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I have two lognormally distributed random variables $Y_i=e^{X_i}$ where $X_i \sim \mathcal{N}\big(\mu_i, \: \sigma_i^2 \big)$ for $i=1,2$, and $X_1$ and $X_2$ are correlated by $\rho_{12}$. Now, Let $Z=\alpha_2 Y_2 - \alpha_1 Y_1$.

Question : What is the covariance between $Y_1$ and $Z$?

What I understand so far are the general results for $Y_1$ and $Y_2$:

$\mathbb{E}[\:Y_i \:]=a_i \:\text{exp}(\mu_i+\frac{\sigma_i^2}{2})$

$\mathbb{Var}[\:Y_i \:]=a_i^2 \: \text{exp}(2\mu_i) \big[\text{exp}(\sigma_i^2) -1 \big] \: \text{exp}(\sigma_i^2)$

$\mathbb{Cov}[Y_1 \: Y_2]=\mathbb{E}[\:Y_1 \: Y_2]- \mathbb{E}[\:Y_1 \:]\mathbb{E}[\:Y_2 \:]\\ \; \;\;\; \: \quad \quad \quad \:= \alpha_1\alpha_2 \:\text{exp}(\mu_1 +\mu_2)\text{exp}(\frac{1}{2}(\sigma_1^2+2\rho_{12}\sigma_1\sigma_2+\sigma_2^2))\\ \; \;\;\; \: \quad \quad \quad \quad \qquad-\alpha_1\text{exp}(\mu_1+\frac{\sigma_1^2}{2}) \:\alpha_2\text{exp}(\mu_2+\frac{\sigma_2^2}{2})\\ \; \;\;\; \: \quad \quad \quad \:= \alpha_1\alpha_2\: \text{exp}(\mu_1+\mu_2)\text{exp}(\frac{1}{2}(\sigma_1^2+\sigma_2^2))\big[\text{exp}(\rho_{12}\sigma_1\sigma_2)-1\big]$

How can I utilise these results to compute $\mathbb{Cov}[Y_1, Z]$?

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You can use what you've already derived to answer your question with just a few more steps: \begin{align} \mathbb{Cov}[Y_1, Z] &= \mathbb{Cov}[Y_1, \alpha_2 Y_2 - \alpha_1 Y_1]\\ &=\mathbb{Cov}[Y_1, \alpha_2 Y_2] - \mathbb{Cov}[Y_1, \alpha_1 Y_1]\\ &=\alpha_2\mathbb{Cov}[Y_1, Y_2] - \alpha_1\mathbb{Var}[Y_1] \end{align} Now just plug in your expressions.

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  • $\begingroup$ Thank you @psboonstra. You made the exact point that I was missing. Truly appreciate your help. $\endgroup$ – yufiP Mar 31 at 15:07

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