I know the following result:

Let (X,Y) be a normal random vector 2-dimensional with mean vector $m=(m_1,m_2)$ and convariance matrix $\Sigma=(\sigma_{ij})$. For the lognormal random vector $(e^X,e^Y)$ we have that $$ E[e^X]=e^{m_1+\frac{1}{2} \sigma_{11}}, \quad E[e^Y]=e^{m_2+\frac{1}{2} \sigma_{22}}.$$ And a similar formula for the covariance: $$ Cov(X,Y)=E[X]E[Y](e^{\sigma_{12}}-1).$$

In particular I know that $$E[e^Xe^Y]=E[e^X]E[e^Y]e^{\sigma_{12}}.$$ What Can I say about $$E[e^X(e^Y-K)^+]$$ where $K>0$ costant and $(e^Y-K)^+=\max (e^Y-K,0) \quad ?$

Thanks to all.


1 Answer 1


(Sorry, this is not a complete answer, but it did not fit into the comments.) Maybe it helps to apply the Law of Iterated Expectations: \begin{align} \mathbb{E}\left[ e^X(e^Y - K)^{+} \right] & = \mathbb{E}\left[\mathbb{E}\left[ e^X(e^Y - K)^{+} |e^Y\right] \right]\\ & =\mathbb{E}\left[ e^X(e^Y - K) |e^Y > K\right]\cdot \mathbb{P}(e^Y>K) + 0\cdot\mathbb{P}(e^Y<K) \\ &= \mathbb{P}(e^Y>K) \left[ \mathbb{E}\left[ e^Xe^Y|e^Y > K \right] - K\mathbb{E}\left[ e^X|e^Y > K \right]\right] \end{align} Now if you can evaluate those two conditional expectations, you can solve the problem.


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