Assume a variable $x_0>0$ with log-normally distributed noise, such that the observation $x$ of $x_0$ has the following PDF:

$$ p(x\mid x_0) = \frac{1}{\sqrt{2\pi}\sigma x}e^{-\frac{\left(\ln{(\frac{x}{x_0})} - \sigma^2\right)^2}{2\sigma^2}} $$

(NB: $x_0$ thus corresponds to the mode of the log-normal distribution, following the reparameterization $\mathrm{mode}=e^{\mu-\sigma^2}$)

As it turns out, I cannot measure $x$ directly, but only a transformation $y(x)$:

$$ y(x) = (1-\delta)\tanh(\beta x) + \delta $$

The parameters $\delta$ and $\beta$ are not relevant to my question, but it can be assumed $0\le\delta\le1$ and $\beta>0$.

Now I have a measurement $y_m$ and I want to compute the probability that a latent variable $x_0$ would generate a $y$ within the window $[y_m-\epsilon; y_m+\epsilon]$.

I have two questions related to this:

  1. What is the PDF for $y$ that would allow me to compute this probability? (Note that for reasons not mentioned here, I want to avoid transforming $y_m$ to $x$-space)
  2. More pragmatically, the scientific programming module I am using (scipy.stats.lognorm) has CDF implementations for the log-normal distribution. If instead of the above $\tanh$ transformation I had the identity transformation $y(x)=x$, I could simply use something along the lines of lognorm(x0,σ).cdf(y_m+ε) - lognorm(x0,σ).cdf(y_m-ε). Can I still use this lognorm CDF implementation after some appropriate transformation?
  • $\begingroup$ If $y(x)$ is observed/known, so is $x$ (for a given $(\delta,\beta)$ $\endgroup$
    – Xi'an
    Sep 7 '20 at 12:56
  • $\begingroup$ Yes, that's correct, but as mentioned in 1., I want to avoid back-transforming $y_m$ to $x$-space (the reason is that I have other y(x) that are not so easily invertible). I'm starting to think though that the PDF of $y$ will require computing $y^{-1}$. $\endgroup$
    – monade
    Sep 7 '20 at 13:04

The hyperbolic tangent function $\tanh$ is a strictly increasing function, so it is quite simple to get the CDF of the random variable $Y$. I am going to give a more general answer that what you are looking for --- specifically, I will not make any assumption about the distribution $X$, and I will not assume that this is a non-negative random variable. Using the stipulated transformation with $\beta$ and $0 \leqslant \delta < 1$ you have:

$$\begin{align} F_Y(y) &\equiv \mathbb{P}(Y \leqslant y) \\[12pt] &= \mathbb{P}((1-\delta) \tanh (\beta X) + \delta \leqslant y) \\[6pt] &= \mathbb{P} \bigg( \tanh (\beta X) \leqslant \frac{y-\delta}{1-\delta} \bigg) \\[6pt] &= \mathbb{P} \bigg( \frac{e^{2 \beta X}-1}{e^{2 \beta X}+1} \leqslant \frac{y-\delta}{1-\delta} \bigg) \\[6pt] &= \mathbb{P} \bigg( e^{2 \beta X}-1 \leqslant \frac{y-\delta}{1-\delta} \cdot (e^{2 \beta X}+1) \bigg) \\[6pt] &= \mathbb{P} \bigg( e^{2 \beta X} \cdot \frac{1-y}{1-\delta} \leqslant 1 + \frac{y-\delta}{1-\delta} \bigg) \\[6pt] &= \mathbb{P} \bigg( e^{2 \beta X} \cdot \frac{1-y}{1-\delta} \leqslant \frac{1+y-2\delta}{1-\delta} \bigg) \\[6pt] &= \mathbb{P} \bigg( e^{2 \beta X} \cdot (1-y) \leqslant (1+y-2\delta) \bigg). \\[6pt] \end{align}$$

Taking account of the limits of the function you then have:

$$\begin{align} F_Y(y) &= \begin{cases} 0 & & \text{if } y \leqslant 2\delta-1, \\[6pt] F_X \bigg( \frac{1}{2 \beta} \cdot \log \Big( \frac{1+y-2\delta}{1-y} \Big) \bigg) & & \text{if } 2\delta-1 < y < 1, \\[6pt] 1 & & \text{if } y \geqslant 1, \\[6pt] \end{cases} \\[6pt] \end{align}$$

It is easily shown that:

$$\begin{align} \frac{d}{dy} \log \Big( \frac{1+y-2\delta}{1-y} \Big) &= \frac{1}{1+y-2\delta} + \frac{1}{1-y} \\[6pt] &= \frac{(1-y)+(1+y-2\delta)}{(1+y-2\delta)(1-y)} \\[6pt] &= \frac{2(1-\delta)}{(1+y-2\delta)(1-y)}. \\[6pt] \end{align}$$

Therefore, differentiating the CDF with respect to $y$ gives the corresponding density:

$$f_Y(y) = \frac{(1-\delta)}{(1+y-2\delta)(1-y) \beta} \cdot f_X \bigg( \frac{1}{2 \beta} \cdot \log \Big( \frac{1+y-2\delta}{1-y} \Big) \bigg) \quad \quad \text{for } 2\delta-1 < y < 1.$$

Substitution of the density function for $X$ will give you the final form. Note that in your problem you have assumed that $X$ is a non-negative random variable, which leads to the effective bound $y \geqslant \delta$. This bound emerges correctly from substitution into the above equation in the case where $f_X$ has support only over positive argument values.

  • 1
    $\begingroup$ Thanks! To confirm my understanding of your answer: this essentially corresponds to solving $y_m = (1-\delta)\tanh(\beta x_m)+\delta$ for $x_m$ and then computing the CDF of $x_m$ in x-space using the known $p(x\mid x_0)$ (using the terminology of my OP)? $\endgroup$
    – monade
    Sep 7 '20 at 15:00
  • $\begingroup$ I could not make sense of your $p(x|x_0)$ so I did not do that last step. But yes, that is essentially the method. $\endgroup$
    – Ben
    Sep 7 '20 at 22:33
  • $\begingroup$ Ok, thanks. $p(x|x_0)$ is the log-normal distribution following the reparameterization $\mu = \ln{(x_0)}+\sigma^2$, where $x_0$ is the mode of the log-normal. $\endgroup$
    – monade
    Sep 8 '20 at 9:36
  • $\begingroup$ Elegant answer (+1) $\endgroup$
    – BruceET
    Sep 9 '20 at 7:43
  • 1
    $\begingroup$ My answer is more general than that --- it does not assume that $X$ is non-negative. However, if you substitute any $y < \delta$ into the final equation you get $\log (\tfrac{1+y-2\delta}{1-y}) < 0$ so the argument for the density $f_X$ is negative. Thus, if $X$ is non-negative then you get zero density for $Y$ for all $y<\delta$. I have updated my answer to specify my assumptions more clearly. $\endgroup$
    – Ben
    Sep 15 '20 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.