Expectation and variance of stock value of company after n days Stock value of company $X$, at the beginning of day 1, is $Q$. After each day, stock value of $X$ either raises by a factor of $(1+\epsilon)$ with probability $p$, or drops by a factor $(1-\epsilon)$ with probability of $(1-p)$ (so it's either $q(1+\epsilon)$ or $q(1-\epsilon)$ where $q$ is stock value at the beginning of the day), where $0\leq\epsilon<1$ is some constant. What is $E[V], Var(V)$, where $V$ is stock value of $X$ after $n$ days?
It's easy to see, that $E[V]=\sum_{k=0}^nQ(1+\epsilon)^k(1-\epsilon)^{n-k}\binom{n}{k}p^k(1-p)^{n-k}$ - from $n$ days we choose $k$ days, after which the value raised, and then, with probability $p^k(1-p)^{n-k}$, the stock value of company is $Q(1+\epsilon)^k(1-\epsilon)^{n-k}$.
This sum, however, seems really hard to simplify, and even if we manage to somehow find the closed form, we will still need to count $E[V^2]$, in order to count the variance - which looks as if it will cause even more pain...
My intuition tells me, we need to somehow cleverly split $V$ into simpler random variables, and only then, start reasoning about $V$. But after thinking for a while I haven't come up with anything clever - I thought of reasoning on raise indicator variables on the $i$-th day, variables that would tell us net change of stock value after $i$-th day, but all these ideas lead to nowhere, as there is a deep dependency on stock value on the $i$-th and $i-1$-th day...
How to reason about such problem? Obviously $V=f(R=Binom(p))$ (if $R=k$, then $f(R)=Q(1+\epsilon)^k(1-\epsilon)^{n-k}$). Can we somehow use this fact?
 A: The expression for the mean you have written is correct. Here is how you simplify it.
$$
\operatorname{E}[V] = Q\sum_{k=0}^n\big((1+\epsilon)p\big)^k\big((1-p)(1-\epsilon)\big)^{n-k}\binom{n}{k}
$$
We recognize the summation as part of the binomial identity, so we know this is
$$
\operatorname{E}[V] = Q\big((1+\epsilon)p+(1-p)(1-\epsilon)\big)^n,
$$
or
$$
\operatorname{E}[V] = Q(1+2p\epsilon-\epsilon)^n.
$$
The summation for the variance goes like this. Take each outcome and subtract the mean outcome, square it, and do a weighted sum over the probabilities. Here is what I get:
$$
\operatorname{Var}[V] = Q\sum_{k=0}^n p^k (1-p)^{n-k}\binom{n}{k} \Big((1+\epsilon)^k(1-\epsilon)^{n-k} - \big((1+\epsilon)p+(1-p)(1-\epsilon)\big)^n\Big)^2
$$
This greatly simplifies too. I'll post a bit more later.
EDIT
The easiest approach to the variance is to directly calculate $\operatorname{E}[V^2]$:
$$
\operatorname{E}[V^2] = \sum_{k=0}^n p^k (1-p)^{n-k} \binom{n}{k} \Big(Q(1+\epsilon)^k(1-\epsilon)^k\Big)^2
$$
Rearranging
$$
\operatorname{E}[V^2] = Q^2\sum_{k=0}^n (p(1+\epsilon)^2)^k ((1-p)(1-\epsilon)^2)^{n-k} \binom{n}{k}
$$
And then recognizing that, again, as a binomial expansion and using the binomial identity:
$$
\operatorname{E}[V^2] = Q^2 (p(1+\epsilon)^2 + (1-p)(1-\epsilon)^2)^n
$$
or
$$
\operatorname{E}[V^2] = Q^2\left(1-2\epsilon-4 p\epsilon +\epsilon^2\right)^n
$$
Now finally, the variance
$$
\operatorname{Var}[V] = \operatorname{E}[V^2] - \operatorname{E}[V]^2
$$
$$
\operatorname{Var}[V] = Q^2\left(1-2\epsilon-4 p\epsilon +\epsilon^2\right)^n - \Big(Q(1+2p\epsilon-\epsilon)^n\Big)^2
$$
With a little simplifying, we get the final result
$$
\operatorname{Var}[V] = Q^2\Big(\left(1-2\epsilon-4 p\epsilon +\epsilon^2\right)^n-\left(1+2p\epsilon-\epsilon\right)^{2n}\Big)
$$
