Distribution function

$F(x) = P\{ X ≤ x\}$

(1) $F(-\infty)=0$, $F(\infty) = 1$

(2) monotone increase

(3) continuous on the right

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e.g.1 $F(x) = \left \{ \begin{array}{rcl} A+Be^{-x}, x>0 \\ 0, x\leq 0 \end{array}\right.$, solve A, B. $F(0) = 0 = F(0+) = A+B$ $F(+ \infty) = A = 1$ $B = 0- A = -1$

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(4) $P\{ a< X \leq b\} = F(b) - F(a)$

(5) $P\{ X<x \} = F(x-), P\{X=x\} = F(x)-F(x-)$

Discrete random variable

Probability mass function

$F(x) = P \{X ≤ x \} = \sum _{x_i \leq x} p_i$

Continuous random variable

probability density function

$f(x)$ that satisfies $F(x) = \int ^{x} _{-\infty} f(t)dt$

(1) $P \{ X=x\} = F(x) - F(x-) = 0$

(2) $\int ^{+\infty}_{-\infty} f(x)dx=1$

(3) $P \{ a< x≤b\} = \int^b_a f(x)dx$, here the $<$ equals $\leq$, from (1)

(4) at continuous points, $F'(x) = f(x)$

Eight major distributions

0-1 distribution

$X \sim B(1,p)$

Binary distribution

$X \sim B(n,p)$

$Y = n-X \sim B(n, 1-p)$

Generally, if $(n+1) p \in N$, the most probable value of $X$ is $(n+1)p$ and $(n+1)p-1$; if not, it is $[(n+1)p]$

<aside> ✔️ e.g.2 $P(X=1) = P(X=2) = 0.5$, when $X = x$, $Y\sim P(x)$, solve: (1) $P(Y≥1)$; (2) $P(X=1|Y≥1)$. $Y|X = x \sim P(x)$, define $Y|X = x$ as $Z_x$. then $P(z_x = k) = \frac{x^k e^{-x}}{k!}, k =0, 1, 2, …$ (1) $P(Y≥1) = P(X=1)P(Y≥1|X=1)+P(X=2)P(Y≥1|X+1) = \frac{1}{2}(P(Z_1\ge 1)+P(Z_2 \ge1)) = ...$ (2) $P(X=1|Y≥1) = \frac{P(X=1)P(Y≥1|X=1)}{P(Y≥1)} = ...$

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Poisson distribution

$X \sim P(\lambda)$

$P \{ X=k\} = \frac{e^{-\lambda}\lambda^{k}}{k!}, k=0,1, 2,...$

Derivation: Poisson distribution could be derived from binary distribution, when $n → +\infty$ $C_n^k p^k (1-p)^{n-k} = \frac{n!}{(n-k)!k!} p^k (1-p)^{n-k} = \frac{1}{k!}(1-\frac{1}{n})…(1-\frac{k-1}{n})(np)^k(1-p)^{n-k} \overset{\text{when }n \rightarrow + \infty, \text{ the fraction part}\rightarrow1 }{=} \frac{1}{k!}(np)^k(1-p)^{n-k}= \frac{1}{k!}(np)^k(1-p)^{-\frac{1}{p}(kp-np)} = \frac{1}{k!}(np)^ke^{(kp-np)}\overset{n \gg k}{=}\frac{1}{k!}(np)^ke^{-np} \overset{\lambda = np}{=} \frac{e^{-\lambda}\lambda^{k}}{k!}$

<aside> ✔️ e.g.3 the number of clients to the store each day is a random variant $X \sim P(\lambda)$, the probability of each client shopping is $p$, and each client is independent of one another. Solve the probability distribution of the number of clients shopping in the store each day( $Y$). $P(Y= k) = \sum ^{+ \infty}{n= k} P(X=n)P(Y=k|X=n) =\sum^{+ \infty}{n=k}\frac{e^{-\lambda}\lambda^{n}}{n!}C^{k}{n}p^k(1-p)^{n-k} = \sum^{+ \infty}{n=k}\frac{e^{-\lambda}\lambda^{n}}{n!}\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}= \frac{(\lambda p)^k e^{-\lambda}}{k!} \sum^{+\infty}{n=k} \frac{(\lambda q)^{n-k}}{(n-k)!}\overset{\sum^{+\infty}{n=k}\frac{\lambda q)^{n-k}}{(n-k)!} = \sum^{+\infty} _{n=0}\frac{\lambda q)^{n}}{n!}= e^{\lambda q}}{=} \frac {(\lambda q-\lambda}{k!}=\frac{(\lambda p)^k e^{-\lambda p}}{k!}$ therefore, $Y \sim P(\lambda p)$

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Geometric distribution