Modeling this stochastic process I'd like to model a process where:


*

*at time $t = 0$, there is one organism with characteristics $X_1$. It chooses another organism. This organism's characteristics $X_2$ are drawn from a distribution $F$, and the expected value of the distribution is $X_1$. 

*at time $t = 1$, these two organisms together select a third individual from a distribution $F$, and the expected value of the distribution is the mean of organism 1 and 2's characteristics, $\frac{X_1 + X_2}{2}$. 

*...

*at time $t$, the existing $t+1$ organisms, who have characteristics $X_1,..,X_{t+1}$, respectively, choose a $(t+2)$th individual from a distribution $F$, and the expected value of the distribution is the mean of existing organism's characteristics, $\frac{\sum_{i=1}^{t+1} X_i}{t+1}$. 


For simplicity we can say $F$ is the normal distribution.
Does this resemble any named stochastic process? 
 A: It's probably easier to look at your process in terms of the means $Y_t = \frac1{t+1}\sum_{i=1}^{t+1} X_i$.  Then $Y_{t+1} = Y_t + \frac1{t+2}W_{t+1}$, where the random variables $W_t$ are i.i.d. with a zero mean.
In particular, this implies that the process $(Y_0, Y_1, Y_2, \dots)$ is a martingale.  In particular, if the distribution of the increments $W_t$ (i.e. your $F$ distribution) is Gaussian, then the only difference between your process and an unbiased Gaussian random walk is the $\frac1{t+2}$ scaling of the increments, which should lead to some kind of anomalous diffusion. 
Specifically, let the variance of the unscaled increments $W_t$ be $\sigma^2$.  Then $Y_t$ has the variance $$\sigma_t^2 = \sum_{\tau=1}^{t+1} \frac{\sigma^2}{\tau^2} = \sigma^2 \sum_{\tau=1}^{t+1} \frac1{\tau^2} = H^{(2)}_{t+1} \sigma^2,$$ where $H^{(2)}_n$ is the $n$-th generalized harmonic number with exponent 2.  In particular, as $t$ increases, the variance of $Y_t$ converges to the limit $$\lim_{t \to \infty} \sigma^2_t = \frac{\pi^2}6 \sigma^2.$$  This implies that the means $Y_t$ will also converge in distribution to some limiting distribution with finite variance (assuming that $\sigma^2$ is finite to begin with).  In particular, if the increments $W_t$ are Gaussian, then $Y_t$ will also have a Gaussian distribution, and thus (since a Gaussian distribution is uniquely defined by its mean and variance), the distribution of $Y_t$ will converge to $\mathcal N(0, \frac{\pi^2}6 \sigma^2)$.
