Solution: $\mathbb{P}\left[0\le X\le 40\right]=\mathbb{P}\left[|X-20|\le 20\right]=\mathbb{P}\left[|X-\mu |\le 20\right]=1-\mathbb{P}\left[|X-\mu |>20\right]\ge 1-20∕20^2=19∕20$.

1. Look up the distribution of a Poisson random variable with parameter $\lambda$, state it and use that to calculate $\mathbb{P}\left[X\ge 1\right]$, and $\mathbb{P}\left[X\ge 2\right]$ where $X$ is a Poisson random variable with parameter $1$.
2. Given that the m.g.f. ${\varphi }_X\left(t\right)$ of a Poisson random variable with parameter $\lambda$ is $e^{\lambda \left(e^t-1\right)}$, show that the sum of independent Poisson random variables $X_1$ with parameter ${\lambda }_1$ and $X_2$ with parameter ${\lambda }_2$ is again Poisson with parameter ${\lambda }_1+{\lambda }_2$.
3. Using the fact that the sum of two independent Poisson random variables with means ${\lambda }_1$ and ${\lambda }_2$ is again Poisson with mean ${\lambda }_1+{\lambda }_2$ find the exact probability that $\mathbb{P}\left[X_1+\cdots +X_{10}>15\right]$. Take ${\lambda }_i=1$ where each $X_i$ is a Poisson random variable with parameter $1$.
4. Use the Markov Inequality to get a bound on $\mathbb{P}\left[X_1+\cdots +X_{10}>15\right]$ where each $X_i$ is a Poisson random variable with parameter $1$.

Solution:

1. $$\mathbb{P}\left[X\ge 1\right]=1-\mathbb{P}\left[X=0\right]=1-\frac{e^{-1}{\left(1\right)}^0}{0!}=1-e^{-1}$$

   and

   $$\mathbb{P}\left[X\ge 2\right]=1-\mathbb{P}\left[X=0,1\right]=1-\frac{e^{-1}{\left(1\right)}^0}{0!}-\frac{e^{-1}{\left(1\right)}^1}{1!}=1-2e^{-1}$$

2. The mgf of $X_1+X_2$ is the product of the respective mgfs of $X_1$ and $X_2$ that is $e^{{\lambda }_1\left(e^t-1\right)}\cdot e^{{\lambda }_2\left(e^t-1\right)}=e^{\left({\lambda }_1+{\lambda }_2\right)\left(e^t-1\right)}$ which is the mgf of a Poisson random variable with parameter ${\lambda }_1+{\lambda }_2$.
3. By part (b), $X_1+\cdots +X_{10}$ is Poisson with parameter 10, so the probability is
   $$\mathbb{P}\left[X_1+\dots X_{10}>15\right]\approx 0.0487404033.$$

4. $X_1,X_2,\dots ,X_{10}$ are all positive (and integer-valued) random variables and $E\left[X_1+\cdots +X_{10}\right]=10$. Then by the Markov inequality bound, $\mathbb{P}\left[X_1+\cdots +X_{10}>15\right]=\mathbb{P}\left[X_1+\cdots +X_{10}\ge 16\right]=10∕16=0.625$. The Markov bound is not sharp, although it is true.

where $X_1,X_2,\dots$ are independent, identically distributed continuous random variables with mean $0$ and variance ${\sigma }^2$. (Note that this is an additive assumption about the change in a stock price. In the binomial tree models, we assumed that a stock’s price changes by a multiplicative factor up or down. We will have more to say about these two distinct models later.) Suppose that a stock’s price today is $100$. If ${\sigma }^2=1$, what can you say about the probability that after $10$ days, the stock’s price will be between $95$ and $105$ on the tenth day?

Solution: Let $X={\sum }_{i=1}^{10}{X}_i$, so $E\left[X\right]=0$ and $Var\left[X\right]=10$ because of the independence. Then by Chebyshev’s inequality

Note that the problem does not assume that the daily changes $X_i$ are normal, so we cannot use the normal distribution for the sum. With only $10$ summands, the Central Limit Theorem cannot be reliably used.

Solutions: The m.g.f. is

Then

Then let

Now apply L’Hopital’s Rule twice to evaluate ${lim}_{n\to \infty }L\left(t,n\right)$. (Details are omitted.) The result is:

Therefore ${\varphi }_X\left(t∕\sqrt{n}\right)\to exp\left(t^2∕2\right)$ as $n\to \infty$.