Example 2.5. Mean and Variance of the Bernoulli.伯努利随机变量的均值和方差
Consider the experiment of tossing a coin, which comes up a head with probability $p$ and a tail with probability $1 - p$. and the Bernoulli random variable $X$ with $PMF$ :
$$ p_X(k)=\cases{p, & if $k=1$\\1-p, & if $k=0$} $$The mean. second moment. and variance of $X$ are given by the following calculations:
$$ E[X]=1\cdot p + 0 \cdot (1-p) = p ,\\ E[X^2]=1^2\cdot p + 0 \cdot (1-p) = p, \\ var(x)=E[X^2]-(E[X])^2=p-p^2=p(1-p) $$Example 2.7. The Mean of the Poisson. 泊松随机变量的均值方差
The mean of the Poisson $PMF$ :
$$ p_X(k)=e^{-\lambda}\frac{\lambda^k}{k!}, k=0,1,2,\ldots $$can be calculated is follows:
$$ \begin{eqnarray} E[X] &=& \sum\limits^{\infty}_{k=0}ke^{-\lambda}\frac{\lambda^k}{k!}\\ &=& \sum\limits^{\infty}_{k=1}ke^{-\lambda}\frac{\lambda^k}{k!} \quad (\text{k=0的项等于0})\\ &=& \lambda\sum\limits_{k=1}^{\infty}e^{-\lambda}\frac{\lambda^{k-1}}{k-1!}\\ &=& \lambda\sum\limits_{m=0}^{\infty}e^{-\lambda}\frac{m^{k-1}}{m!} \quad (\text{让m=k-1})\\ &=& \lambda \end{eqnarray} $$The last equality is obtained by noting that is the normalization property for the Poisson $PMF$.
Example 2.20. Variance of the Binomial and the Poisson. 二项随机变量和泊松随机变量的方差
We consider $n$ independent coin tosses, with each toss having probability $p$ of coming up a head. For each $i$, we let $X_i$ be the Bernoulli random variable which is equal to $1$ if the $i$th toss comes up a head, and is 0 otherwise. Then, $X = X_l + X_2 + . . . + X_n$ is a binomial random variable. Its mean is $E[X] = np$. as derived in Example 2. 10. By the independence of the coin tosses. the random variables $X_1 , . . . . X_n$ are independent, and
$$ var(X)=\sum\limits_{i=1}^{n}var(x_i)=np(1-p) $$As we discussed in Section 2.2. a Poisson random variable $Y$ with parameter $\lambda$ can be viewed as the “limit” of the binomial as $n\rightarrow \infty, p\rightarrow 0$. while $np = \lambda$. Thus, taking the limit of the mean and the variance of the binomial. we informally obtain the mean and variance of the Poisson: $E[Y] = var(Y) = \lambda $ . We have indeed verified the formula $E[Y] = \lambda$ in Example 2.7. To verify the formula $var(Y) = \lambda$, we write
$$ \begin{eqnarray} E[Y^2] &=& \sum\limits_{k=1}^{\infty}k^2e^{-\lambda}\frac{\lambda^k}{k!} \\ &=& \lambda\sum\limits_{k=1}^{\infty}k\frac{e^{-\lambda}\lambda^{k-1}}{(k-1)!} \\ &=& \lambda\sum\limits_{m=0}^{\infty}(m+1)\frac{e^{-\lambda}\lambda^{m}}{m!}, m = k-1 \\ &=& \lambda[\sum\limits_{m=0}^{\infty}m\frac{e^{-\lambda}\lambda^{m}}{m!}+\sum\limits_{m=0}^{\infty}\frac{e^{-\lambda}\lambda^{m}}{m!}], \\ &=& \lambda(E[Y]+1) , \\ &=& \lambda(\lambda+1) \\ \end{eqnarray} $$from which
$$ var(Y)=E[Y^2]-(E[Y])^2=\lambda(\lambda+1)-\lambda^2=\lambda $$