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Assume that $\mathrm{d}S = \sigma \, \mathrm{d}W$ with initial level $S(0)$ and where $\mathrm{d}W$ is usual Brownian motion. Now

$$A(T) = \frac{1}{T} \int_0^T S(t) \, \mathrm{d}t.$$

What is the distribution of $A(T)$?

Solution:

$$S(t) = S(0) \, \sigma \, W(t) \sim \mathcal{N}(0, \sqrt{S(0)}\sigma^2t)$$

so

$$A(T) = \frac{1}{T} \int_0^T S(0) \, \sigma \, W(t) \, \mathrm{d}t$$

.... and now I'm confused. Any help appreciated.

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Your "solution" (to the SDE for $S(t)$) is wrong: it should be $S(t) = S(0) + \sigma W(t)$.

In any case, your problem reduces to computing the distribution of $\int_0^T W(t)\,dt$. One way to do so is by writing it as a double integral and using Fubini's theorem (which continues to hold in the stochastic context) to exchange the order of integration: \begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*} The term $\int_0^T u\, dW(u)$ is a standard Ito integral, so its distribution is normal with mean $0$ and variance $\int_0^T u^2\, du = \frac{T^3}{3}$. The final answer should be easy to compute from this.

(Another way to rearrange the integral would be to use an "integration by parts" formula for $d[t W(t)]$.)

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  • $\begingroup$ This supposes to compute the covariance of W(T) and the last integral term--a task at least as complicated as the original question. $\endgroup$ – Did Nov 18 '13 at 15:44
  • $\begingroup$ Computing the covariance of $\int_0^T dW(t)$ and $\int_0^T t\, dW(t)$ is pretty standard... $\endgroup$ – Jan Ladislav Dussek Nov 18 '13 at 15:59
  • $\begingroup$ As I said: this is roughly as standard as the direct approach. $\endgroup$ – Did Nov 18 '13 at 17:01
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Since $S(t)=S(0)+\sigma W(t)$, $A(T)=S(0)+\frac1T\sigma C(T)$ where $C(T)=\int\limits_0^TW(t)\mathrm dt$. Since $C(T)$ is a barycenter of the centered normal family $(W(t))_{0\leqslant t\leqslant T}$, $C(T)$ is centered normal. Furthermore, $E[C(T)^2]=\int\limits_0^T\int\limits_0^TE[W(t)W(s)]\mathrm dt\mathrm ds=\int\limits_0^T\int\limits_0^T\min(t,s)\mathrm dt\mathrm ds=\frac13T^3$.

Thus, $A(T)$ is normal with mean $S(0)$ and variance $\frac13\sigma^2T$.

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Let $\mathcal{F}$ be the natural filtration of $W$. Define $$\begin{align} B &= \int_0^T W_tdt\\ Y_t &= \mathbb{E}[B|\mathcal{F}_t]\\ &= \int_0^tW_udu + \mathbb{E}\left[\int_t^TW_udu\bigg|\mathcal{F}_t\right]\\ &= \int_0^tW_udu + (T-t)W_t. \end{align}$$

Then, by Ito's formula $$\begin{align}dY_t &= W_tdt+TdW_t-tdW_t-W_tdt\\ &= (T-t)dW_t.\end{align}$$

In particular the martingale $Y$ is a deterministic time change of Brownian motion and hence a Gaussian process. So $Y_u$ is normally distributed for each $u\in[0,T]$. Note that $B = Y_T$.

As $Y$ is a martingale, $\mathbb{E}[B]=0$.

By the Ito Isometry, $$\begin{align}\text{var}(B) &= \int_0^T(T-t)^2dt \\ &= \frac{T^3}3.\end{align}$$

$A(T) = S(0) + \sigma B /T$, so $$A(T) \sim N\left(S(0), \frac{\sigma^2T}{3}\right).$$

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