### uniform by three

Here is a problem recently described to me. Apparently there is a more elegant solution (which may give more insight), but I don’t see it yet.

The problem: $$X, Y, Z$$ are independent random variables uniformly distributed over [0,1]. What is the distribution of $$(XY)^Z$$?

The conventional solution is to find the distribution of $$XY$$ first, then of $$(XY)^Z$$.

The distribution of $$XY$$ can be derived from its CDF $$F_{XY}(xy \le k)$$, which is the total shaded area shown. The red curve is the function $$y=\frac{k}{x}$$. This area is thus given by:

$$\int_k^1 \frac{k}{x} dx + k = k \ln(x) \vert_k^1 + k = -k \ln(k) + k$$, for $$k>0$$.

The PDF is the derivative of the above:

$$f_{XY}(k) = -\ln(k)$$, for $$k>0$$. The density is not well defined at $$k=0$$.

The second part is to find the distribution in question. Here, the red curve is the function $$z=\frac{\ln(k)}{\ln(xy)}$$. The CDF $$F_{(XY)^Z}((xy)^z \le k)$$ is the total area to the left of the red curve. A column slice shaded in blue has probability per unit of $$z$$ as labeled. Thus, the CDF is:

$$\int_0^k f_{XY}(v) dv (1-\frac{\ln(k)}{\ln(v)}) = \int_0^k -\ln(v) dv (1-\frac{\ln(k)}{\ln(v)})$$
$$= \int_0^k -\ln(v) dv + \int_0^k \ln(k) dv$$
$$= -v \ln(v) + v \vert_{\downarrow 0}^k + k \ln(k) = -k \ln(k) + k + k \ln(k) = k$$, for $$k>0$$.

For $$k=0$$, we can fill in 0, because the minimum value of $$(XY)^Z$$ is 0, so it is the minimum of the support of its CDF.

So amazingly, $$(XY)^Z$$ has the CDF of (and is) a uniform distribution. How does that happen? What’s the intuition?

1. November 24th, 2008 | 22:49

You mean the area to the right of the left curve in the 2nd image.

2. November 25th, 2008 | 15:25

oops, no you don’t