Solve equation for a function as solution

Question

I need to solve the following function:

$$ h(z) = \sum^{40}_{k=0} \frac{k \, p(k)}{z- k \, h(z)} $$ where $p(k)$ is a probability distribution (in this case binomial with n = 40). Note that $h(z)$ is also inside the sum and also that $z$ is inside. I've been struggling to solve this function. I'm sortof interested in the case in which this sum is solvable but I am sure of certain cases in which the expression is exactly solvable. Also Mathematically one can deduce that the solution will have an imaginary part. An example could be for a binomial distribution as following. I think I didn't define well here that h[z] is actually a function.

p[n_, k, pkans_] := Binomial[n - 1, k]*pkans^k*(1 - pkans)^(n - 1 - k);
s = Solve[h[z] == Sum[(k*p[40, k, 0.6])/(z - k*h[z]), {k, 0, 40}],h[z]]

My question would be how I could try to use Mathematica to solve this equation? I am not sure how to use the solve command for functions. Also I would like to extend solving these equations to other probability distributions and to continuous distributions as well.

Regards,

Note further that $h(z)$ is in the denominator dependent on $z$ as well as $h(z)$ .

Answer 1

$$ h(z) =\sum _{k=0}^n \frac{k\, p(k)}{z-k\, h (z)}$$

Observe that the rhs of the equality is the expectation of the random variable $u = k/ (z- k\, h(z))$, where $k$ is a Binomial random variable with parameters $n$ and $ρ$. So we can try TransformedDistribution and Expectation or NExpectation to see if we get anything useful:

n = 10;
ρ = 6/10;

tdist = TransformedDistribution[k/(z - k h[z]), 
           Distributed[k, BinomialDistribution[n, ρ]]];

rhs = Simplify[Expectation[u, Distributed[u, tdist]]]

sols = h[z] /. FullSimplify[Solve[h[z] == rhs, h[z], Reals], Assumptions->{z > 0}]

Plot[Evaluate[sols], {z, 0, 10}]