Find the PDF of W = X + Y + Z on a Uniform Distribution

[Dwolfson]

I am stumped.

I have that W=X+Y+Z and that S=X+Y

These are all X, Y, & Z and Independent and Uniformly Distributed on (0,1)

I found the pdf of S to be (Assume all these < rep. less than or equal to):

S when 0<S<1
2-S when 0<S<1

So I continued:

To do pdf of S+Z=W

I figured there will be 3 intervals:

when 0<W<1, 1<W<2, and 2<W<3:

I Have figured out the one from 0<W<1

to be integral from 0 to W pdf(w)=S(pdf(W-S))ds

= W^2/2

For the other two intervals I am struggling on which pdf of S to use and what is the interval of integration..

Thank you in advance for your help,
--Derek

 

[lanedance]

so the sum of 2 RVs is given by their convolution, in particular the square pulse integral
http://en.wikipedia.org/wiki/Convolution

so for S = X+Y, with $p_X(X=x), \ p_Y(Y=y)$
$$p_S(s) = \int dx p_X(x) p_Y(s-x)$$

similarly, it should just follow that for W = S + Z
$$p_W(w) = \int dz p_Z(z) p_S(w-z)$$

 

[lanedance]

I am stumped.

I have that W=X+Y+Z and that S=X+Y

These are all X, Y, & Z and Independent and Uniformly Distributed on (0,1)

I found the pdf of S to be (Assume all these < rep. less than or equal to):

S when 0<S<1
2-S when 0<S<1

and i assume you mean
2-S when 1<S<2

 

[Dwolfson]

Please look here -- I did some further work and plugged this thread into the proper message board:

https://www.physicsforums.com/showthread.php?p=3019646#post3019646

 

[blue_raver22]

Hello Derek,

Thank you for providing your progress and thoughts on this problem. I can see that you are on the right track, but there are a few things that need to be clarified and corrected.

First, let's define our variables. You have correctly stated that W=X+Y+Z and S=X+Y. However, it would be better to use different letters for the variables to avoid confusion. Let's use U for W and V for S. So, U=X+Y+Z and V=X+Y.

Next, let's define the ranges for each variable. You have stated that X, Y, and Z are uniformly distributed on (0,1). This means that their values can range from 0 to 1. Therefore, the ranges for U and V would be 0<U<3 and 0<V<2.

Now, let's find the pdf of V. We can do this by finding the cdf of V first and then differentiating it. Since X, Y, and Z are independent, we can use the product rule for cdfs. So, the cdf of V would be:

Fv(v) = P(V<v) = P(X+Y<v) = integral from 0 to v integral from 0 to v-x 1 dx dy = integral from 0 to v (v-x) dx = v^2/2

Differentiating Fv(v) with respect to v, we get the pdf of V as:

fV(v) = dFv(v)/dv = v

Now, let's find the pdf of U. We can do this by using the convolution formula, which states that the pdf of the sum of two independent random variables is the convolution of their individual pdfs. So, the pdf of U would be:

fU(u) = integral from 0 to u fV(v)fZ(u-v) dv

Since Z is also uniformly distributed on (0,1), its pdf would be 1. Substituting this in the above formula, we get:

fU(u) = integral from 0 to u v dv = u^2/2

Therefore, the pdf of U is a triangular distribution with a range of 0<U<3.

To find the pdf of U+V, we can use the same convolution formula. So, the pdf of U+V would be:

fU+V(u+v) = integral from