# Given a product of these two functions, can I recover the factorization?

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.

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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.

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Thanks. I should have noticed $1/\sigma$ was additive, but I was thinking in my head of functions other than the logistic as well. – guy Jan 25 '13 at 22:08
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. – dexter04 Jan 27 '13 at 7:51