I hope someone can help cross validate the answer to this. This is not homework; I graduated from university a long time ago.

Let $x$ be a normally distributed RV i.e. $N(m,\sigma^2)$ where $m$ and $\sigma^2$ are the mean and variance respectively.

Apply the transformation $y = \alpha / [1 + \beta e^{-k(x-x_o)}]$ to $x$.

What is the probability density distribution of $y$?

This question has risen from our need to do a sensitivity analysis, where we want to understand the degree of changes in $x$ affecting $y$. Experimentally, we have proven $x$ is normally distributed. We also proved experimentally the transformation.

I had already followed the standard technique for solving this which uses differentiation, which is

$f_Y(y)=f_X(g^{-1}(y)) {d \over dy} g^{-1}(y)$

where $g()$ is the transformation function.

And the answer I get is

$f_Y(y)={1\over \sqrt{s\pi\sigma^s}} e^{- {({1\over k} (\ln{{\beta y}\over {\alpha - y}}) + x_o - m)^2}\over{2\sigma^2}} {{\alpha}\over{ky(\alpha-y)}}$

It would be helpful if someone could check the answer I have developed.

  • 2
    $\begingroup$ This reads as a homework style question, which get slightly different treatment see help center. There's no difficulty here. If you don't already have a formula for dealing with monotonic transformations, then the argument from basic principles is straightforward: converting your random variables to the more conventional upper-case, consider $P(Y\leq y) = P( {\alpha \over 1 + \beta e^{-k(X-x_o)}})\leq y)$. Convert to a statement in $P(X\leq ...)$, which means $F_Y(y)=F_X(\text{...})$. Differentiate back to a density. .... alternatively, edit your question to ask something more particular. $\endgroup$ – Glen_b -Reinstate Monica Sep 27 '17 at 22:35
  • $\begingroup$ Many thanks for your kind assistance. I am familiar with the method you describe which is available in standard text books. $\endgroup$ – BigGuyNeedsHelp Sep 28 '17 at 6:48
  • $\begingroup$ This is what I had done... and I ve expanded my original question to include the answer I had gotten. However, I would very much appreciate it someone can verify the results. $\endgroup$ – BigGuyNeedsHelp Sep 28 '17 at 7:18
  • 1
    $\begingroup$ What is this $s$ in the density formula? Are you confusing $s$ and $2$??? $\endgroup$ – Xi'an Sep 28 '17 at 16:42
  1. Generate $10^4$ simulations from a $\mathcal{N}(.3,.2)$, $x_1,\ldots,x_{10^4}$ ($m=.3,\sigma^2=.2^2$)
  2. Apply the transform $y_i=.2/\{1+.4\exp[-.5(x_i-.6)]\}$ ($(\alpha=.2$, $\beta=.4$, $k=.5$, $x_0=.6$)
  3. Check the histogram of the $y_i$'s fits the density curve$$f_Y(y)= {1\over \sqrt{2\pi\sigma^2}} {\alpha\over ky(\alpha-y)} \exp\left\{- {({1\over k} (\ln{{\beta y}\over {\alpha - y}}) + x_0 - m)^2 \over 2\sigma^2}\right\}$$

result of one draw from the above algorithm

  • 2
    $\begingroup$ +1 There's an art to selecting values for parameters such as $\alpha,\beta,\sigma$, etc. But the simulation can be done so quickly that one can play with any parameters whose effects are not fully understood. Note, too, that apparent agreement between simulation and theory is satisfying, but not conclusive: it is possible for a formula to be incorrect, but nevertheless come so close to being correct for a large range of situations that the error will not be detected by simulation. It helps to have a good idea of the right form the answer should take: any errors are then usually blatant. $\endgroup$ – whuber Sep 28 '17 at 17:06

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