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Suppose $X_1$ and $X_2$ are taken at random from a uniform distribution on the interval $[\theta - 1/2, \theta + 1/2]$, where $\theta$ is unknown $(-\infty < \theta < \infty)$. Let $Z = Y_2 - Y_1$, where $Y_1 = min(X_1, X_2)$ and $Y_2 = max(X_1, X_2)$.

How do I calculate the conditional distribution of $X = 0.5(X_1 + X_2)$ given $Z=z$? Specifically, how do I show that this conditional distribution is uniform on the interval $[\theta - 1/2(1 - z), \theta + 1/2(1 - z)]$?

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Looks like homework. Homework problems should be tagged as such. Then the answers are supposed to be hints to make it easier for you to get the solution on your own. –  Michael Chernick Sep 27 '12 at 16:26
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1 Answer 1

One can assume without loss of generality that $\theta=\frac12$, hence $X_1$ and $X_2$ are i.i.d. uniform on $[0,1]$, and compute the distribution of $(X,Z)$. Since $Z=|X_1-X_2|$, for every test function $u$, $$ \mathrm E(u(X,Z))=\iint u\left(\tfrac12(x_1+x_2),|x_1-x_2|\right)\,[0\leqslant x_1,x_2\leqslant 1]\,\mathrm dx_1\mathrm dx_2. $$ The change of variable $x=\tfrac12(x_1+x_2)$, $z=|x_1-x_2|$, yields $x_1=x\pm \frac12z$, $x_2=x\mp\frac12z$, $dx_1dx_2=2dxdz$ (the factor $2$ for the fact that two points $(x_1,x_2)$ correspond to the same point $(x,z)$), and $$ \mathrm E(u(X,Z))=\iint u(x,z)\,2\,[0\leqslant 2x\pm z\leqslant2,\,z\geqslant0]\,\mathrm dx\mathrm dz. $$ The indicator function is $[z\leqslant 2x\leqslant2-z,\,0\leqslant z\leqslant1]$, hence $(X,Z)$ is uniform on this set (this is the triangle in the $(x,z)$ plane with vertices $(0,0)$, $(1,0)$ and $(\tfrac12,1)$) and, conditionally on $[Z=z]$ for some $z$ in $[0,1]$, $X$ is uniform on the set $\{x\mid z\leqslant2x\leqslant2-z\}$, which is the interval $[\frac12z,1-\tfrac12z]$.

If $\theta\ne\frac12$, $X_1$, $X_2$, $Y_1$, $Y_2$ and $X$ are shifted by $\theta-\frac12$ and $Z$ is unchanged hence, conditionally on $[Z=z]$ for some $z$ in $[0,1]$, $X$ is uniform on the interval $[\theta-\frac12+\frac12z,\theta+\frac12-\tfrac12z]$.

This, or draw a picture.

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WOW! Simply amazing. Thanks so much. I will never ceased to be amazed that there are people out there with that special combination of generosity and intellect necessary to leave the above post. thanks again. This problem comes from a discussion in Degroot's Statistics and Probability textbook about the shortcomings of confidence interval in which he asserts the statement above without proof. I suppose it is clear if one draws a picture but I was just not seeing it. Thanks for clearing this up for me!!! –  Will Sep 28 '12 at 1:38
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