# Call Option on the Square of a Log-Normal: Process of Underlying under Stock Measure and Risk Neutral Measure

I'm working on some quant interview questions from the book called Quant Job Interview Questions And Answers (by Mark Joshi and other authors).

Here are the questions from the bookd, and the answers come up by myself. Since the book does not have answers for those questions, I want to confirm with you if my idea is correct(ex. if I used the correct theorem and if the overall process is correct), but there is no need to check if my computation is correct since I don't want to waste you too much time on those:)

Suppose $$S$$ follows a log-normal Brownian motion

1. What is the process of $$S^2$$ in the stock measure?

To begin with, I think the stock measure of $$S$$ is: $$dS = μSdt + σSdW$$ In order to get stock measure for $$S^2$$, we need to apply Ito's formula to compute $$d(S^2)$$, here I omit the procedure of applying Ito's formula since I'm confident about the result (actually it is an answer provided by the interview book): here I use $$Y$$ to denote $$S^2$$, so $$dY = (2μ+σ^2)Ydt + 2σYdW$$

1. What is the process of $$S^2$$ in risk neutral measure?

Since a risk-neutral probability is a probability measure under which the discounted risky asset $$S/B$$ is a martingale, we derive Girsanov Kernal $$φ$$ for Girsanov transformation $$dW(stock measure) = φdt + dW(risk neutral measure)$$, which is $$φ = (r-μ)/σ$$ (here again I omit the derivation process), so for $$dY = (2μ+σ^2)Ydt + 2σYdW$$, we replace its $$dW$$ under stock measure with $$dW(risk neutral measure)$$: $$dY = (2μ+σ^2)Ydt + 2σYdW = (2μ+σ^2)Ydt + 2σY(φdt+dW(risk neutral measure))$$, simplify it, we get: $$dY = (2r+σ^2)Ydt + 2σYdW(risk neutral measure)$$. Do you think this is the correct way to get the process of $$S^2$$ under risk neutral measure?

1. Suppose you have a call option on the square of a log-normal asset $$S$$. What equation does the price satisfy?

Based on the result from the above question, I think we can use Feynman-Kac theorem directly and get the result: $$dV/dt+(2r+σ^2)*dV/dY + (1/2)*(2σY)^2*dV^2/dS^2 = rV$$