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The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim \mathrm{Bin}(n,y)$. Put $\Delta =W/n-y$ and $q=y(1-y)$. Direct binomial moments giveThe fourth centered moment of a binomial distribution implies $\mathbb{E}|\Delta {|}^3\le (\mathbb{E}{\Delta }^2\mathbb{E}{\Delta }^4{)}^{1/2}\le C{n}^{-3/2}$. The accelerated generator of the Wright–Fisher binomial sampling chain isFor $f\in C^3([0,1])$, Taylor's formula uniformly yieldsThe initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.

We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded martingale, with predictable square compensator $V_t^n=n^{-1}{\sum }_{k<\lfloor nt\rfloor }{X}_{k/n}^n(1-X_{k/n}^n)$. For bounded stopping times ${\tau }_n\le {\sigma }_n\le {\tau }_n+{\delta }_n$, optional sampling gives $\mathbb{E}|X_{{\sigma }_n}^n-X_{{\tau }_n}^n{|}^2=\mathbb{E}(V_{{\sigma }_n}^n-V_{{\tau }_n}^n)\le ({\delta }_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most ${\delta }_n$. Compact containment is automatic. The Aldous tightness criterion therefore proves tightness.

Furthermore, the fourth-moment bound and a union bound giveEvery subsequential limit is continuous. For each smooth $f$, the discrete compensated processis a martingale. Generator convergence and continuity of any limiting path turn the sum into ${\int }_0^t\frac{1}{2}{X}_s(1-X_s)f^{\prime \prime }(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their martingale identities pass to the limit against bounded continuous functions of past coordinates. A Monotone class theorem argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the martingale problem for $Lf(y)=\frac{1}{2}y(1-y)f^{\prime \prime }(y)$ on $[0,1]$.

To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous martingale with bracket ${\int }_0^t{X}_s(1-X_s)ds$. On an enlargement carrying independent Brownian motion $W$, setIts bracket is $t$, and the Lévy characterization of Brownian motion makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $d{X}_t=\sqrt{X_t(1-X_t)}d{B}_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting martingale problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. ConsequentlyThe limit is the Wright–Fisher diffusion. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded martingale identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.

An urn with $j$ red balls among $n$ is sampled independently $n$ times with replacement, and the next generation's red count is the number of red draws. Its proportion is a bounded discrete-time martingale. Accelerating generation time by $n$ gives a Wright–Fisher diffusion limit: the one-step variance of the proportion is $x(1-x)/n$, and the higher moments make Taylor remainders negligible.