Stock increase multiplicatively, expected value and variance Assume that a stock starts with price $1$. Each day, if the stock starts with $q$, then with probability $p$ it increases to $qr$, and with probability $1-p$ it decreases to $q/r$. What is the expected value and variance of the price of stock on $d$th day?
The expected value is by definition $\sum_i i\cdot Pr[X=i]=\sum_{i=0}^d\dbinom{d}{i}p^i(1-p)^{d-i}r^{2i-d}$. Is there any way to simplify this sum? (Similar question for the variance, which is calculated using $E[X^2]-E[X]^2$)
 A: For the expected value I have the formula:
$q\cdot  \sum_{i=0}^{d} {d \choose i} (r\cdot p)^i \cdot \left( \frac{1}{r}(1-p)\right)^{d-i}$
$\sum_{i=0}^{d} {d \choose i} (r\cdot p)^i \cdot \left( \frac{1}{r}(1-p)\right)^{d-i}$ can be simplified by using the binomial theorem.
For $r=1.25,d=2,p=0.7$ and $q=100$ the expected value is $124.3225$.
A: A far easier approach seems to be using a recursive method, combined with the linearity of conditional expectation. Let X be a binary random variable that dictates whether a stock price at day (d+1) is the result of an increase or decrease of a stock price at day (d). Then:
$$ \mathbb{E}[sp_d] = p(\mathbb{E}[sp_{d-1}|X = 0]) + (1-p)(\mathbb{E}[sp_{d-1}|X = 1) $$ 
This gives us:
$ \mathbb{E}[sp_d] = pr(\mathbb{E}[sp_{d-1}]) + \frac{1-p}{r}(\mathbb{E}[sp_{d-1}]) $
Noticing that this is an easy power-recursion with 
$\mathbb{E}[sp_0] = 1$ tells us that:
$ \mathbb{E}[sp_d] = (pr + \frac{1-p}{r})^d $
In your case, $\mathbb{E}[sp_0] = q$, which simply scales the above as:
$ \mathbb{E}[sp_d] = q(pr + \frac{1-p}{r})^d $
Seems far more elegant than working out an ugly binomial sum, only to apply the binomial formula to simplify said sum later.
