Consider the following exercise (ex. 3.2.5 from Calin's book An Informal Introduction to stochastic Calculus with Applications):
If $X_t=e^{W_t}$, find $\mathrm{Cov}(X_s, X_t)$ a. by direct computation; b. by using the previous exercise.
The previous exercise is:
Exercise 3.2.4. Let $X_t = e^{W_t}$. $\\$
$(a)$ Show that $X_t$ is not a martingale.
$(b)$ Show that $e^{-\frac t2}X_t$ is a martingale.
$(c)$ Show that for any constant $c\in\mathbb R$, the process $Y_t= e^{cW_t - \frac12 c^2t}$ is a martingale.
Supposing $s<t$, by direct computation I find: $$ \begin{array} \mathbb{E}[e^{W_t}e^{W_s}] &= \mathbb{E}[e^{W_t - W_s} e^{2W_s}] \\ &= \mathbb{E}[e^{W_t - W_s}]\mathbb{E}[e^{2W_s}] \\ &= e^{\frac12(t-s)} e^{2s} \\ &= e^{\frac12 t + \frac32 s}. \end{array} $$ So the covariance would be: $$ \begin{array} \mathrm{Cov}(X_s, X_t) &= \mathbb{E}[X_s X_t] - \mathbb{E}[X_s]\mathbb{E}[X_t] \\ &= e^{\frac12 (t + s)}(e^s - 1). \end{array} $$
Using the previous exercise, I would do: $$ \begin{array} \mathbb{E}[e^{-\frac12 t}e^{W_t}e^{-\frac12 s}e^{W_s}] &= e^{-\frac12(t+s)} \mathbb{E}[e^{W_t}e^{W_s}]. \end{array} $$ Note that on the right hand side we have exactly what we need for computing the covariance. Now, ideally I would like to split that left hand side into the product of the expectations of the terms in $t$ and of those in $s$, apply the martingale property and conclude. However I do not see how this is justified ($e^{W_t}$ surely depends on $e^{W_s}$), and indeed even if I did, I would not recover the result obtained by explicit computation.
Assuming that that $W_0 = 0$,
\begin{align} E[e^{W_s}e^{W_t}] &= e^{\frac{t}{2}}E[e^{W_s}e^{W_t - \frac{t}{2}}] \\ &= e^{\frac{t}{2}}E[e^{W_s}E[e^{W_t - \frac{t}{2}}|\mathcal{F}_s]] \quad \text{$W_s$ is $\mathcal{F}_s$-mesurable and thanks to b)}\\ &= e^{\frac{t}{2}}E[e^{2W_s - \frac{s}{2}}] \\ &= e^{\frac{t+3s}{2}}E[e^{2W_s - 2s}] \quad \text{thanks to c) with $\lambda=2$}\\ &= e^{\frac{t+3s}{2}}\\ \end{align} Now : \begin{align} E[e^{W_s}]E[e^{W_t}] &= e^{\frac{t+s}{2}}E[e^{W_s- \frac{s}{2}}]E[e^{W_t - \frac{t}{2}}] \\ &= e^{\frac{t+s}{2}}\\ \end{align} You have all the ingredients to compute the covariance.