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Is there any easy way to see that the map $f(x,y,z) = xy+(1-x)z$ pushes forward the (three-dimensional) Lebesgue measure on the unit cube to the (one-dimensional) Lebesgue measure on the unit interval?

I'm pretty sure I can prove it by showing directly that $\lambda^{3}\circ f^{-1}[0,b) = b$ for every $b \in [0,1]$, but the calculuation is a bit tedious, involving cross-sectional areas, Fubini, and whatnot. I suspect there's an elegant way of doing it.

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There isn't, because it doesn't. You are taking three uniformly distributed independent random variables $X,Y,Z$ and forming a weighted average of $Y$ and $Z$ using $X$ as a weight. The result of averaging will fall near $1/2$ more often than near $0$ or $1$. I took $10^6$ samples for illustration:

histogram

And here is a mathematical derivation of the pushforward density. By symmetry, it suffices to consider $0<x<1/2$ and double the result. For a fixed $x\in (0,1/2)$ the probability density function of $xY$ is $\frac{1}{x}\chi_{[0,x]}$, because $xY$ is uniformly distributed on $[0,x]$. Similarly, the pdf of $(1-x)Z$ is $\frac{1}{1-x}\chi_{[0,1-x]}$. Since the two random variables are independent, the pdf of their sum $xY+(1-x)Z$ is the convolution of pdfs: $$p_x(t)=\frac{1}{x(1-x)}\chi_{[0,x]}*\chi_{[0,1-x]} = \frac{1}{x(1-x)} \times\begin{cases} t,\quad &0\le t\le x \\ x,\quad &x\le t\le 1-x \\ 1-t,\quad &1-x\le t\le 1 \end{cases} $$ This is a trapezoidal function which stays flat on $[x,1-x]$ and drops to $0$ at $t=0,1$.

We get the pdf of $XY+(1-X)Z$ by integrating $p_x$ over $x$: $$p(t) = 2\int_0^{1/2} p_x(t)\,dx = (2t-2)\log(1-t)-2t\log t $$ Looks pretty neat; I never saw this distribution but this does not say much. Here is the plot, for comparison with the above histogram.

nice curve, much logs

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  • $\begingroup$ I don't see where the characteristic functions $\chi_{[0,x]}$ and $\chi_{[1-x,1]}$ come in. Would you please elaborate on that? $\endgroup$ Mar 24 '14 at 17:21
  • $\begingroup$ Okay, so did you mean $\chi_{[0,1-x]}$ instead of $\chi_{[1-x,1]}$? $\endgroup$ Mar 24 '14 at 23:18
  • $\begingroup$ @QuinnCulver You are right; I fixed this mistake and also edited my previous comment in. $\endgroup$
    – user127096
    Mar 24 '14 at 23:26

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