# Sum involving hypergeometric function

I think that the following equality holds:

$$\alpha=\sum_{x=0}^\infty \frac{x-C}{y-C}\binom{x+y}{x}\frac{(1-\alpha)^{x+y}(1+C)^yC^x}{(1+(1-\alpha)C)^{x+y+1}} {}_2F_1\left[-x;-y;-(x+y);1-\frac{\alpha}{(1-\alpha)^2 (1+C)C}\right]$$

where $y\neq C$ is a non-negative integer, $C>0$ is a real number, $0<\alpha<1$, and ${}_2F_1[a;b;c;z]$ is a hypergeometric function.

Note that $p(x)=\binom{x+y}{x}\frac{(1-\alpha)^{x+y}(1+C)^yC^x}{(1+(1-\alpha)C)^{x+y+1}} {}_2F_1\left[-x;-y;-(x+y);1-\frac{\alpha}{(1-\alpha)^2 (1+C)C}\right]$ is a probability distribution. The expression I have is essentially an expectation of an estimator $\hat{\alpha}=\frac{x-C}{y-C}$ for $\alpha$. I have a reason to believe that it is unbiased, however, I am having hard time proving it.

Using the definition of the hypergeometric function, I can get rid of hypergeometric function by re-writing the expression for $p(x)$ as follows:

$$p(x)=\frac{\sum_{i=0}^{\min(x,y)}\left(\frac{(1-\alpha)C}{1+(1-\alpha)C}\right)^{x-i}\left(\frac{(1-\alpha)(1+C)}{1+(1-\alpha)C}\right)^{y-i}\left(\frac{\alpha-(1-\alpha)^2(1+C)C}{(1+(1-\alpha)C)^2}\right)^{i}\binom{(x-i)+(y-i)+i}{(x-i),~(y-i),~i}}{1+(1-\alpha)T}$$

where $\binom{a+b+c}{a,b,c}$ is a multinomial. This seems to be a kind of a marginal over a multinomial distribution, however, I can't find intuition for it... and it doesn't seem helpful in showing the equality above.

I am thus stuck and would appreciate any pointers.

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