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Let $X$, $Y$ and $Z$ be random variables with probability distributions $f_X, f_Y, f_Z$ taking values in the unit interval $[0,1]$. I'm interested in the distribution of $$X\cdot Y + (1-X)\cdot Z,$$ which always again takes values in $[0,1]$.

Below we see a histogram of samples for $X,Y,Z$ uniformly distributed (and it reminds me of some Shannon entropy function).

enter image description here

It turns out that the distribution of $X\cdot Y$ with two variables is already tricky enough to compute, see Wikipedia, and no literature I found was for the unit interval. Okay so now let's first focus on that two-variable question and even assume uniform distributions of $X$ and $Y$, then I expect I gotta compute something along the lines of

$$\int^x 1 \int^{z/x} 1 \,{\mathbb d}y \,{\mathbb d}x $$

I left out the lower bound $0$ because this makes problems with the log. I suppose my first much reduced question would thus be this one:

Let $X$ and $Y$ be random variables with probability distributions $f_X, f_Y$ taking values in the interval $[\epsilon, 1]$. I'm interested in the distribution of $$X\cdot Y,$$ which takes values in $[\epsilon^2, 1]$?

In this thread such a question is asked an Mathematica is used to give a few expressions involving logs of $z$. I can imagine the result to be something along the lines of $a-b\,\log(z)$ but I'm not sure why $z$ would make it into the nonlinear expression given the above kind of integral.

Assuming we find this, how do deal with the limit $\epsilon \to 0$? And then, for the 3 variable case, we might have to go back to the definition for the mulivariate distribution case (Wikipedia) but I first need to get the $\log$-issue above.

This StackExchange question discusses the case of $n$ uniformly distributed variables, with values in $[0,1]$,for large $n$, in which case we get a Gaussian.

Nikolaj-K
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1 Answers1

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If $X,Y$ are independent random variables uniformly distributed on $[0,1]$, then for every positive measurable function $f$ we have
$$\eqalign{E(f(XY))&=\int_0^1\int_0^1f(xy)dxdy= \int_0^1\frac{1}{y}\int_0^yf(t)dtdy\cr &= \int_0^1f(t) \int_t^1\frac{dy}{y}dt\cr &=\int_0^1f(t)\ln\left(\frac{1}{t}\right)dt }$$ So, the desity function of $T=XY$ is $f_T(t)=\textbf{1}_{(0,1)}(t)\ln(1/t)$.

Similarly, if $X,Y Z$ are independent random variables uniformly distributed on $[0,1]$, then for every positive measurable function $f$ we have
$$\eqalign{E(f(XY+(1-X)Z))&=\int_0^1\int_0^1\int_0^1f(xy+(1-x)z)dxdydz= \int_0^1\int_0^1\frac{1}{y-z}\int_z^yf(t)dtdydz\cr &=2 \int_{y=0}^1\int_{z=0}^y\frac{1}{y-z}\int_z^yf(t)dtdydz\cr &=2 \int_{t=0}^1f(t)\int_{z=0}^t\int_{y=t}^1\frac{dy}{y-z}dydzdt\cr &=2 \int_{t=0}^1f(t)\int_{z=0}^t\big(\ln(1-z)-\ln(t-z)\big)dzdt\cr &=2 \int_{t=0}^1f(t)\Big[-(1-z)\ln(1-z)+(t-z)\ln(t-z)\Big]_{z=0}^{z=t}dt\cr &=2 \int_{t=0}^1f(t)\big(-(1-t)\ln(1-t)-t\ln(t)\big)dt\cr &=2 \int_{t=0}^1f(t)\ln\frac{1}{(1-t)^{1-t}t^t}dt\cr }$$ So, the desity function of $T=XY+(1-X)Z$ is $$f_T(t)=2\textbf{1}_{(0,1)}(t)\ln((1-t)^{t-1}t^{-t}).$$ A plot of this density function can be found here.

Omran Kouba
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