Let $X$ be a random variable with support {1,2,3,5,15,25,50}. Each point of $X$ has the same probability of $\frac{1}{7}$. Find the value of c $\in \mathbb{R}$ which minimizes $h(c)=E(|X-c|)$.
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A well known result is that any point $c$ that minimises $\sum_k |x_k -c|$ is a median (see http://math.stackexchange.com/q/318381/27978, for example) – copper.hat Feb 25 '16 at 06:39
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The expectation is the sum of 7 terms. The sign of each term is unchanged when $c$ is in between two support points. So the expectation is piecewise linear in $c$, with the support point as knots. When there are odd numbers of support points, the minimum will occur at the median; when there are even number of support points, the function is a constant over the two middle support points. – BGM Feb 25 '16 at 06:45
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Functions having an equation of the form $f(x)=\sum_{k=1}^n p_k|x-a_k|$ (with all $p_k \geq 0$) are continuous and piecewise affine on intervals $(-\infty,a_1],\cdots [a_k,a_{k+1}],\cdots [a_n,+\infty)$. As their limit at $-\infty$ and $+\infty$ is $+\infty$ and that their are bounded below by zero, they have at least a minima. These minima can occur only at points $(a_k,f(a_k))$.
In our case where the $a_k$ are the points in the support and the $p_k=1/7$, a rapid inspection of values $f(a_k)$ show that it is at point $c=a_4=5$ that the minimum of $E(|X-c|)$ occurs.
Jean Marie
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