Sorry for the vague question, I'm not sure how to make this more specific.

Let $\sigma(x)$ denote the logistic function $\frac{1}{1 + e^{-x}}$, and $f(y)$ denotes the density of some random variable. I am capable of determining the following function: $$ h(y) = \sigma(\alpha + \beta y) f(y). $$ That is, I know $h(y)$ for all $y$. I am interested in whether I am capable of recovering $(\alpha, \beta, f)$ from this information.

My gut says "no" but I'm not strongly convinced either way. I think one approach is to ask if $$ \sigma(\alpha + \beta y) f(y) = \sigma(a + by) g(y), $$ implies that $\alpha = a, \beta = b, f = g$. Since $g$ must integrate to $1$, what I need is $$ \int \frac{\sigma(\alpha + \beta y)}{\sigma(a + by)} f(y) \ dy = \int \frac{h(y)}{\sigma(a + by)} \ dy= 1, $$ to have multiple solutions to show that I can't recover $(\alpha, \beta, f)$; by assumption, I know it has at least one. Since this seems like it might depend on the form of $h(y)$, conditions under which this is possible are also desirable.


Let $e^{-\alpha} = A$. Then

$$f(y) = (1+Ae^{-\beta y})h(y)$$ So, $$\int_0^{\infty}f(y)dy = \int_0^{\infty}(1+Ae^{-\beta y})h(y) dy = 1 $$

This will give you a relation of the form $H_1(A,\beta) = 1$. In answer to your original query, whether this will have multiple solutions, consider any general $\beta >0$ $$\int_0^{\infty}h(y)dy = I_h, \int_0^{\infty}e^{-\beta y}h(y)dy = I_{b}$$

So, $$I_h+e^{-\alpha}.I_b = 1$$ $$\alpha = \ln\left(\frac{I_b}{1-I_h}\right)$$

It is easy to see that $I_h<1$. So, for every possible $\beta$, there will be an $\alpha$ such that $(\alpha,\beta)$ satisfies the relation. So, there are infinite solutions.

If you have any other information about $f(y)$, say the probability of the r.v. lying between $y_1$ and $y_2$ is $K$, then you can evaluate both $\alpha,\beta$.

$$\int_{y_1}^{y_2}f(y)dy = \int_{y_1}^{y_2}(1+Ae^{-\beta y})h(y) dy = H_2(A,\beta)= K $$

A good approximation can be obtained if you have some upper limit on $\alpha,\beta$. In that case, select $y_1,y_2$ such that $Ae^{-\beta y}$ is negligible in the interval $[y_1,y_2]$. Then, $$\int f(y)dy \approx \int h(y)dy$$ in $[y_1,y_2]$. This way, you can get a good approximation of $\alpha,\beta$.

Even if the $(1+Ae^{-\beta y})h(y)$ is hard to analytically evaluate, you can frame it as a numerical minimization problem $$\min_{(A,\beta)\in D} (H_1(A,\beta)-1)^2 +(H_2(A,\beta)-K)^2$$ where $D$ is a suitable region for $(A,\beta)$. Once $\alpha,\beta$ is known, $f(y) = \frac{h(y)}{\sigma(\alpha,\beta)} $.

If you have no other information about $f$, the best you can do is to determine it to any degree of precision by a polynomial approximation. Let $$f(y) = \sum_0^N a_iy^i$$ for some large $N$.

Then, $$h(y) = \frac{\sum a_iy^i}{1+Ae^{-\beta y}}$$ Evaluating the above relation at a suitably large number of points will give you a system of equations in $A,\beta,a_i$ which can be numerically solved.

Hope it helps.

  • $\begingroup$ Thanks. I should have noticed $1/\sigma$ was additive, but I was thinking in my head of functions other than the logistic as well. $\endgroup$ – guy Jan 25 '13 at 22:08
  • $\begingroup$ I don't think there can be a general way to determine general functions from their product, at least in polynomial time. The best you can do is to find their approximation ny polynomials or some other basis. $\endgroup$ – dexter04 Jan 27 '13 at 7:51

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.