Pricing a Jointly Conditioned Call Payoff with a Gaussian Copula
Summary
The document considers a payoff that pays the positive amount by which one correlated lognormal stock exceeds a strike, provided a second stock finishes between two bounds. It starts from the discounted expectation and explains how to split the payoff into a stock-value term and a strike times a joint event probability. The latter can be evaluated from a bivariate Gaussian distribution after expressing the stock thresholds in terms of correlated Brownian increments.
For the stock-value term, the proposed method changes the pricing measure to use the first stock as numéraire. Under that measure, the joint threshold probability can be computed using shifted Brownian drifts, and the two terms combine into a semi-analytic price without Monte Carlo. The exposition assumes a risk-neutral setting and simplifies initially to zero interest rates. Its displayed threshold expressions and drift changes merit careful checking before implementation; it does not provide a worked numerical valuation or full derivation of the bivariate probabilities.
Key ideas
- Decompose the conditional call payoff into a joint probability and a stock-weighted joint probability.
- Map stock thresholds to regions for correlated Brownian increments.
- Use a bivariate normal probability to evaluate the event component.
- Change to the first stock numéraire to express the stock-weighted component as another joint probability.
- Check the measure-change details and threshold formulas when implementing the method.
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Full text
# Copula analytic formula for $max(S_T^1 - K, 0) 1_{\{L<S_T^2<U\}}$
# Copula analytic formula for $max(S_T^1 - K, 0) 1_{\{L<S_T^2<U\}}$
Consider the payoff function $$ V_T = max(S_T^1 - K, 0) 1_{\{L<S_T^2<U\}} = (S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}$$
where $S_T^1$ and $S_T^2$ are two GBM distributed stocks with correlation, $\rho$. How would you find $V_t$ without the use of MC simulation (hint: use a Gaussian copula)?
My attempt: $$V_t = e^{-r(T-t)}E[(S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}]$$ $$V_t = e^{-r(T-t)}\int_{L}^{U}\int_{K}^{\infty}(x-K)f_{x,y}(x,y)dxdy$$
where $f_{x,y}$ is the joint pdf coming from the Guassian copula. How do I go further now? If I can't, how would I implement the double integral on a computer?
## Answer by NN2 (score 2, accepted)
https://quant.stackexchange.com/a/59951
For simplicity, we suppose $r = 0$. $$V_t = E^Q((S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) = E^Q(S_T^11_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) - KE^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$$ For the second term: $$E^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})=P(\{S_T^1 > K\}\cap \{ L<S_T^2<U \}) =P(\{(W_T^1-W_t^1) > d_1 \}\cap \{ d_2<(W_T^2-W_t^2))<d_3 \}) $$ with $$d_1 = \frac{\ln (K/S_t^1)+(1/2\sigma^1)(T-t))}{\sigma^1}$$ $$d_2 = \frac{\ln (L/S_t^2)+(1/2\sigma^2)(T-t))}{\sigma^1}$$ $$d_3 = \frac{\ln (U/S_t^2)+(1/2\sigma^2)(T-t))}{\sigma^1}$$ Because $(W_T^1-W_t^1,W_T^2-W_t^2)$ follows the bidimensional gaussian distribution, you can easily obtain the analytic formula for $E^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$.
Now, for the first term $E^Q(S_T^11_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$, we change the mesure to the $S_t^1$ as numéraire. $$E^Q(S_T^1 1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) = S_t^1 E^{Q_{S^1}}(1_{\{S_T^{1} > K\}}1_{\{L<S_T^{2}<U\}})$$ In the new mesure, we have $$dW_t^{1'} = dW_t^1 -\sigma^1dt$$ $$dW_t^{2'} = dW_t^2 -\sigma^1\rho dt$$ with $$\frac{dS_t^1}{S_t^1} = \sigma^1 dW_t^1 = \sigma^1 (dW_t^{1'} + \sigma^1dt)$$ $$\frac{dS_t^2}{S_t^2} = \sigma^2 dW_t^2 = \sigma^2 (dW_t^{2'} + \sigma^2 \rho dt)$$ By the same argument we did with the first term, we obtain the analytic formula for $E^{Q_{S^1}}(1_{\{S_T^{1} > K\}}1_{\{L<S_T^{2}<U\}})$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.