Let $F(t,s)$ be the ordinary Black-Scholes price of the terminal claim $G(S_T)$. Denote the new asset by $U_t=a(t)S_t^2$. Since it is strictly increasing in $S_t>0$, invert it as $s=\sqrt{U/a(t)}$ and write $H(t,U)=F(t,\sqrt{U/a(t)})$. Its derivative isThe replicating portfolio in the new asset and bank account therefore holdsThe wealth is $F$. The diffusion coefficient of $\Gamma dU$ is $2\sigma U\Gamma =\sigma S{F}_s$, agreeing with that of $dF$. Its drift together with the bank holding iswhich equals the drift of $F$ by its Black-Scholes equation. This verifies self financing and terminal replication. Thus the square-payoff claim has nonzero volatility and can replace the original stock as the traded risky asset.

The stochastic differential computed above isThus this positive no-dividend traded claim behaves as a new geometric Brownian motion asset withThe transformed drift need not be positive, even though the three original market parameters were assumed positive. Under the risk-neutral measure it becomes $r$, as it must for a non-dividend traded asset.

Under the risk-neutral measure, the same second-moment formula uses drift $r$. Discounting the power option payoff therefore givesLet $a(t)=e^{(r+{\sigma }^2)(T-t)}$ and $B_t=e^{rt}$. The delta hedging portfolio holdsIts wealth is ${\Delta }_t{S}_t+{\beta }_t{B}_t=a(t)S_t^2=V_t$. To verify self financing, Itô's formula givesAt maturity the wealth is $S_T^2$, completing replication.

Applying Itô's formula to $\log S$ gives the geometric Brownian motionThe Gaussian exponential moment $\mathbb{E}{e}^{a{W}_t}=e^{a^{2}t/2}$ then yieldsThis expectation is under the physical measure and therefore uses drift $\mu$.

The zero means and orthogonality give the exact second momentThus $A_N\to 0$ in $L^2$. Chebyshev's inequality gives the requested weak law of large numbers:Independence is unnecessary; the stated second-moment orthogonality is sufficient.

For an explicit almost surely convergent subsequence take $N_j=2^j$. For each positive integer $m$,The First Borel-Cantelli lemma makes only finitely many such exceedances occur. Intersecting the resulting probability-one events over $m$ gives $A_{2^j}\to 0$ almost surely.