Let me clarify my needs. The PDF must comply to: 1. The mean is always in the shorter tail 2. Should have an inverse function 3. Be defined in the interval [0, 1] 4. Should have a shape parameter that allows the choice of the distance between the Mode and the Mean 5. 1st derivative must be continuous And, if it is a common use function the better
closed as too localized by Robin Chapman, Charles Siegel, Loop Space, S. Carnahan♦, Gjergji Zaimi Jun 18 '10 at 16:22
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$\begingroup$ What do you mean? It is easy to construct such a PDF. But are you looking for one that is in common use and has the desired property? $\endgroup$ – Harald HancheOlsen Jun 13 '10 at 23:42

4$\begingroup$ Sorry, I'm voting to close. It is straightforward to specify a piecewise linear function that does the job. $\endgroup$ – S. Carnahan♦ Jun 14 '10 at 1:58

$\begingroup$ I included more detail on the PDF I need. I do not think it is trivial. Will you look a little deeper? Thanks $\endgroup$ – Paulo Andrade Jun 20 '10 at 15:42
OK, the cumulative distribution function (CDF) is $$ P(X \leq x) = F(x) = \begin{cases} 0 & \text{for } x < 0, \\ 9x^2 & \text{for }0 \le x \le 1/3, \\ 1 & \text{for }x > 1/3. \end{cases} $$ and the probability density function is $$ f(x) = F'(x) = \begin{cases} 0 & \text{for }x < 0, \\ 18x & \text{for }0 < x < 1/3, \\ 0 & \text{for }x > 1/3. \end{cases} $$ This is continuous in the sense that the CDF is continuous, and also in the stronger sense that 100% of the probability comes from integrating a density function (absolute continuity with respect to Lebesgue measure). The mode is 1/3. The mean is less than 1/3. The right "tail" runs from 1/3 to 1 only because you've declared (0,1) the interval of interest by fiat. Or if you like, take $(1  10^{100})$ times this density and add $10^{100}$ times the uniform density on the interval $(1/3,1)$. Then the support of the distribution is $[0,1]$ so it's not just by fiat.

$\begingroup$ SorryI meant $10^{100}$, not $10^{+100}$. $\endgroup$ – Michael Hardy Jun 14 '10 at 2:32

$\begingroup$ You can make the PDF continuous (as Paulo Andrade wanted) by smoothing the jump discontinuity slightly, e.g., by convolving your function with a continuous bump of mass 1 supported on $(0,\epsilon)$. $\endgroup$ – S. Carnahan♦ Jun 14 '10 at 4:03

$\begingroup$ Oh. I assumed he meant a continuous distribution, which usually means either that the CDF is continuous, or that the CDF is absolutely continuous (in which last case the PDF is all you need). But he did say a continuous PDF. $\endgroup$ – Michael Hardy Jun 14 '10 at 14:04