Covariance of geometric Brownian motion via first and second moments and MGFs

Question

My attempts: Substitute in the $Z(t)= X(t)Y(t)$ in $\mathrm{Cov}(X(t),Z(t))$. From that, I tried to split it into smaller covariances. Also trying to figure out how can the moment generating functions can come into play here.

Answer 1

$$\textrm{Cov}[X_t,Z_t]=E[X_tZ_t]-E[X_t]E[Z_t]$$ $$E[X_tZ_t]=E[X_t^2Y_t]=E[X_t^2]E[Y_t]$$ $$\textrm{Cov}[X_t,Z_t]=E[X_t^2]E[Y_t]-E[X_t]^2E[Y_t]=(E[X_t^2]-E[X_t]^2)E[Y_t]=\textrm{Var}[X_t]E[Y_t]$$ Let $f_{X_t}$ be the lognormal distribution of $X_t$ and $f_{\ln(X_t)}$ the normal distribution of $\ln(X_t)$. Then if $e^z=x \sim \textrm{Lognorm}$ then $z \sim \textrm{Norm}$ so

$$E[X_t^n]=\int_{(0,\infty)}x^nf_{X_t}(x)dx=\int_\mathbb{R}e^{zn}f_{\ln(X_t)}(z)dz$$ which is the moment generating function $\varphi(n)$ of a Gaussian with mean $\ln(X_0)+(\mu_X-0.5\sigma_X^2)t$ and variance $\sigma^2_Xt$, so now we know that $$E[X_t^n]=e^{n(\ln(X_0)+(\mu_X-0.5\sigma_X^2)t)+n^20.5\sigma^2_Xt}$$ Therefore $$E[X_t]=X_0e^{\mu_X t}$$ $$E[X_t^2]=X_0^2e^{2\mu_X t+\sigma^2_Xt}$$