# Is the definite time integral of a Brownian Motion a Markov process and a martingale?

I am trying to figure out the below scenario:

Assume that $(B_t)_{t>=0}$ is a Brownian Motion. Is the below Stochastic process

$X(t) =\int_{0}^tB_sds$

a) Gaussian?

b) a Markov Process?

c) a Martingale?

d) Find the distribution of the process and the auto-covariance function.

What I have figured out by reading some links online and perusing books on Stochastic Calculus is that we have two different definitions for this kind of integral. One corresponding to Strantonovich and the other one to Itô Integral. And the texts say that the definition by Strantonovich is not well suited for the Markov and Martingale properties of the stochastic integral while the one by Itô integral is more martingale friendly.

And so I am confused on how should I be answering this question.

For part a) the justification that I can think of is that I would expect $X(t)$ to be a Gaussian because the integral is a linear functional of the given Brownian motion path , B. And then I can answer the part d) very easily.

But again, I am not very sure if I am heading in the correct direction to answer this question in the best way possible.

Any help on this is much appreciated!

• Stratonovich vs. Ito has nothing to do with this question because you are integrating $ds$ not $dB$.
– Ian
Apr 25, 2017 at 12:08
• Thanks @Ian. I am sorry, but I am very new to Stochastic Processes and Stochastic Calculus. What is your suggestion on how I should be approaching this problem? Using the Reimann sum approximation perhaps?
– user397601
Apr 25, 2017 at 12:11

1. It is Gaussian, in particular, $X(t) \sim \mathcal{N}\left(0, \int_0^t (s-t)^2 ds\right)$ for each $t > 0$. The idea is to write $X(t)$ as the limit of Riemann sums, the solution has been posted by @saz in Integral of Brownian motion is Gaussian?, alternatively you may want to use stochastic Fubini theorem.
2. It is not a Markov process. A Gaussian process is Markovian, if and only if , its covariance, $\mbox{Cov}(X(t), X(s))$ satisfies the condition $$\mbox{Cov}(X(t), X(s)) = \frac{\mbox{Cov}(X(t), X(r))\mbox{Cov}(X(r), X(s))}{\mbox{Cov}(X(r), X(r))}$$ for all $t < r < s$.
1. It is not a Martingale w.r.t the natural filtration $(\mathcal{F}_t)$ of $(B_t)$. Namely, $$\mathbb{E}\left[\int_0^t B_s ds \ | \ \mathcal{F}_r\right] = (t-r)B_r + \int_0^r B_s ds.$$ The whole solution which I have written some time ago can be found here process with integral is martingale
2. For the distribution see 1. Now we find the covariance \begin{align*}\mbox{Cov}(X(t), X(s)) =& \mathbb{E}\left[ X(t)X(s) \right] = \int_0^t \int_0^s \mathbb{E}(B_uB_v)dudv =\int_0^t \int_0^s \mbox{Cov}(B_u,B_v)dudv \\ =& \int_0^t \int_0^s \min(u,v) du dv = \int_0^t\left( \int_0^v udu + \int_v^s vdu \right)dv\\ =& \frac{t^3}{6}+\frac{t^2}{6}(3s-2t) = \frac{t^2}{6}(3s-t) \end{align*}.