Pricing of European Power Call Option via Black-Scholes formula: reasoning?

I want to price a European Power Call option (without dividend yield) with payoff $\max\{S_T^2-X, 0\}$, where $T$ is the maturity and $X$ the strike. Let $(S_t)_{t\ge 0}$ be the price process of an underlying with dynamics

$dS_t=S_t(r dt + \sigma dW_t)$.

So first I observed that $dS_t^2=2S_tdS_t+d\langle S\rangle_t=S_t^2\left((2r+\sigma^2)dt+2\sigma dW_t\right)$

So $\tilde{S}_t$ is a BS-stock with volatility $\tilde{\sigma}=2\sigma$ and interest rate $\tilde{r}=2r+\sigma^2$. Then I applied the BS-formula for option pricing which gives a call price.

But this leads to the wrong result. As i looked it up the right way is to choose $\tilde{\sigma}=2\sigma$, $\tilde r = r$ and a dividend $q=-(\sigma^2+r)$.

Now my question is: What is the reasoning behind this choice? Why is the price of a European Power Call option without dividend yield derived by using Black-Scholes formula with dividends? That doesn't make sense to me. And why did my straightforward approach lead to a wrong result?

• Possible duplicate of How to derive the price of a square-or-nothing call option? – LocalVolatility Jul 15 '17 at 16:20
• I think my question is not a duplicate of this link. I modified my question maybe now my problem is a bit clearer. – lemontree Jul 16 '17 at 15:59
• Your approach is wrong, because when directly applying the B/S formula with $\tilde{r}$ and $\tilde{\sigma}$, you use $\tilde{r}$ not only for the drift but also for discounting which should still be $r$. Don't get hung up on the term "dividend yield". It just happens that the correction that you need to get the both the correct drift and discount factor is equivalent to a B/S formula with the given dividend yield. – LocalVolatility Jul 16 '17 at 16:22