Pricing European Power Calls in the Black–Scholes Model

# Power call option pricing problem


# Power call option pricing problem












Q: The payoff of a European ’power call’ is given by $max(S^\alpha - K,0)$. Derive the price of a European power call option.

I've keep arriving at a different solution compared to standard references. Specifically, my confusion lies in (II) of the expectation calculation. Here’s the setup and where my derivation diverges: Under $Q$, the Black-Scholes framework, the dynamics of $S_t$ follow a Geometric Brownian Motion (GBM):

$$ dS_t = r S_t \, dt + \sigma S_t \, dW_t $$

$$ d(\ln S_t^\alpha) = \alpha \left( r - \frac{1}{2} \sigma^2 \right) dt + \alpha \sigma dW^Q_t $$

$$ S_t^\alpha = S_0^\alpha \exp \left( \alpha \left( r - \frac{1}{2} \sigma^2 \right) T + \alpha \sigma W_T^Q \right) $$

The power call option price is:

$$ C_0 = \underbrace{e^{-rT} \mathbb{E}^Q[S_T^\alpha \mathbf{1}_{S_T^\alpha > K}]}_{(II)} - \underbrace{e^{-rT} K \mathbb{E}^Q[\mathbf{1}_{S_T^\alpha > K}]}_{(I)} $$

#### I:

$$ e^{-rT} K \mathbb{E}^Q[\mathbf{1}_{S_T^\alpha > K}] = e^{-rT} K Q(S_T^\alpha > K) $$ $$ = e^{-rT} K Q \left( S_0^\alpha \exp \left( \alpha \left( r - \frac{1}{2} \sigma^2 \right) T + \alpha \sigma W_T^Q \right) > K \right) $$

$$ = e^{-rT} K \Phi \left( \frac{-\ln K + \ln S_0^\alpha + \alpha (r - \frac{1}{2} \sigma^2) T}{\alpha \sigma \sqrt{T}} \right) $$

$$ = e^{-rT} K \Phi \left( \frac{-\ln X + \ln S_0 + (r - \frac{1}{2} \sigma^2) T}{\sigma \sqrt{T}} \right): X = K^{\frac{1}{\alpha}} $$

#### II:

$$ e^{-rT} \mathbb{E}^Q[S_T^\alpha \mathbf{1}_{S_T^\alpha > K}] = e^{-rT} \mathbb{E}^Q \left[ S_0^\alpha \exp \left( \alpha \left( r - \frac{1}{2} \sigma^2 \right) T + \alpha \sigma W_T^Q \right) \mathbf{1}_{S_T^\alpha > K} \right] $$

After simplifying, I arrive at:

$$ S_0^\alpha \exp \left( (\alpha - 1) \left( r + \frac{\alpha \sigma^2}{2} \right) T \right) Q(S_T^\alpha > K) $$

Issue: The $N(d_1)$ in Part II doesn't match the typical solutions I find online. Can someone point out where my derivation might be going wrong?

The specific Black-Scholes solution I'm referencing is detailed here: https://demonstrations.wolfram.com/PricingPowerOptionsInTheBlackScholesModel/ - where part II. looks like the following:

$$ S_0^\alpha \exp \left( (\alpha - 1) \left( r + \frac{\alpha \sigma^2}{2} \right) T \right) \Phi \left( \frac{-\ln K + \ln S^\alpha_0 + \alpha(r + (\alpha - \frac{1}{2}) \sigma^2) T}{\sigma \alpha \sqrt{T}} \right) $$

Any insights would be greatly appreciated!

