## Martingales – Problem (29/365)

Show that every martingale $\theta = (\theta_i, \mathcal{D}_k)$ has uncorrelated increments: if $a < b < c < d$ then

$\displaystyle \text{cov}(\theta_d - \theta_c, \theta_b - \theta_a)$

Proceed as follows.

$\displaystyle \mbox{cov}(\theta_{d}-\theta_{c},\theta_{b}-\theta_{a}) \\ = E\left[(\theta_{d}-\theta_{c}-E(\theta_{d}-\theta_{c}))(\theta_{b}-\theta_{a}-E(\theta_{b}-\theta_{a}))\right] \\ = E\left[(\theta_{d}-\theta_{c}-E\theta_{d}+E\theta_{c})(\theta_{b}-\theta_{a}-E\theta_{b}+E\theta_{a})\right] \\ = E\left[(\theta_{d}-\theta_{c})(\theta_{b}-\theta_{a})\right]\mbox{ since }E\theta_{k}=E\theta_{1}\mbox{ for all }k=0 \\ = E\left[(\theta_{d}-E(\theta_{d}|\mathcal{D}_{c}))(\theta_{b}-E(\theta_{b}|\mathcal{D}_{a}))\right]\mbox{ by martingale definition} \\ = E\left[(E(\theta_{n}|\mathcal{D}_{d})-E(\theta_{n}|\mathcal{D}_{c}))(E(\theta_{n}|\mathcal{D}_{b})-E(\theta_{n}|\mathcal{D}_{a}))\right] \\ \mbox{ by martingale definition} \\ = E\left[E(\theta_{n}|\mathcal{D}_{d})E(\theta_{n}|\mathcal{D}_{b})\right]-E\left[E(\theta_{n}|\mathcal{D}_{b})E(\theta_{n}|\mathcal{D}_{c})\right]-E\left[E(\theta_{n}|\mathcal{D}_{d})E(\theta_{n}|\mathcal{D}_{a})\right]+E\left[E(\theta_{n}|\mathcal{D}_{c})E(\theta_{n}|\mathcal{D}_{a})\right] \\ = EE(\theta_{n}^{2}|\mathcal{D}_{d})-EE(\theta_{n}^{2}|\mathcal{D}_{c})-EE(\theta_{n}^{2}|\mathcal{D}_{d})+EE(\theta_{n}^{2}|\mathcal{D}_{c}) \\ \mbox{ because the finer decomposition subsumes the coarser one} \\ = 2\theta_{n}^{2}-2\theta_{n}^{2}\mbox{ by total probability formula} \\ = 0$

As a quick note as to why $E\theta_k = E\theta_1$,

$\displaystyle \theta_{1} \\ = E(\theta_{2}|\mathcal{D}_{1}) \\ = E(E(\theta_{3}|\mathcal{D}_{2})|\mathcal{D}_{1})\mbox{ by definition of martingales} \\ = E(\theta_{3}|\mathcal{D}_{1})\mbox{ since }\mathcal{D}_{1}\le\mathcal{D}_{2} \\ = E(\theta_{k}|\mathcal{D}_{1})\mbox{ continuing the same way} \\ E\theta_{1} = EE(\theta_{k}|\mathcal{D}_{1})=E\theta_{k}\mbox{ by total probability formula}$