 to 1. However, the book gives an answer of 1/4 ... which can be
obtained by integrating the triangular region under y=(1-x)/2 from x=0 to
x=1 (this is the region for which 0<x+2y<=1).
Can someone tell me where I went wrong?
Thanks in advance!
Al
PS: Here are the other f ((z-x)/2)'s I have tried:
Y
I had a hard time picking what f ((z-x)/2) should look like so I tried the
Y
"picture" approach to get a rectangle with height 1/2 and base 2. When I
took the "mathematical" approach the base of the rectangle became 4 which
left me wondering if I should adjust the height to be 1/4 instead of 1/2
(area should be 1?).
With f ((z-x)/2) taken to be a rectangle with base of 4 and height of 1/4:
Y
[ 0 z <= 0
[ (1/4)z**2 0 < z <= 1
f (z) = [ 1/4 1 < z <= 4
Z [ (1/4)-(1/4)(z-4)**2 4 < z <= 5
[ 0 z > 5
(this also integrates to 1 over -oo to +oo)
With f ((z-x)/2) taken to be a rectangle with base of 4 and height of 1/2:
Y
[ 0 z <= 0
[ (1/2)z**2 0 < z <= 1
f (z) = [ 1/2 1 < z <= 4
Z [ (1/2)-(1/2)(z-4)**2 4 < z <= 5
[ 0 z > 5
Neither of the above f(z)'s gives an answer of 1/4 for P[0<Z<=1].
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1135.1 | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Oct 08 1989 00:39 | 22 | |
With probability 1/4, Y will be less than -1/2, and so 2Y
will be less than -1 and Z=X+2Y will be less than 0. Yet
your solution for Z has Z always at least zero.
Therefore your solution for f (z) is incorrect.
Z
A correct solution for Z will have nonzero probability
for -2 < Z < 3 and zero probability outside that range,
with an overall probability of at least 1/4 that Z is
negative (the case of the first paragraph occurs with
probability 1/4 but does not exhaust all possibilities of
negative Z).
The probability that 0 < Z <= 1 will be the integral as y
goes from -1/2 to 0 of (1/2)dy times Prob(x > 2|y|),
plus the integral as y goes from 0 to 1/2 of (1/2)dy
times Prob(x <= 1-2y). Plugging in Prob(x > 2|y|) = 1 -
4y^2 for y in -1/2 to 0, and Prob(x <= 1-2y) = (1-2y)^2
for y in 0 to 1/2, gives Prob(0<Z<=1) = 1/6 + 1/12 = 1/4.
Dan
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| 1135.2 | My mistake | WFOVX5::PLANTE | Sun Oct 08 1989 14:52 | 31 | |
RE -.1,
Thank you Dan.
When I flipped the rectangle, f (y) --> f ((z-x)/2), I came up with
Y Y
the right edge being located at z+2, the left edge at z-2, and z
in the middle. But, in evaluating the convolution integral, I used
"z" to represent the leading (right) edge and "z-4" to represent the
left edge -- that's where I was wrong.
I had to use the rectangle with base=4, height=1/4. The
change in height was justified by noting that since the function
should be a pdf, then the height is related to the base such that
the overall area = 1.
My final answer was:
[ 0 z <= -2
[ (1/4)(z+2)^2 -2 < z <= -1
f (z) = [ 1/4 -1 < z <= 2
Z [ (1/4)-(1/4)(z-2)^2 2 < z <= 3
[ 0 z > 3
This integrates to 1 over (-oo, +oo) and also gives P[0<Z<=1]=1/4.
Also, P[Z<=0]=1/3 > 1/4, and the pdf is nonzero only on (-2, 3] as
you mentioned.
Thanks again.
Al
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| 1135.3 | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Oct 08 1989 18:24 | 6 | |
That satisfies two other tests: it's weighted a little
heavier at the top (like the X distribution is), and it
combined linear and constant pdf's and came out
quadratic. So that's "probably" :-) it.
Dan
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