From Mark Joshi, The Use of Power Numeraires in Option Pricing, 2003: Powers of the stock price, $S^\alpha_T$, are also not representable as self-financing contracts for $\alpha \ne 0, 1$, since their discounted value is not a martingale in the risk-neutral measure. We therefore adopt the solution of taking a contract that pays a given power of the stock price at a fixed time $T$. Its price process is then trivially valid as a numeraire whilst having similar but not identical properties to $S^\alpha_T$. We can let $N^{\alpha}_{t,T}$ be the value at time $t$ of a derivative that pays $S^{\alpha}_T$ at time $T$. It follows by risk-neutral valuation that: $$ N^{\alpha}_{t,T} = e^{-r(T-t)} \mathbb{E} \left[ S^{\alpha}_T \middle| \mathcal{F}_t \right] = S^{\alpha}_t exp \left( (T-t)(r(\alpha-1) + 0.5\sigma^2 (\alpha^2 - \alpha)) \right). $$ $$ dN^{\alpha}_{t,T} = d \left( S^{\alpha}_t exp \left( (T-t)(r(\alpha-1) + 0.5\sigma^2 (\alpha^2 - \alpha)) \right) \right) = r N^{\alpha}_{t,T} dt + \sigma N^{\alpha}_{t,T} \alpha dW_t. $$ The process for $N^{\alpha}_{t,T}$ - using itô's lemma - is thus log-normal with drift $r$ and volatility $\sigma\alpha$ in the risk-neutral measure. The discounted value of $N^{\alpha}_{t,T}$ is a martingale in the risk-neutral measure - and used as numeraire. The Radon–Nikodym derivative of the $\alpha$-measure change - and the $S^\alpha_t$ dynamics in $Q_\alpha$ $$ \frac{dQ_\alpha}{dQ} \big|_{\mathcal{F}_T} = exp \left( -\frac{1}{2}\alpha^2 \sigma^2T + \alpha \sigma W^Q_T \right) $$ $$ S_T^\alpha = S_0^\alpha \exp \left( \alpha \left( r - \frac{1}{2} \sigma^2 \right) T + \alpha \sigma (W^{Q_\alpha}_T + \sigma \alpha T ) \right) $$ Going back to the (in)famous last step $$ S_0^\alpha \exp \left( (\alpha - 1) \left( r + \frac{\alpha \sigma^2}{2} \right) T \right) \mathbb{E}^Q \left[\mathbf{1}_{S_T^\alpha > K} \right] $$ $$ = N^{\alpha}_{0,T} \mathbb{E}^Q \left[ \color{red}{\frac{dQ_\alpha}{dQ}} \mathbf{1}_{S_T^\alpha > K} \right] = N^{\alpha}_{0,T} \mathbb{E}^{Q_\alpha} \left[\mathbf{1}_{S_T^\alpha > K} \right] = N^{\alpha}_{0,T} Q_\alpha(S_T^\alpha > K) $$ $$ = N^{\alpha}_{0,T} Q_\alpha( S_0^\alpha \exp \left( \alpha \left( r - \frac{1}{2} \sigma^2 \right) T + \alpha \sigma (W^{Q_\alpha}_T + \sigma \alpha T ) \right) > K) $$ $$ = N^{\alpha}_{0,T} Q_\alpha \left( \frac{W^{Q_\alpha}_T}{\sqrt{T}} > \frac{lnK - lnS_0^\alpha - \alpha (r - \frac{1}{2}\sigma^2)T - \alpha^2 \sigma^2 T}{\alpha \sigma\sqrt{T}} \right) $$

Q.1: In the above derivation, where does $\frac{dQ^{\alpha}}{dQ}$ come from? It seems intuitive and perhaps trivial, but I’m struggling to grasp it at the moment. For a vanilla payoff $(S_T - K)^+$, we typically introduce the Radon-Nikodym derivative by substituting $ S_T$ inside the expectation to switch to the stock measure. However, in this case, it appears to emerge unexpectedly. What am I missing?

