Find the PDF of W when W= X + Y + Z. Random Varibles, Uniform Distrubutions.

[Dwolfson]

Homework Statement

1. Let X , Y and Z be independent random variables, uniformly distributed on the interval
from 0 to 1. Use Theorem 3.8.1 twice to find the pdf of W = X + Y + Z .

Thm. 3.8.1 States: If X & Y are continuous random varibles wth pdfs f_x(x) and f_y(y), respectively then

f_w(w)=[tex]\int[/tex][tex]^{INF}_{-INF}[/tex] f_x(x)f_y(w-x) dx

Homework Equations

When setting up the problem I understand how to do it once.. I set S=X+Y and get the PDF of s to be a piece wise defined function:

Keep in mind that I am completely unsure of my bounds of integration -- I do not have any idea if these bounds are correct -- But I tried to use my best logic and intuition to figure them out.

PDF of s = s when 0[tex]\leq[/tex]s<1
PDF of s = 2-s when 1[tex]\leq[/tex]S[tex]\leq[/tex]2

The Attempt at a Solution

I figured that there will be three intervals for the final solution:

0[tex]\leq[/tex]w<1

1[tex]\leq[/tex]w<2

2[tex]\leq[/tex]w[tex]\leq[/tex]3

Some of these may be strictly < or > but that is part of my confusion...

So my intuition tells me:

I should calculate:

[tex]\int[/tex][tex]^{W}_{0}[/tex] S ds
Giving PDF of W = w^2/2 for 0[tex]\leq[/tex]w<1

After this I am very lost:

I Imagine the interval from 1 to 2 will be the sum of two integrals... [tex]\int[/tex][tex]^{w}_{1}[/tex] 2-S ds +
[tex]\int[/tex][tex]^{w-1}_{0}[/tex] S ds
which gives : W+1 as the PDF ( I Think this is incorrect by the way).

I justified my bounds by this logic:

If S is between 0 and 1 we'd use the pdf(s) that = S meaning that if W is between 1 and 2 and S is between 0 and 1 we'd integrate from S=0 to S=w-1

If S is between 1 and 2 we'd use the pdf(s) that = 2-S from 1 to W becuase S has to be at least one and less than W.

For the third interval of W is between 2 and 3:

I used this integral:

[tex]\int[/tex][tex]^{w-1}_{1}[/tex] 2-S ds

I justified my bounds by this logic:

Since W=X+Y+Z and S=X+Y S would have to be atleast 1 for this interval to make sense... Since if S=1 and Z=1 then W=2. The upper bound of this integral would be W-1 because if W is maximized at 3 then S would be 2. So for any number between 2 and 3 the upper bound is W-1.Thank You for your help,
-Derek

 

[matiasmorant]

the solution:

[tex]f_w(w)=\begin{cases}
\frac{w^2}{2} & 0<w\leq 1 \\
-\frac{3}{2}+3 w-w^2 & 1<w\leq 2 \\
\frac{1}{2} (-3+w)^2 & 2<w\leq 3 \\
0 & \text{Everywhere } \text{else}
\end{cases}
[/tex]

 

[Dwolfson]

Thanks... I like to work backwards.. now I can figure out the bounds.

Appreciated.

 

[matiasmorant]

did you get it? need some more help?

 

[Dwolfson]

I did eventually figure it out.. Thank you for the followup.