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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.

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  • $\begingroup$ It was unclear to me why you said the distribution of $S$ is $\min(s,1-s)$. Later you say $s$ has a uniform distribution, which I believe is correct, but I think it is uniform on $[0,\frac12]$ since you are always cutting $X$ and $Y$ from the smaller stick created by the first cut. $\endgroup$
    – David K
    Jan 19, 2016 at 21:54
  • $\begingroup$ Sorry, I meant the length function of $S$. But why is it uniform on [0,1/2]? $\endgroup$
    – nomadicmathematician
    Jan 19, 2016 at 21:56
  • $\begingroup$ When you cut a stick of length $1$ in two pieces, the longer piece is at least $1/2$ and the shorter piece is at most $1/2$. $\endgroup$
    – David K
    Jan 19, 2016 at 22:01
  • $\begingroup$ Yes you're right, just got it. Then if I integrate the above conditional expectations over $[0,1/2]$, would I get the correct answer? $\endgroup$
    – nomadicmathematician
    Jan 19, 2016 at 22:02

1 Answer 1

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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.

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  • $\begingroup$ can you just assert that the shorter stick is U(0,1/2)? I tried working out min(X,1-X) it doesn't give me U(0,1/2). Also should the length of Y be U(s/2,1/2)? $\endgroup$
    – yoshi
    May 20, 2016 at 3:59
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    $\begingroup$ @yoshi: Sorry, $Y$ is uniform on $\left[\frac s2,s\right]$, the upper bound $1$ was wrong (but yours, $\frac12$, is also wrong). I've fixed that; it didn't affect the calculation. About $S$: The situation is symmetric. If the cut is in $\left[\frac12,1\right]$, just reflect it to the other half; that doesn't change the lengths. Thus each half of the uniform distribution on $[0,1]$ is mapped to a uniform distribution on $\left[0,\frac12\right]$. (By the way, $X$, $Y$ are the lengths, not the segments, so it's "the length $Y$", not "the length of $Y$" as you wrote.) $\endgroup$
    – joriki
    May 20, 2016 at 4:08
  • $\begingroup$ Could you elaborate on how to obtain the $E[XY]$ integral? I tried expanding to $E[E[XY|S]]$ but I'm getting stuck on how to interpret/evaluate $E[XY|S]$. The $X$ and $Y$ expectations went through okay. $\endgroup$
    – yoshi
    May 20, 2016 at 14:43
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    $\begingroup$ @yoshi: In that integral, you're treating $x$ and $y$ as independent. They're not, in fact they're as dependent as they can be; knowing the value of one gives you the value of the other. They don't have a joint density function; you could describe their joint distribution by a delta "function" density, $\frac1s\delta(x+y-s)$ on $[0,s]^2$. Their joint cumulative distribution function is $F(X\le x,Y\le y)=\frac1s\min(x,y)$. $\endgroup$
    – joriki
    May 20, 2016 at 16:27
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    $\begingroup$ @yoshi: I edited the joint cumulative distribution function into the comment, don't know if you saw that, too. And I've been notified of your comments because they were under my answer; if you ever want to have me notified elsewhere, you'll need to spell my username right :-) $\endgroup$
    – joriki
    May 20, 2016 at 16:31

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