## Answer by solid (score 0, accepted)

https://quant.stackexchange.com/a/81508

A.1: $$ S_0^\alpha \exp \left( (\alpha - 1) \left( r + \frac{\alpha \sigma^2}{2} \right) T \right) \mathbb{E}^Q \left[\mathbf{1}_{S_T^\alpha > K} \right] $$ $$ = N^{\alpha}_{0,T} \mathbb{E}^{Q_\alpha} \left[\frac{dQ}{dQ_\alpha} \mathbf{1}_{S_T^\alpha > K} \right] = N^{\alpha}_{0,T} \mathbb{E}^{Q_\alpha} \left[\frac{N^{\alpha}_{0,T}B_T}{N^{\alpha}_{T,T}} \mathbf{1}_{S_T^\alpha > K} \right] $$ $$ = N^{\alpha}_{0,T} \mathbb{E}^{Q_\alpha} \left[\frac{N^{\alpha}_{0,T}B_T}{S^{\alpha}_T } \mathbf{1}_{S_T^\alpha > K} \right] = N^{\alpha}_{0,T} Q_\alpha(S_T^\alpha > K) $$

A.2: $$ e^{-rT} \mathbb{E}^Q[S_T^\alpha \mathbf{1}_{S_T^\alpha > K}] = e^{-rT} \mathbb{E}^{Q_\alpha} \left[\frac{dQ}{dQ_\alpha} S_T^\alpha \mathbf{1}_{S_T^\alpha > K} \right] $$ $$ = e^{-rT} \mathbb{E}^{Q_\alpha} \left[ \frac{N^{\alpha}_{0,T}B_T}{S^{\alpha}_T} S_T^\alpha \mathbf{1}_{S_T^\alpha > K} \right] = N^{\alpha}_{0,T} Q_\alpha(S_T^\alpha > K) $$

## Answer by Andrea (score 2)

https://quant.stackexchange.com/a/81307

I think there is a simpler approach.

$Y_t = S_t^{\alpha}$

$dY_t = \alpha S_t^{\alpha-1} dS_t + \frac{1}{2} \alpha (\alpha - 1) S_t^{\alpha -2} <dS_t, dS_t>$

which simplifies as

$dY_t = \alpha S_t^{\alpha} r dt + \alpha S_t^{\alpha} \sigma dW_t + \frac{1}{2} \alpha (\alpha - 1) S_t^{\alpha} \sigma^2 dt$

$dY_t = \alpha Y_t r dt + \alpha Y_t \sigma dW_t + \frac{1}{2} \alpha (\alpha - 1) Y_t \sigma^2 dt = \alpha (r + \frac{1}{2} (\alpha - 1) \sigma^2 ) Y_t dt + \alpha \sigma Y_t dW_t$

Which means you can use the standard BS formula with a different forward and volatility, but the same discount factor.

The forward is $S_0^{\alpha} e^{\alpha (r + \frac{1}{2} (\alpha - 1) \sigma^2) T}$, while the volatility is $\alpha \sigma$.

EDIT: after reading your comment.

$d_2$ is correct, you can just use the normal one with $K^{\frac{1}{\alpha}}$. This is under the bank account measure which has not changed.

$d_1 = \frac{\log F - \log K + \frac{1}{2} \text{vol}^2 T}{\text{vol} \sqrt{T}}$ or, with our values

$d_1 = \frac{\alpha ( \log S + (r + \frac{1}{2} (\alpha-1) \sigma^2 ) T) - \log K + \frac{1}{2} \alpha^2 \sigma^2 T}{\alpha \sigma \sqrt{T}}$

which becomes

$d_1 = \frac{\log S -\log X + (r + \frac{1}{2} (\alpha-1) \sigma^2 + \frac{1}{2} \alpha \sigma^2) T}{\sigma \sqrt{T}}$

$d_1 = \frac{\log S - \log X + (r + (\alpha - \frac{1}{2}) \sigma^2 ) T}{\sigma \sqrt{T}}$

I see you added the link, where $d_1$ looks like mine. I am a bit lost in what your 2 alternative expressions are. You should probably write them next to each other to avoid confusion.