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For $v>0$ and $m>0$, let $\Phi$ and $\varphi$ be the standard normal distribution function and density, and putThe exercise event is $Z>(\log m+v/2)/\sqrt{v}=-d_2$. Complete the square in the first truncated expectation:The probability of exercise is $\Phi (d_2)$. Hence the normalized Black-Scholes call function isAt $v=0$ it is $(1-m{)}^{+}$. If $m\le 0$, the positive underlying always exceeds $m$, so $F(v,m)=1-m$; the reciprocal identity below is understood on its natural domain $m>0$.

For the reciprocal strike, $d_1(v,1/m)=-d_2(v,m)$ and $d_2(v,1/m)=-d_1(v,m)$. ThusThereforeThe $v=0$ case follows either by continuity or by checking the two positive parts directly.

For the stock model, the original probability measure is already risk-neutral: $e^{-rt}{S}_t=S_0{e}^{\sigma W_t-{\sigma }^{2}t/2}$ is a true martingale. Write $\tau =T-t$. Conditional on ${\mathcal{F}}_t$, the stock at maturity iswith $Z$ independent of current information and standard normal distribution. ThereforeEquivalently,which is a true martingale, since the payoff is integrable. The sufficient direction of the fundamental theorem of asset pricing says that an equivalent measure making all discounted traded prices local martingales excludes arbitrage by admissible trading strategies. The original measure works simultaneously for the bank account, stock and call. Thus the call-augmented market has no arbitrage.

For the put, hold one call, short one share, and hold $K{e}^{-rT}$ units of the bank account. This self-financing portfolio has current value $C_t-S_t+K{e}^{-r\tau }$ and terminal payoffIts discounted value is also the conditional expectation of that payoff. Thus the put-call parity priceadds another discounted martingale to the same market, proving no arbitrage with both call and put included.

Now put $m=K{e}^{-r\tau }/S_t$. The reciprocal identity givesThe resulting Put-call symmetry in the Black-Scholes model isThe final formula printed in the PDF has two errors: its discount-rate symbol is undefined, and the reciprocal argument has the wrong sign in the interest-rate exponential. Both are resolved by the displayed calculation. Even replacing the undefined symbol by $r$ leaves an incorrect expression with $e^{-r\tau }$ inside $F$. For $\tau >0$, $r>0$ and $\sigma >0$, $F(v,m)$ is strictly decreasing in $m$, because ${\partial }_{m}F=-\mathbb{P}(e^{-v/2+\sqrt{v}Z}>m)<0$. The printed argument is smaller than the reciprocal argument and hence gives a strictly larger put price. Thus the literal printed symmetry cannot be established; the boxed expression is the corrected identity.

Write $h=T-t$. Conditional on ${\mathcal{F}}_t^S$, independent increments of the Brownian motion givewhere $Y$ has the standard normal distribution and is independent of ${\mathcal{F}}_t^S$. Factoring $S_t$ out of the positive part identifies the normalized Black-Scholes call function:The European call option payoff is integrable, because $0\le (S_T-K{)}^{+}\le S_T$ and $\mathbb{E}{S}_T=S_0$. The tower property of conditional expectation therefore proves that $C$ is a true martingale. At $t=T$, the boundary value $F(0,m)=(1-m{)}^{+}$ gives $C_T=(S_T-K{)}^{+}$. The answer is the conditional payoff martingale

For the normalized Black-Scholes call function, completing the square gives $F(v,m)=\Phi (d_1)-m\Phi (d_2)$, with $d_1=(-\log m+v/2)/\sqrt{v}$ and $d_2=d_1-\sqrt{v}$. The identities $d_1(v,1/m)=-d_2(v,m)$ and $\Phi (-x)=1-\Phi (x)$ implyFor a stock with risk-neutral drift $r$, maturity difference $\tau$, and positive strike $K$, put-call parity then givesThe positive sign of $r\tau$ in the reciprocal argument is forced by taking the reciprocal of $K{e}^{-r\tau }/S_t$. The identity includes $v=0$ by continuity and requires $m>0$.