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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.