Interview Questions VII: Integrated Brownian Motion

 
For a standard Weiner process denoted W_t, calculate

    \[{\mathbb E} \Bigl[ \Bigl( \int^T_0 W_t dt \Bigr)^2 \Bigr]\]

This integral is the limit of the sum of W_t at each infinitessimal time slice dt from 0 to T, it is called Integrated Brownian Motion.




We see immediately that this will be a random variable with expectation zero, so the result of the squared expectation will simply be the variance of the random variable, which is what we need to calculate. Here I show two ways to solve this, first by taking the limit of the sum and second using stochastic integration by parts

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1) Limit of a sum

(1)   \begin{align*} I &= \int^T_0 W_t dt \nonumber \\ &= \lim_{n \to \infty}{\frac T n}\sum^n_{i=1} W_{\frac {Ti} n} \nonumber \end{align*}

This sum of values along the Wenier process is not independent of one-another, since only the increments are independent. However, we can re-write them in terms of sums of these independent increments

(2)   \begin{align*} W_{\frac {Ti} n} &= \sum^i_{j=1} ( W_{\frac {Tj} n} - W_{\frac {T(j-1)} n} ) \nonumber \\ &= \sum^i_{j=1} \overline{W}_{\frac {Tj} n} \nonumber \end{align*}

where \overline{W}_{\frac {T(j-1)} n} \sim {\mathbb N}(0,{\frac T n}) are the individual independent increments of the brownian motion. Substituting into our previous equation and reversing the order of the summation

(3)   \begin{align*} \lim_{n \to \infty}{\frac T n}\sum^n_{i=1} W_{\frac {Ti} n} &= \lim_{n \to \infty}{\frac T n} \sum^n_{i=1} \sum^i_{j=1} \overline{W}_{\frac {Tj} n} \nonumber \\ &= \lim_{n \to \infty}{\frac T n} \sum^n_{j=1} \sum^n_{i=j} \overline{W}_{\frac {Tj} n} \nonumber \\ &= \lim_{n \to \infty}{\frac T n} \sum^n_{j=1} (n-j+1) \overline{W}_{\frac {Tj} n} \nonumber \\ \end{align*}

which is simply a weighted sum of independent gaussians. To calculate the total variance, we sum the individual variances using the summation formula for \sum_1^N n^2

(4)   \begin{align*} {\mathbb V}[I] &= \lim_{n \to \infty} {\frac {T^2} {n^2}} \sum^n_{j=1} (n-j+1)^2 {\frac T n} \nonumber \\ &= \lim_{n \to \infty} {\frac {T^3} {n^3}} \Bigl( {\frac 1 3} n^3 + O(n^2) \Bigr) \nonumber \\ &= {\frac 1 3} T^3 \end{align*}

which is the solution.

2) Stochastic integration by parts

The stochastic version of integration by parts for potentially stochastic variables X_t, Y_t looks like this:

    \[X_T \cdot Y_T = X_0 \cdot Y_0 + \int^T_0 X_t dY_t + \int^T_0 Y_t dX_t\]

Re-arranging this gives

    \[\int^T_0 Y_t dX_t = \Bigl[ X_t \cdot Y_t \Bigr]^T_0 - \int^T_0 X_t dY_t\]

Now setting Y_t = -W_t and X_t = (T-t) we have

(5)   \begin{align*} \int^T_0 W_t dt &= \Bigl[ -W_t \cdot (T-t) \Bigr]^T_0 + \int^T_0 (T-t) dW_t \nonumber \\ &= \int^T_0 (T-t) dW_t \nonumber \end{align*}

We recognise this as a weighted sum of independent gaussian increments, which is (as expected) a gaussian variable with expectation 0 and variance that we can calculate with the Ito isometry

(6)   \begin{align*} {\mathbb E}\Bigl[ \Bigl( \int^T_0 (T-t) dW_t \Bigr)^2 \Bigr] &= {\mathbb E}\Bigl[ \int^T_0 (T-t)^2 dt \Bigr] \nonumber \\ &= \Bigl[ T^2t - Tt^2 + {\frac 1 3} t^3 \Bigr]^T_0 \nonumber \\ &= {\frac 1 3} T^3 \end{align*}

which is the solution.

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