# Model for stock market changes over the day

A common model for stock returns is as follows: the number of trades $N$ of stock XXX in a given day has a Poisson distribution with parameter $\lambda$. At each trade, say the $i$’th trade, the change in the price of the stock is $X_i$ and has a normal distribution with mean $0$ and variance $\sigma^2$ , say and these changes are independent of one another and independent of $N$. Find the moment generating function of the total change in stock price over the day. Is this a distribution that you recognise? What is its mean and variance?

My textbook is quite terse and actually does not have any detailed examples on deriving mgf's, let along (apparently) messy ones like a huge sum of poissons and normals. How would I approach this question?

-
this is a compound poisson process and you can find exactly what you want here en.wikipedia.org/wiki/Compound_Poisson_process – mike Apr 24 '13 at 11:58

## 1 Answer

Notation:
- $X$-Normal random variable mean 0 variance $\sigma^2$. - $N$-Poisson random variable mean $\lambda$. - $Z$-Sum of $N$ independent normal variables $X_1,\dots,X_N$

You need two things to start off with

First the probability generating function of a Poisson $$\mathbb E(s^N) = e^{\lambda(s-1)}$$ And the moment generating function of the Gaussian $$\mathbb E(e^{tX}) = e^{\frac{\sigma^2t^2}2}.$$

There's no trick to working these out you just have to work out the expectation the same way you would with any other random variable. I looked them up on wikepedia because I'm lazy.

The change in price over the day is going to be the sum of $N$ independent copies of $X$. Remember that if $X_1$ and $X_2$ are independent random variables then $\mathbb E(X_1X_2) = \mathbb E(X_1)\mathbb E(X_2)$.

So if I told you the value of $N$ you could work out $$\mathbb E(e^{tZ}) = \mathbb E\left( e^{t(X_1+X_2+\dots X_N)}\right) = \mathbb E\left( e^ {t(X_1)}e^ {t(X_2)}\dots e^ {t(X_N)}\right) = \left(\mathbb E\left(e^ {t(X)}\right)^N\right)$$

Now I know what $\mathbb E\left(e^ {t(X)}\right)$ is so I can write down the conditional expectation of $e^{t(X_1+X_2+\dots X_N)}$ given $N$.

$$\mathbb E \left(\left. {Z}\right | N\right) = \left(e^{\frac{\sigma^2t^2}2}\right)^N$$.

For fixed $t$ the expression above is just a function of $N$, in other words it's a random number. This is a very important point about conditional expectation and leads to what's called the tower property.

$$\mathbb E( \mathbb E(e^{tZ}|N)) = \mathbb E(e^{tZ})$$.

You can think about this in terms of the stocks. If we were trading some complex derivitive, which meant that I'd promised to give you $e^{tZ}$ at the end of the day, that contract is "worth" $\mathbb E(e^tZ)$ to you. If you somehow find out how many trades there were, but not the values of the trades then the stock is "worth" $\mathbb E(e^{tZ}|N)$ after you've found that out. If you wanted you could sell it at this point (to somebody who also knows what $N$ is because otherwise you go to jail). If you decide at the beginning of the day that you're going to sell as soon as you find out $N$ then the stock is "worth" $\mathbb E( \mathbb E(e^{tZ}|N))$ as this is an idealized model the two prices must be the same.

So the moment generating function of $Z$ is $$\mathbb E(e^{tZ}) = \mathbb E\left(\left(e^{\frac{\sigma^2t^2}2}\right)^N\right) = e^{\lambda\left(e^{\frac{\sigma^2t^2}2}-1\right)}$$

Not something I recognise, but you can calculate the mean and variance by taking the first two derivatives at $t=0$.

-