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Theory tells me that random variable $Z$ distributes with known mean $\mu$, over support $(1,\infty)$. Unfortunately, I cannot observe $Z$. However, I do observe $X$, which sample mean is $\bar u$, over support $(0,1)$. Importantly, theory tells me that when $X=0$, $Z=1$, and that when $X=1$, $Z=\infty$.

I am looking for a function $f(X)$ which does two things, simultaneously:

  1. Its range is the support of $Z$. For example:

$$f(X) = \frac{1}{1-X}$$

where it holds that $f(0)=1$ and $f(1)=\infty$.

  1. Transforms the mean of $X$ into that of $Z$. For example:

$$ g(X) = \frac{\mu}{\bar u}X $$

where it holds that

$$ E(g(X) = \frac{\mu}{\bar u}E(X) = \mu = E(Z) $$

I am struggling to find a transformation that does both things simultaneously. Clearly $f(X)$ is not a solution because $E[f(X)]\neq \mu$ (except by chance). Similarly, $g(X)$ has range $(0,\dfrac{\mu}{\bar u})$.

Such transformation (which must be non-linear) will change the distribution of $X$. Because I do not know the distribution of $Z$, I am for the moment not concerned about this.

Any ideas? The combination of functions 1 and 2 cannot do the trick, because to keep the range it must be that, for an arbitrary function

$$ h(X) = \frac{c}{c-dX}$$

in order to keep the range fixed, $d=c$, thereby reducing $h(X)$ to $f(X)$, which we know is not a solution.

I am looking at logarithms, exponentials, and so on, but it seems impossible to get both things at the same time. Maybe I am hitting a wall that has been systematised in a theorem?

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    $\begingroup$ Wikipedia has a detailed page on log and logit transformations of the Beta distribution. While these map [0, 1] to [0, infinity] There's no reason you can't shift the resulting distribution, e.g. the 1+log(U) with U a uniform RV generates an shifted exponential RV with mean 2. $\endgroup$ – AdamO Mar 22 '18 at 16:57
  • $\begingroup$ @AdamO Thanks, but log(x) map into [0,infinity] with the wrong limits. This is, 0 goes into infinity, and 1 into 0. This is contrary to my theoretical restriction. The logit maps into all the real line, so not helpful either. $\endgroup$ – luchonacho Mar 22 '18 at 17:37
  • $\begingroup$ Then do log(1-X) $\endgroup$ – AdamO Mar 22 '18 at 17:40
  • $\begingroup$ @AdamO 1-log(1-X) works in terms of having the same limits than Z, but its mean is fixed. I need to transform it to that of Z. Without a further parameter in the transformation, this is not possible. But any parameter you add will break the limits. Back to nothing. $\endgroup$ – luchonacho Mar 22 '18 at 17:59
  • $\begingroup$ @AdamO Error! Adding a C in front of log gives the constant I need. Then I can adjust C to match any mean. Amazing! Thanks! Do you want to post the answer? $\endgroup$ – luchonacho Mar 22 '18 at 18:15
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Probability theory may lead to a quicker solution than pure mathematics here. Begin with simpler requirements by taking a known random variable and transformation. For example, a uniform random variable $U$ has a [0-1] support, and $-log(1-U)$ is exponentially distributed as $E$ with a [0, infinity] support. Note here that 0 maps to 0 and 1 maps to infinity. The mean of an exponential random variable is 1. As exponentials are part of a scale family, $cE$ has a mean of $c$. A shifted exponential is generated by $1+cE$. Inverting this distribution arrives at $1-\exp(-cE)$ which takes a $\text{Beta}(c, 1)$ distribution. Inverse mapping theorems will allow you to map any RV distributed on [0,1] to the quantiles of the Beta RV with (c,1) parametrization, then after applying the transform, generate a targeted outcome with [1, $\infty$] support and known mean.

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