I do not quite agree with the answer of Wulf Rehder. He has computed only the cumulative distribution function(CDF). The question wants the probability distribution function(PDF), so we need to take a derivative over the CDF.

In the first case, PDF = 1/(1+x)^2 = (1-u)^2

In the second case, PDF = e^x/(1+e^x)^2 = u(1-u)

Hello,

Distribution of log(U/1-U) is the logistic distribution.

I do not quite agree with the answer of Wulf Rehder. He has computed only the cumulative distribution function(CDF). The question wants the probability distribution function(PDF), so we need to take a derivative over the CDF.

In the first case, PDF = 1/(1+x)^2 = (1-u)^2

In the second case, PDF = e^x/(1+e^x)^2 = u(1-u)

Yes, I was not familiar with that nomenclature .. Thanks a lot for pointing out ...

If X = U/(1-U), U ~ unif(0,1), then the pdf of X is given by

f_X(x)=\frac{1}{(x+1)^2}, x>0.

The cdf of X is given as

F_X(t) = \frac{t}{t+1}, t> 0.

dx/(1+x)2 for x belongs to R+

Just another (more nice?) form of the answers:

for the cumulative df we have: G(x) = ( 1 + tanh(x/2) )/2 , x \in R,

for the density df we have: g(x)=G'(x) = ( 2 cosh(x/2) )^{-2}, x \in R.

Just an illustration of probability density function of log (u/(1-u)). Obtained after 10 000 random draws from uniform distribution.

@Marek

It's almost a standard normal, but it has a larger spread.

ANOTHER INTERESTING SIMPLE IDENTITIES (for excercising probability calculus):

Since U is in distribution equal to the ratio e1/(e1 + e2), where e1, e2 are iid exponential, hence U/(1-U) is equal in distribution to the ratio e1/e2.

Since log(e1), log(e2) are iid with double exponential, hence

log(U/(1-U)) is equal in distribution to de1 - de2, where de1, de2 are iid of double exponential distribution.

Since e1 and e2 are equal in distribution to -log(U1) and -log(U2), respectively, hence log(U/(1-U)) is equal in distribution to the difference

log(-log(U1)) - log(-log(U2))=[ - log(-log(U2))] - [ - log(-log(U1))], where U1, U2 are iid uniform on (0,1).

From the above identities some simple formulas for convolution of double exponential distribution are obtainable.

In more general terms, you have an expression (either u/(1-u) or Log[u/(1-u) in this case) that contains a random variable u with a certain distribution, (u is distributed as UniformDistribution[{0,1}] in this case). The expression creates another random variable X that you want to describe with a probability distribution (either a PDF or CDF).

In Mathematica, this can be done in one line using the function TransformedDistribution which takes an expression and a specified distribution for random variables, and returns a distribution for which we can evaluate the CDF at X=x:

CDF[TransformedDistribution[u/(1 - u),

u \[Distributed] UniformDistribution[{0, 1}]], x]

which outputs: x/(1+x) for x>0

So differentiating this CDF gives the PDF as 1/(1+x)^2

Similarly, for the Log version:

CDF[TransformedDistribution[Log[u/(1 - u)],

u \[Distributed] UniformDistribution[{0, 1}]], x]

which outputs: e^x/(1+e^x)

So the PDF is e^x/(1+e^x)^2

We can change the expression and/or the type of distribution to derive new distributions. For example, suppose you independently resample for the two occurrences of the variable u in Log[u/(1-u)]. I.e., the expression is Log[u1/(1-u2)] where u1 and u2 are uniformly distributed. In this case, the CDF is given by:

CDF[TransformedDistribution[

Log[u1/(1 - u2)],

{u1 \[Distributed] UniformDistribution[{0, 1}],

u2 \[Distributed] UniformDistribution[{0, 1}]}], x]

which outputs (1/2)e^x for x0. As we would expect the CDF for resampling is different than the previous CDF that does not resample even though the distributions still uniform. Indeed, the resampling results in lower mean and quantiles for Log[u1/(1-u2)] than those for Log[u/(1-u)]. Intuitively, this makes sense because large absolute values occur when u is close to 0 or close to 1 and these large values would be generally reduced when the two values of u are independently sampled.

@Naresh M. Patel:

The question was about derivation but, of course, the questioner was also interested in final result. Unfortunately, your tool (Mathematica) does not explain how it appears at the final result. Yet this is especially interesting when you allow two (presumably) independent random variables u1 and u2, both uniformly distributed on [0,1].

Thanks to all for the nice discussion.

For U uniform on (0,1), and x >= 0, the distribution function of V :=U/(1-U) is P{U/(1-U)

Your solution satisfies the functional equation f(z) = 1/(z^2)*f(1/z) and is also a density function, but it is not the solution of the original problem. That solution, f(z) = 1/(1+z)^2, satisfies the same functional equation, whose solution is evidently not unique.

Yes, indeed all probability distributions (p.d.-s) of positive random variables Z satisfying the property Z=d 1/Z, i.e. with equality of distribution of Z and 1/Z form a huge family obtainable as follows:

Take ANY p.d. concentrated on (0,1), call it Q1.

Build a p.d. Q2 defined by the condition Q2(a,b) = Q1(1/b, 1/a), for

arbitrary 0

Wulf > In my second offering, above, I wanted to avoid integration and use a more algebraic argument, via a functional equation.

Yes, the integration has been avoided, and the functional equation for the density of U1/U2 is ok, but the density (f(x)=1/2 for 0

Wulf > A polynomial p(z) is often the first trial, but it’s obvious from (#) that any positive power z^n, n > 0, will not work.

But if restricted to (0,1), then it works! Simply apply the last formulas with a=n+1.:

f(x)= (n+1) * xn/2 for 0 You correctly give a whole family of functions, depending on a parameter a, that fulfills (#)

Let me correct this claim, by repeating that the full family of solutions is parametrized by the set of all functions f1 which are a density of a p.d. concentrated on (0,1) [devided by 2]. An example known from the answer to the main question of this thread is given by g(x):=1/(1+x)2 which is not in the family of densities parametrized by a. And it satisfies (#). Indeed:

x-2 * g(1/x) = x-2 * 1/[ 1 + 1/x]2 = x-2 * x2/ (x + 1)2= g(x), for all x>0.

@ Wulf

Thanks for recognizing math! Indeed, I am a mathematician with emphasis upon probability, analysis, approximation and functional equations and their application. But still I cannot indicate one book with something like tabel with types. Seemigly those found by google under "functional equations pdf" would be sufficient. The one you have refered t is a good introduction to the subject. The advanced works are easily found via popular browsers. One of a good survey paper is the following

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.467.1047&rep=rep1&type=pdf. A very very special pack of results on regulaly varying solutions to linear equations are put in my dissertation "Regularly varying .." available among my contribution listed on RG.

You have written> ... you gave a whole family of solutions parameterized by a

I had decided to correct it publicly as it could be understood wrongly, that this is the whole family. What you wanted to say was probably the following: ... you gave a whole family of solutions of the form xp for 0

This follow logistic distribution......