# Distribution of the sum of three random variables

1. Apr 26, 2009

### TeXfreak

Hi everyone. I have this problem. Given three random variables X, Y, Z with joint pdf (probability density function)

f(x,y,z)=\exp(-(x+y+z)) if x>0, y>0, z>0; 0 elsewhere

find the pdf of U (f_U), where U is the random variable given by U=(X+Y+Z)/3.

Now I know how to find the joint pdf of a random vector of equal dimension as that of the original vector (via the Jacobian of the inverse transformation, that is, when the transformation is from R^n to R^n, but in this case it is from R^3 to R), or how to find the pdf of the sum of two independent random variables (via the convolution of the two pdfs), but I can't figure out how to do this one.

One could set the transformation to be g : R^3 \to R^3 defined by g(x,y,z)=((x+y+z)/3,y,z) (though I am not sure whether that would be right), so as to find the pdf of g(X,Y,Z) and then find the marginal density function of U, but then the integral does not converge.

And trying convolutions, something like f_U = f_X * (f_Y * f_Z) ---and here I am less sure if it's ok--- the integral doesn't converge either.

Could anybody can help me with this problem, please? Thanks in advance.

2. Apr 26, 2009

### mXSCNT

First find the cumulative distribution function, F_U(u) = P((X+Y+Z)/3 < u), by integrating the joint density function f(x,y,z) over the tetrahedron satisfying x > 0, y > 0, z > 0, (x+y+z)/3 < u. Then differentiate F_U to get the density function for U.

3. Apr 28, 2009

### gel

A general method I often use is as follows. The probability density function fU of a random variable U is defined by the following expression for the expected value of h(U), for any function h

$$E[h(U)] = \int f_U(u) h(u)\,du.$$

Just substitute in U=(X+Y+Z)/3

$$\int f_U(u) h(u)\,du=E[h((X+Y+Z)/3)]=\int_0^\infty \int_0^\infty\int_0^\infty f(x,y,z)h((x+y+z)/3) \,dx\,dy\,dz$$

change variables x = 3u - y - z in the inner integral

$$\int f_U(u) h(u)\,du=\int_0^\infty \int_0^\infty\int_{(y+z)/3}^\infty f(3u-y-z,y,z)h(u) 3\,du\,dy\,dz$$

Change the order of integration

$$\int f_U(u) h(u)\,du=\int_0^\infty\int_0^{3u} \int_0^{3u-z} f(3u-y-z,y,z)h(u) 3\,dy\,dz\,du$$

from which you can read off the density

$$f_U(u)=3\int_0^{3u} \int_0^{3u-z} f(3u-y-z,y,z) \,dy\,dz$$

This is virtually the same as calculating the cumulative distribution (by taking $g(u)=1_{\{u>K\}}$), but without the differentiation step to convert to the density at the end. So, whichever method you prefer.