Let's say stock price follows following process:
$$dS(t) = \sigma dW(t)$$
where $W(t)$ is Standard Brownian motion. The initial level for the stock is $S(0)$. Define the average of stock price $Z(t)$ as:
$$Z(t) = \frac{1}{T}\int_0^T S(t)dt$$
What is the distribution of $Z(t)$?
Note: this was asked in one of quant interviews and I could not find any reference to it on the web and on this form.
Could you please help me on how to get started?
The model for the stock is the Bachelier model with the solution $$ S(t) = S(0) + \sigma W(t) $$ Thus the law of the stock $S(t)$ is Gaussian with mean $S(0)$ and variance $\sigma^2 t$. For average process $Z(T)$ is thus the average of linear Brownian motion, we can rewrite this as $$ Z(T) = \frac{1}{T} \int_0^T S(0) + \sigma W(t) dt = S(0) + \frac{\sigma}{T}\int_0^T W(t) dt $$ Thus all you need is the law of the average of Brownian motion. Is is clearly Gaussian. The mean is $S(0)$ and all you need is the variance. Using integration by parts you get the following expression for the integral of Brownian motion w.r.t. time. You can Google this on the web and find e.g. this document where it says that $\int_0^T W(t) dt $ is Gaussian with mean $0$ and variance $T^3/3$.
Finally, the distribution of the $Z(T)$ is Gaussian with mean $S(0)$ and variance $\sigma^2 T/3$ (as we devide by $T$ and it enters the variance with a square). Note that the variance of the average of Brownian motion is a third of the variance of BM itself.
I like Richard's answer, but I think we can compute the mean and the variance of $\int_0^T W_t dt$ by ourselves using Ito's lemma. Let $f(W_t, t) = t W_t$. $$ d( t W_t ) = W_t dt + t dW_t . $$ Integrating both sides, and re-arranging the terms, we get $$ \int_0^T W_t dt = T W_T - \int_0^T t dW_t \, . $$ We'll be using Ito's isometry formula $\mathbb{E} \left[ \int_0^T f_t dW_t \int_0^T g_t dW_t \right] = \int_0^T \mathbb{E} \left[f_t g_t \right] dt$.
The integral $\int_0^T W_t dt$ is a Gaussian random variable with zero mean $$ \mathbb{E} \left[ \int_0^T W_t dt \right] = T \mathbb{E} \left[ W_T \right] - \mathbb{E} \left[ \int_0^T t dW_t \right] = 0, $$ and variance $$ \mathbb{E} \left[ \left(\int_0^T W_t dt \right)^2 \right] = T^2 \mathbb{E} [W_T^2] - 2 T \mathbb{E} \left[W_T \int_0^T t dW_t \right] + \mathbb{E} \left[ \left(\int_0^T t dW_t \right)^2 \right] $$ $$ = T^3-2 T \int_0^T t dt + \int_0^T t^2 dt = \frac{T^3}{3}. $$
Hence, continuing with Richard's derivations, $Z(T)$ is a Gaussian random variable with mean $S(0)$ and variance $\sigma^2 T/3$.
