Continuous Probability Distribution

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Basic Concept

Unlike discrets probability distribution, continuous probability distribution is continuous. The variables of the former can only be integers, while the latter is the entire field of real numbers.

Similar to discrete notation, the probability is usually written as $P(X \leq x)$. However, it should be noted that for a continuous probability distribution, $P(X=x)=0$, that is, for a continuous probability distribution, the probability of obtaining a particular value is 0. Notes: the reference to “probability 0” here DOES NOT mean that it is impossible (it can be analogous to “taking a random point on the field of real numbers, the probability that the point is rational is 0”).

Note that, as with a discrete probability distribution, for a continuous probability distribution, the sum of all its events must be 1, i.e
$$\int_{-\infty}^{\infty}f(x)=1$$
This is often used as a way of checking whether a probability distribution is valid.

We usually write the Probability distribution function as $f(x)$. So, for such a distribution, $P(a \leq x \leq b)=\int_{a}^{b} f(x) \ \mathrm{d}x$. In fact, the same conclusion can be drawn from this, $P(X=x)=0$; after all, for an integral, when the upper and lower limits are equal, it must be equal to 0.

According to the Probability distribution function, We get the Cumulative distribution function: $F(x)= P(X \leq x)=\int_{-\infty}^{x} f(t) \ \mathrm{d}t$

Expactation: $\mu = \mathbb{E}[x]=\int_{-\infty}^{\infty}xf(x) \mathrm{d}x$

Varience: $Var(x) = \int_{-\infty}^{\infty}x^2 f(x) \mathrm{d}x - \mu^2$

Median: $m$ satisfies $P(x \leq m)=P(x \leq m) = 1/2$. It is usually written in the Cumulative distribution function form, such as
$$\int_{-\infty}^{m} f(x)\ \mathrm{d}x \ = \ \int_{m}^{\infty}f(x) \mathrm{d}x = 1/2$$

Mode: where the probability distribution has a maximum

Normal Distribution

Symbol: $X \sim N(\mu, {\sigma}^2)$, 其中$\mu$代表Expectation $\mathbb{E}(x)$, ${\sigma}^2$ 代表Variance $Var(x)$

Definition: $f(x) = \frac{1}{\sqrt{2 \pi \sigma}} e^{-\frac{(x-\mu)^{2}}{2{\mu}^{2}}}$

It is important to note that for this function, its integral (i.e., Cumulative distribution function) cannot be expressed as an elementary function. The questions usually give, for example, $\phi(z)=\int_{-\infty}^{t} f(x) \mathrm{d}x$

Facts:

• For a normal distribution, its mode is $\mu$(that is, the x value corresponding to the highest point of the image), and the function image is symmetric along $x=\mu$.

• If $x \sim B(u, p)$ (Notes：$B(u, p)$ is Binomial distribution), and n is “large”, p is “close to” 1/2, then $x$ can be approximated as a normal distribution $x \sim N(np, np(1-p))$ (类似于“抛硬币”).

• If $x \sim Po(\lambda)$ (Notes: $Po(\lambda)$ is Poisson distribution), and $\lambda$ is large, the $x$ can be approximated by a normal distribution $x \sim N(\lambda, \lambda)$.