Geometric distribution

When trials are repeated independently, the number of trials until just before the event of interest first occurs is $X$ , and if $p$ is the probability of the event of interest occurring, The probability of the event of interest $i+1$ st time is

$\displaystyle P_{r}(X = i) = (1-p)^{i}p$

Then he random variable $X$

should follow the geometric distribution $G_{e}(p)$ and denoted

$\displaystyle X \sim G_{e}(p)$

Also,

$\displaystyle E(X) = \frac{q}{p},\ V(X) = \frac{q}{p^{2}}$

One way to find the expected value $E(X)$ is to use a generating function.

Discrete case

Let $\eta(t) = E(t^{X}) = \sum_{k=0}^{\infty}t^{k}P(X = k)$ . Then $\eta'(t) = E(Xt^{X-1})$ . Thus for， $t=1$ ， $\eta(1) = E(X)$ . Therefore, the expectation can be found if $\eta(t)$ is found.

Continuous case

Let $\phi(t) = E(e^{tX}) = \int_{-\infty}^{\infty}e^{tx}dF(x)$ . Then express $e^{tx}$ by the power series. Then

$\displaystyle \phi(t)$	$\displaystyle =$	$\displaystyle \int_{-\infty}^{\infty}e^{tx}dF(x)$
	$\displaystyle =$	$\displaystyle \int_{-\infty}^{\infty}\sum_{k=0}^{k}\frac{(tx)^{k}}{k!}dF(x)$
	$\displaystyle =$	$\displaystyle \sum_{0}^{k}\frac{t^{k}}{k!}\int_{-\infty}^{\infty}x^{k}dF(x)$
	$\displaystyle =$	$\displaystyle \sum_{0}^{k}\frac{t^{k}}{k!}E(X)$

Thus

$\displaystyle \phi'(t) = \sum_{1}^{k}\frac{kt^{k-1}}{k!}E(X)$

Now set，

. Then

$\displaystyle \phi'(0) = E(X)$

Hypergeometric distribution There are $N_{1}$ white balls and $N_{2}$ black balls in the jar. Let $X$ be the number of white balls when taking out $n$ balls one by one without undoing (non-restoration extraction). The probability that the number of white balls is $i$ is given by

$\displaystyle P_{r}(X = i) = \frac{\binom{N_{1}}{i} \binom{N_{2}}{n-i}}{\binom{N}{n}}$

Then we say that the random variable $X$

follows the hypergeometric distribution and denoted by

$\displaystyle X \sim H_{g}(N,N_{1},n)$

Also,

$\displaystyle E(X) = \frac{n N_{1}}{N}, \ V(X) = \frac{n N_{1}}{N} \cdot \frac{n - N_{1}}{N} \cdot \frac{N - n}{N - 1}$

Exercise11

(a)

For $X \sim G_{e}(p)$ , show

$\displaystyle E(X) = \frac{q}{p}, \ V(X) = \frac{q}{p^2}$

Here,，

(b)

There is a team with a winning percentage of $30\%$ . On average, find out how many games you need to play to win for the first time.

(c)

There are 10 lots, 4 of which are winning and the remaining 6 are out. Now draw 3 lots and find the probability of winning 2 of them.．

Answer

(a)

$\displaystyle \eta(t)$	$\displaystyle =$	$\displaystyle E(t^X) = \sum_{k=0}^{\infty}t^{k}P(X=k)$
	$\displaystyle =$	$\displaystyle \sum_{k=0}^{\infty}t^{k}(1-p)^{k}p$

Then differentiate both sides with respect to $t$

$\displaystyle \eta'(t) = \sum_{k=1}^{\infty}kt^{k-1}(1-p)^{k}p$

Set

. Then

$\displaystyle \eta'(0) = E(X) = \sum_{k=1}^{\infty}k(1-p)^{k}p = p\sum_{k=1}^{\infty}kq^{k}$

Let $S = \sum_{k=1}^{\infty}kq^{k}$ . Then find $S - qS = S(1-q) = Sp$

$\displaystyle Sp$	$\displaystyle =$	$\displaystyle S - qS = \sum_{k=1}^{\infty}kq^{k} - \sum_{k=1}^{\infty}kq^{k+1}$
	$\displaystyle =$	$\displaystyle \sum_{k=1}^{\infty}q^{k} = \frac{q}{1-q} = \frac{q}{p}$

Thus, $S = \frac{q}{p^{2}}$ and

$\displaystyle E(X) = p \frac{q}{p^{2}} = \frac{q}{p}$

(b) Let $X$ be the numbr of games to play to win the first time. Then $X \sim G_{e}(0.3)$ ．Thus the number of games to play to win the first time is $E(X) + 1$

$\displaystyle E(X) + 1 = \frac{q}{p} + 1 = \frac{0.7}{0.3} + 1 = 3.33 .$

(c) If $X$ is the number of winning lottery, $X \sim H_{g}(10,4,3)$ . Therefore, if you draw 3 lots, the probability of winning 2 of them is

$\displaystyle P_{r}(X = 2) = \frac{{4 \choose 2}{6 \choose 1}}{{10 \choose 3}} = \frac{3}{10}$