Given $Y_t$ price of a stock (no-dividents), and a derivative paying $Y_T^2$ at maturity $T$, evaluate the price of the instrument now using risk-neutral approach and check that it satisfies Black&Scholes PDE.

My attempt $$V_0 = \exp(-rT)\cdot E(Y_T^2|F_0) = ... = \exp(-rT)\cdot(\sigma^2 + \mu^2)$$

Unfortunately, this result doesn't satisfy the PDE.

  • 2
    $\begingroup$ Under BS the price is then given by $$Y_t = S_0 e^{rt} \exp( -\frac{1}{2}\sigma^2 t + \sigma W_t) := F(0,t) \mathcal{E}[ \sigma W_t ]$$ where $\mathcal{E}(X_t)$ figures the stochastic exponential of the process $X_t$. Therefore, $$Y_t^2 = F(0,t)^2 \exp( -\sigma^2 t + 2 \sigma W_t ) = F(0,t)^2 e^{\sigma^2 t} \mathcal{E}(2 \sigma W_t) $$ by completing the squares. At the end of the day, the price is then (cf. arb-free pricing theory + properties of the stochastic exponential) $$V_0 = e^{-rT} \Bbb{E}_0[ Y_T^2 ] = S_0^2 e^{(r+\sigma^2)T}$$ $\endgroup$
    – Quantuple
    Oct 20 '21 at 11:53
  • $\begingroup$ @Quantuple, I decided to check the solution and substitute it into the PDE (I'm not sure if it's correct but I replaced $V, S, t$ by $V_0, S_0, T$ in the classic formula). The equality holds only if there is a minus in the power - didn't you forget it? $\endgroup$
    – student
    Oct 21 '21 at 16:35
  • 1
    $\begingroup$ Hi @student, the extension of $$V(t=0,S(0)=S_0) = S_0^2 e^{(r+\sigma^2)T}$$ to the BS PDE quantity $V(t, S(t) = S)$ is rather $$ V(t, S) = S^2 e^{(r+\sigma^2)(T-t)} $$ The "minus" sign comes from there. Plugging this into the PDE shoudl be OK $\endgroup$
    – Quantuple
    Oct 22 '21 at 9:21

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