Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
Thanks a lot for your answer!. I got it now. –  Prakhar Mehrotra Dec 4 '12 at 9:47
    
@Prakhar Mehrotra Wait, I forgot the $\sigma$. I will put into the right places. –  Richard Dec 4 '12 at 9:53
    
@Prakhar Mehrotra Now it is there. –  Richard Dec 4 '12 at 9:55
add comment

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 $\sigma^2 T/3$.

share|improve this answer
1  
A good idea to derive the formula. –  Richard Dec 8 '12 at 19:09
    
+1 for invoking Ito isometry :] –  Veeken May 11 '13 at 15:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.