Is my proof correct?

**joestevens** · October 17th 2011, 05:54 PM

I'm not sure if I proved the following correctly. (It is in a chapter on expectation, and I didn't use anything about expectation.) Any feedback would be greatly appreciated.

Problem:
Suppose that $X, Y,$ and $Z$ are nonnegative random variables such that $P(X+Y+Z \leq 1.3)=1$ . Show that $X, Y,$ and $Z$ cannot possibly have a join distribution under which each of their marginal distributions is the uniform distribution on the interval $[0,1]$ .

Work:
$\int_0^1\int_0^1 f(x,y,z)dxdy=z$
$\int_0^1\int_0^1 f(x,y,z)dxdz=y$
$\int_0^1\int_0^1 f(x,y,z)dydz=x$

$\int_0^1 xdx=\frac{1}{2}$
$\int_0^1 ydy=\frac{1}{2}$
$\int_0^1 zdz=\frac{1}{2}$

Answer:
Integrating $f(x,y,z)$ over the cube gives $p=\frac{1}{2}$ instead of $1$ , therefore we have a contradiction. So $X, Y,$ and $Z$ cannot possibly have a join distribution under which each of their marginal distributions is the uniform distribution on the interval $[0,1]$ .

**matheagle** · October 17th 2011, 11:30 PM

I don't understand your integrals.

IF you integrate out the other variables you obtain the marginal density.
Hence

$\int_0^1\int_0^1 f(x,y,z)dxdy=f_Z(z)=1$ IF you think that Z is U(0,1)

**joestevens** · October 18th 2011, 06:10 AM

You're right. I had done $\int_0^z\int_0^1\int_0^1 f(x,y,z)dxdydz=F_Z(z)=z$ but wrote $\int_0^1\int_0^1 f(x,y,z)dxdy=z$ . So $\int_0^1dz=1$ .

Now I'm at a loss on how to prove this.

**joestevens** · October 19th 2011, 06:16 PM

$P(X+Y+Z \leq 1.3)=1 \Rightarrow E(X+Y+Z) \leq 1.3$
If X, Y, and Z are $U(0,1)$ then $E(X+Y+Z)=E(X)+E(Y)+E(Z)=1.5$
Contradiction. Therefore at least one of X, Y, or Z is not $U(0,1)$ .

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