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Let $(\Omega, \Sigma, \mathbb{P})$ be a probability space, $X$ be a random variable, and $E \in \Sigma$ be an event with $\mathbb{P}(E) = P$. Then $P - P \log(P) \in [0, 1]$, for all $P \in (0, 1]$, and so can be taken as a probability (the $\log$ is natural). Is there an intuitive interpretation for how such a probability arises?

Here are some things I have already noticed:

Let $F : (0, 1] \to \mathbb{R}$ be such that $F(P) = P - P \log(P)$. Then $F$ is a cumulative distribution function (see a plot), with a probability density function $f : (0, 1] \to \mathbb{R}$ such that $f(P) = -\log(P)$. Such a distribution is called by Mathworld a continuous logarithmic distribution (not to be confused with the discrete log-series distribution).

Here are some of my attempts using the Taylor-series expansion of $\log$ around $1$. First,

$F(P) = P + (1 - P) \sum_{i = 1}^{\infty} \frac{1}{i} P (1 - P)^{i - 1} = P + (1 - P)\mathbb{E}(1/Y)$,

where $Y \sim \textrm{Geometric}(P)$. I have trouble interpreting this because I don't know how to interpret the inverse of $Y$.

On the other hand,

$F(P) = P + (1 - P) \sum_{i = 1}^{\infty} \frac{1}{i^2} \binom{i}{1, i-1} P (1 - P)^{i - 1} = P + (1 - P) \mathbb{E}\left[\binom{I}{1, I - 1} P (1 - P)^{I - 1}\right]$,

where $I \sim \textrm{Zeta}(2)$ and $\binom{i}{1, i-1} P (1 - P)^{i - 1}$ is the probability that in a sample $X_1, \dots, X_i$ of $X$ the event $E$ is realized exactly once. I have trouble interpreting this because I don't know how the zeta distribution arises and why it is connected to this probability.

Any insight appreciated!

Edit:

Having thought about the problem for some days now, I have formed the following initial conjecture:

A function $f : S \to [0, 1]$, where $S \subset [0, 1]$, can be given an intuitive interpretation if and only if $f$ is a polynomial with integer coefficients. This is equivalent to that $f(P)$ can be computed by a finite number of $P$-coin tosses. A special case of this is that if $f(\mathbb{Q} \cap S)$ contains an irrational number, then $f$ does not have an intuitive interpretation.

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  • $\begingroup$ I think I have understand your question in a wrong manner. How do you say that $P-P\log{P}$ is positive and less than $1$? for example for $P=.75$, it is equal to $-0.31$. $\endgroup$ – CLAUDE Oct 19 '14 at 17:02
  • $\begingroup$ It should be in the range [0,1]: see the plot in wolframalpha.com/input/?i=P-P*log%28P%29+on+%5B0%2C+1%5D $\endgroup$ – kaba Oct 19 '14 at 17:29
  • $\begingroup$ Isn't $0.75$ in that interval? $\endgroup$ – CLAUDE Oct 19 '14 at 18:20
  • $\begingroup$ @Amir: According to Mathematica, 0.75 - 0.75*Log[0.75] = 0.965762. Could you have calculated 0.75 * Log[2, 0.75] = -0.311278? $\endgroup$ – kaba Oct 19 '14 at 18:27
  • $\begingroup$ What is the base of the logarithm that you are calculating? if it is in base 2 then it is equal to $.75-(-.31)=1.06$ which is outside of that interval $\endgroup$ – CLAUDE Oct 19 '14 at 18:33

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