# Probability distribution needed [closed]

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 ZaimiJun 18 '10 at 16:22

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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? – Harald Hanche-Olsen Jun 13 '10 at 23:42
Sorry, I'm voting to close. It is straightforward to specify a piecewise linear function that does the job. – S. Carnahan Jun 14 '10 at 1:58
I included more detail on the PDF I need. I do not think it is trivial. Will you look a little deeper? Thanks – 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.
Sorry---I meant $10^{-100}$, not $10^{+100}$. – Michael Hardy Jun 14 '10 at 2:32
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 $(0,\epsilon)$. – S. Carnahan Jun 14 '10 at 4:03