Could anyone tell me any asymmetric distribution whose mode=median? Thanks in advance.
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One can produce examples, though not necessarily interesting ones. Let $X$ have density function $f(x)=0$ for $x\lt 0$, $f(x)=x$ for $0\le x\le 1$, and $f(x)=e^{-2(x-1)}$ for $x\gt 1$. The median and the mode are at $x=1$. The distribution is very much not symmetric about $x=1$.
We can also produce simple discrete examples. Let $\Pr(X=0)=\frac{1}{2}$. Let $\Pr(X=1)=\Pr(X=2)=\frac{1}{8}$ and $\Pr(X=-47)=\Pr(X=-99)=\frac{1}{8}$.
If you feel like it, you can modify the discrete example and get an asymmetric distribution where mean, median, and mode are all equal.
André Nicolas
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Slightly related is the "Mean-Median-Mode inequality" (which is false in general), which details are given in Purdue University's Statistics department, TR-92-40. – Batman Jun 26 '14 at 05:08
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Lots of distributions do. Any continuous distribution which is symmetric about its largest pdf value will satisfy this.
Student T and Laplacian are two simple examples
Batman
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Sorry I think I just asked a wrong question. I mean any asymmetric distribution. – monkinsane Jun 26 '14 at 03:42