Here is a problem that I was expected to solve using normal approximation but I misread it and completely solved the problem. Let me know if the approach is correct; it's too hard to compute binomial coefficients for a number of terms.
The problem Statement
A stock trader has 100 shares of a company and adopts a random strategy for buying/selling of shares based on the outcomes of a die roll at each trading session. The strategy is to roll a die and make a choice based on the outcome:
- Outcome =1 , then Do Nothing
- 2 or 4 , Buy one share
- 3 or 5, Sell one share
- 6 , Buy two shares
Calculate an approximate probability that after 18 independent trading sessions the trader will have more than 110 shares of the company.
The Solution This is the solution that I have attempted. The probabilities for the following events are
- Event = N ,Do nothing. Probability 1/6
- Event = BO, Buy One. Probability 2/6
- Event = SO,Sell One. Probability 2/6
- Event = BT, Buy Two. Probability 1/6
The solution would be coefficient of $x^{10}$ (110 shares -100 shares) in the expansion of the following multinomial
$$\Big( \frac{1}{6} + \frac{2x}{6} + \frac{2}{6x} + \frac{x^{2}}{6} \Big)^{18}$$
$$= \frac{1}{6^{18}} \Big( 1 + 2x + \frac{2}{x} + x^{2}\Big)^{18}$$
$$ = \frac{1}{6^{18}} \Big( 1 + x^{2} + 2\frac{(x^{2} +1)}{x} \Big)^{18}$$
$$ = \frac{1}{6^{18}} \frac{1}{x^{18}} (1+x^{2})^{18} (x+2)^{18} $$
Coefficients of $x$ of this expansion are $$\binom{18}{r} (x^{2})^{r} \binom{18}{k} x^{k} 2^{18-k}$$
We are interested in coefficients of $x^{m}$ where $m \ge 10$. Hence we need to find integer pairs $(r,k)$ such that $$2r + k -18 = 10$$ $$2r + k = 28$$ $$r = 14 - \frac{k}{2}$$
Since $max(k)=18$ and $max(r)=18$ , the acceptable range of values for $k$ are $0,2,4,6,8,10,12,14,16,18$ and $14 \ge r \ge 5 $. The solution is
$$ \frac{1}{6^{18}} \sum_{k,r} \binom{18}{r} \binom{18}{k} 2^{18-k}$$
The normal approximation only requires us to calculate the mean and variance of the change in the number of shares for a single session, then use the Central Limit Theorem to obtain an approximate probability. That is to say, let $X_i \in \{-1, 0, 1, 2\}$ be the random number of shares traded in session $i$. We have $$\operatorname{E}[X_i] = 1/3, \quad \operatorname{Var}[X_i] = 11/9.$$ Therefore, under the normal approximation, $$S = 100 + \sum_{i=1}^{18} X_i \sim \operatorname{Normal}(\mu = 106, \sigma^2 = 22),$$ and the rest is straightforward. Since $S$ is integer, it may be a good idea to use continuity correction in this case.
Whether this is a good approximation is a question that would be answered by computing the exact probability.