# Fredholm Integral in Bayesian Appliation

Let $X = x_1, x_2, \ldots, x_n$ be a sequence of Bernoulli random variables with $k$ successes. Suppose that, given $X$, the posterior predictive probability of $x_{n+1} = x$ is known to be $g(x)$ where $$g(x = 1) = \int_0^1 \theta^{k+1}(1-\theta)^{n-k} f(\theta) \ d\theta$$ for an unknown prior $f(\theta)$.

If we define $K(X,\theta) = \theta^{k+1}(1-\theta)^{n-k}$, then we can write the above integral equation as $$g(X) = \int_0^1 K(X,\theta)f(\theta) \ d\theta.$$ At least superficially, this looks like a Fredholm equation of the first kind whose solution is the unknown prior $f(\theta)$.

Such an integral equation has arisen in some work I'm doing and I'd like to be able to solve it. I know how to solve this inverse problem in the case that $K$ is an operator on $L^2([0,1])$ by a felicitous choice of wavelet basis and some regularization procedure. However, it is not at all clear to me that my particular operator is this nice.

For example, $g$ is a function on stochastic processes $X$. It is clear to me that $X$ lives in some product measure space. But, from my understanding, Fredholm operators map between Banach spaces and I do not know whether $g$ lives in such a space.

Is this technically a Fredholm integral? Can I solve it using the method I mentioned above? Any help would be greatly appreciated.

• I don't think this works because densities are not Hilbert or Banach spaces. This is because set of probability densities does not form a vector space; they are not closed under linear combinations because it is possible that $h(x) = f(x) - g(x)$ has negative values for some $x$. Edit: That is also true of distributions, which is more relevant to you question. – Ryan Warnick Dec 16 '17 at 5:37