Continuous random variables (Further) · 连续随机变量(Further)
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| continuous variable/kənˈtɪnjuːəs ˈveərɪəbl/ | 连续变量 | lián xù biàn liàng |
| probability density function/ˌprɒbəˈbɪlɪti ˈdensɪti ˈfʌŋkʃn/ | 概率密度函数 | gài lǜ mì dù hán shù |
| expectation/ekspɪkˈteɪʃn/ | 期望 | qī wàng |
| cumulative distribution function/ˈkjuːmjʊlətɪv ˌdɪstrɪˈbjuːʃn ˈfʌŋkʃn/ | 累积分布函数 | lěi jī fēn bù hán shù |
| median/ˈmiːdiːən/ | 中位数 | zhōng wèi shù |
| percentile/pəˈsentaɪl/ | 百分位数 | bǎi fēn wèi shù |
| quartile/ˈkwɔːtaɪl/ | 四分位数 | sì fēn wèi shù |
Probability as area
- For a continuous variable 连续变量 — a height, a waiting time — no single exact value has a probability; there are infinitely many.
- Instead, probability is the area under a curve, the probability density function 概率密度函数 $f(x)$. The whole area is $1$.
概率作为面积
- 对一个连续变量——一个身高、一个等待时间——没有单个精确的值有一个概率;有无穷多个。
- 相反,概率是一条曲线下的面积,概率密度函数(probability density function)$f(x)$。整个面积是 $1$。
The density function and expectation 期望
- $P(a < X < b) = \displaystyle\int_a^b f(x)\,dx$, and the total area $\int f = 1$.
- The mean of any function of $X$:
Probabilities are areas under f(x); the total area over the whole range is exactly 1.
密度函数和期望
- $P(a < X < b) = \displaystyle\int_a^b f(x)\,dx$,而总面积 $\int f = 1$。
- $X$ 的任何函数的均值:
概率是 f(x) 下面的面积;整个范围上的总面积恰好是 1。
Continuous random variables · 连续随机变量
P(a < X < b) = ∫ f(x) dx
For a continuous variable, probability · 概率 is the area · 面积 under the density curve. · 对一个连续变量,概率是密度曲线下面的面积。
What must the total area under a probability density function equal? · 一个概率密度函数下面的总面积必须等于多少?
A pdf integrates to 1 over its whole range. · 一个概率密度函数在它整个范围上积分为 1。
The cumulative distribution function 累积分布函数
- The CDF accumulates probability from the left:
- It runs from $0$ to $1$, and differentiating recovers the density: $f(x) = F'(x)$.
The CDF $F(x)$ rises from 0 to 1; the median is read off where it reaches a height of 0.5
累积分布函数
- CDF 从左边累积概率:
- 它从 $0$ 到 $1$,而求导恢复密度:$f(x) = F'(x)$。
CDF $F(x)$ 从 0 升到 1;中位数在它达到高度 0.5 的地方读出
The cumulative distribution function F(x) gives: · 累积分布函数 F(x) 给出:
F(x) = P(X ≤ x), the accumulated area up to x. · F(x) = P(X ≤ x),到 x 为止累积的面积。
Differentiating the CDF gives the density: f(x) = F-(x). · 对 CDF 求导给出密度:f(x) = F-(x)。
f(x) = F′(x): the density is the derivative of the CDF. · f(x) = F′(x):密度是 CDF 的导数。
Worked example — the median 中位数
- The median $m$ splits the area in half: $F(m) = 0.5$.
- For $f(x) = \tfrac12 x$ on $[0, 2]$: $F(x) = \tfrac14 x^2$, so $\tfrac14 m^2 = 0.5$ gives $m = \sqrt2 \approx 1.41$.
Percentiles 百分位数 work the same way. The lower quartile 四分位数 solves $F(x) = 0.25$; the 90th percentile solves $F(x) = 0.9$. Always set $F$ equal to the fraction you want.
例题——中位数
- 中位数(median)$m$ 把面积分成两半:$F(m) = 0.5$。
- 对 $[0, 2]$ 上的 $f(x) = \tfrac12 x$:$F(x) = \tfrac14 x^2$,所以 $\tfrac14 m^2 = 0.5$ 给出 $m = \sqrt2 \approx 1.41$。
百分位数以同样的方式工作。 下四分位数解 $F(x) = 0.25$;第 90 百分位数解 $F(x) = 0.9$。总是把 $F$ 设为你想要的分数。
For f(x) = ½x on [0,2], F(x) = ¼x². The median solves ¼m² = 0.5. What is m? (2 dp) · 对 [0,2] 上的 f(x) = ½x,F(x) = ¼x²。中位数解 ¼m² = 0.5。m 是多少?(2 位小数)
¼m² = 0.5 → m² = 2 → m = √2 ≈ 1.41. · ¼m² = 0.5 → m² = 2 → m = √2 ≈ 1.41。
The median m of a continuous distribution satisfies: · 一个连续分布的中位数 m 满足:
The median splits the area in half: F(m) = 0.5. · 中位数把面积分成两半:F(m) = 0.5。
Density is not probability
$f(x)$ can exceed 1. A density isn't a probability — only the area is. A tall narrow peak can have $f(x) > 1$ as long as the total area stays $1$.
- Use cumulative distribution functions (CDFs) to find probabilities and percentiles, and the CDF of a related variable.
密度不是概率
$f(x)$ 可以超过 1。 一个密度不是一个概率——只有面积是。一个又高又窄的峰可以有 $f(x) > 1$,只要总面积保持 $1$。
- 用累积分布函数(cumulative distribution functions,CDF)求概率和百分位数,以及相关变量的 CDF。
A probability density f(x) can be greater than 1. · 一个概率密度 f(x) 可以大于 1。
Only the area must be ≤ 1; the density itself can exceed 1 over a narrow interval. · 只有面积必须 ≤ 1;密度本身在一个窄区间上可以超过 1。
You've got it
- probability = area under $f(x)$; total area $= 1$ (density itself can exceed 1)
- the CDF $F(x) = P(X \leq x) = \int f$, and $f = F'$
- the median solves $F(m) = 0.5$; percentiles solve $F(x) =$ that fraction
你掌握了
- 概率 = $f(x)$ 下面的面积;总面积 $= 1$(密度本身可以超过 1)
- CDF $F(x) = P(X \leq x) = \int f$,而 $f = F'$
- 中位数解 $F(m) = 0.5$;百分位数解 $F(x) =$ 那个分数