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
3
-
$\begingroup$ 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 Hanche-OlsenCommented Jun 13, 2010 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 ♦Commented Jun 14, 2010 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 AndradeCommented Jun 20, 2010 at 15:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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.
-
$\begingroup$ Sorry---I meant $10^{-100}$, not $10^{+100}$. $\endgroup$ Commented Jun 14, 2010 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 ♦Commented Jun 14, 2010 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. $\endgroup$ Commented Jun 14, 2010 at 14:04