A stick of unit length is cut into three pieces of lengths $X, Y, $and $Z$, with $X \le Y \le Z$. First the stick of length $1$ is cut into two pieces at a randomly chosen point uniformly distributed along its length. Then the smaller of these two sticks is also cut into two pieces at a randomly chosen point uniformly distributed along its length. These two cuts are conducted independently. Find Cov(X,Y).
I need to get $E(XY)-EXEY$. First, I write $S=X+Y$. Then the length function of $S$ is $\min(s, 1-s)$. Given $S=s$, I can find the covariance of $X,Y$ as follows. The distribution of $X$ is $\min(x, s-x)$, so $E[X|S]=\int_0^{s/2} xdx + \int_{s/2}^s (s-x) dx=s^2/4.$ Likewise, $E[Y|S]=\int_0^{s/2} (s-y)dy + \int_{s/2}^s y dy=3s^2/4.$ Now $E[XY|S]=\int_0^s x(s-x)dx=s^2/6$,since $y$ is just $s-x$(I'm not sure about this part). Now to get the expectation, I just need to integrate them over $s$, which has the uniform distribution, so that would be integrating them over $[0,1/2]$. Is this correct? I would greatly appreciate any comment.
Best Answer
The length $S$ of the shorter piece is uniform over $\left[0,\frac12\right]$. Given $S=s$, the length $X$ of the shorter piece in the second cut is uniform over $\left[0,\frac s2\right]$, and the length $Y$ of the longer piece in the second cut is uniform over $\left[\frac s2,s\right]$. Thus we have
\begin{align} \mathbb E[X]&=2\int_0^\frac12\mathrm ds\,\frac14s=\frac1{16}\;,\\ \mathbb E[Y]&=2\int_0^\frac12\mathrm ds\,\frac34s=\frac3{16}\;,\\ \mathbb E[XY]&=2\int_0^\frac12\mathrm ds\,\frac2s\int_0^\frac s2\mathrm dx\,x(s-x)\\ &=2\int_0^\frac12\mathrm ds\,\frac2s\left(\frac18s^3-\frac1{24}s^3\right)\\ &=2\int_0^\frac12\mathrm ds\,\frac{s^2}6\\ &=\frac1{72}\;, \end{align}
and the covariance is
\begin{align} \operatorname{Cov}(X,Y)&=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]\\ &=\frac1{72}-\frac1{16}\cdot\frac3{16}\\ &=\frac5{2304}\\ &\approx0.002\;. \end{align}
So the correlation from the first cut slightly outweighs the anticorrelation from the second cut.