probability problem

John · science forum beginner Joined: 12 Sep 2005 Posts: 12

Hello

I am doing a 'home assigned' homework, I need to be guided in-order to
find the solution of the this problem:

**********************************************************************
Let A=(X,Y,Z) be the coordinate of a point selected from the region
defined by x>=0, y>=0, z>=0, x+y+z <=1.

Let I_(x,y,z) be the box containing (x,y,z) with sides of length
delta_x, delta_y, delta_z and assume
P(A element of I_(x,y,z) ) = c(x+y+z) delta_x delta_y delta_z

(a) Find the density for A
(b) Evaluate P(X<1/2)
(c) Evaluate P(X+Y >= 1/4)
**********************************************************************

I understand that (a) is asking the joint density, (b) and (c) are
asking for the marginal density.

(a)
the density for A would be the
(probability law / delta_x delta_y delta_z), using the probability law
from the assumption in the problem, then the density for A should be
c*(x+y+z) is that correct?

the book answer to (a) is c=8
the density function takes the form of f(A) = whatever, is the book
answer correct? does it imply f(A) = the constant c = 8?
if so, how did it come about?

integral of c*(x+y+z) over x>=0, y>=0, z>=0, x+y+z <=1. is equal to
1, thus

c = the above integral / (x+y+z)
I am not able to come up with the integration formula. could some one
help

thanks

MDCovney · science forum beginner Joined: 07 Jun 2005 Posts: 30

The probabilities are more easily viewed and understood in terms of
(relative) volumes within the domain. The first task is to visualize the
(x,y,z) sample space as a right pyramid. Once you see that then you
translate the required probabilities to specific 3D regions within the 3D
domain. Volumes can be determined using techniques from calculus.

I'm a bit rusty at this but I calculated the volume of the domain by
integrating the area of the x-y plane within the domain [ 1/2 * (1 - z) *
(1 - z) ] from z=0 to z=1.
However, this produces 1/6 (vs your book's answer of 1/ Cool

.

Re (b), thanks to the symmetry of the problem, the volume can be determined
by integrating the same x-y "area" function as above but integrating from z
= 0 to z = 1/2. Prob (x < 1/2) = 7/8. Note that the region's volume is
7/56 but the probability is determined by relating the volume of this region
to the volume of the entire domain (1/6).

For part (c), relying again on the symmetry of x, y and z, I modified the
area function to remove the section corresponding to x+y< 1/4 (area = 1/32)
and then integrated the area function from z = 0 to z = 3/4.
P(x+y>= 1/4) = 54/64.

Note that since I didn't get the book's nswer for (a) (1/6 vs 1/ Cool

I don't
imagine that my answers for (b) or (c) will agree either.
If your teacher confirms the book's answers can you please post them again.
Thanks.

"John" <falseaddress@nospam.nomore> wrote in message
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Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

Took another look at this and am convinced that the "volume" of the domain
of points (x,y,z) selected at random within the region (x>=0, y>=0, z>=0 and
x+y+z<=1) is 1/6. Another way of looking at this is to check the
probability that x+y+z < 1 when each of x, y and z are selected
independently and randomly from the interval (0,1).

But how do you account for the book's answer of c=8 (vs 6)? Books are not
often wrong. But then again, the book or maybe you, John, expressed the
density as c*(x+y+z) which is not correct, nevermind the fact that c<>8.

"John" <falseaddress@nospam.nomore> wrote in message
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Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

Further thought: Although one can figure the probabilities without
explicitly calculating the marginal densities (as I did) I can see you want
to figure them using marginals which is fine. In fact, it may be better than
trying to visualize multi-dimensional domains. Anyway, the joint and
marginal density functions of a point (x,y,z) selected at random in the
region x >=0, y >=0, z >=0 and x+y+z<=1 are:
f(x,y,z) = 6; g(x,y) = 6*(1 - x - y); h(x) = 3*(1 - x)^2. (The marginals
produce the same probabilities for X and X+Y as I posted earlier.)

"John" <falseaddress@nospam.nomore> wrote in message
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Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

Not that it matters any more but I just noticed an error in a statement in
my earlier post regarding part (b). The "volume" of the region (the portion
of the domain for which 0 < z < 1/2) is not 7/56 at stated but 7/48. The
required probability
P(X<1/2) = 7/8 was correct as stated.

"MDCovney" <mdcovney@optonline.net> wrote in message
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