2022-12-23

Joint probability distribution

What is joint probability distribution

A joint probability distribution is a distribution of the probability of two or more events occurring simultaneously.

For example, the joint probability of two events XX and YY is represented as P(X,Y)P(X,Y) or P(XY)P(X \cap Y), and the joint probability of three events XX, YY and ZZ is represented as P(X,Y,Z)P(X,Y,Z) or P(XYZ)P(X \cap Y \cap Z).

When the random variables are discrete, the distribution is a discrete joint probability distribution; when they are continuous, the distribution is a continuous joint probability distribution.

Discrete joint probability distribution

Suppose that the blood types of a certain elementary school class have the following distribution:

X\Y Type A Type B Type O Type AB
Boy 0.25 0.10 0.10 0.05
Girl 0.20 0.20 0.05 0.05

Considering event XX as boy or girl and event YY as blood type, joint probability that event XX is a girl and event YY is type O is 0.10.

Continuous joint probability distribution

The joint probability distribution of continuous random variables XX and YY is expressed as follows:

P(aXb,cYd)=abcdf(x,y)dxdy P(a \leq X \leq b, c \leq Y \leq d) = \int^b_a \int^d_c f(x, y)dxdy

Since the sum of the probabilities is 1, the following equation holds

f(x,y)dxdy=1 \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} f(x, y)dxdy = 1

As an example, consider the following probability density function:

f(x,y)={x+y(0x1,0y1)0(otherwise) f(x, y) = \left\{ \begin{array}{ll} x+y & (0 \leq x \leq 1, 0 \leq y \leq 1) \\ 0 & (otherwise) \end{array} \right.

The probability P(0x12,0y12)P(0 \leq x \leq \frac{1}{2}, 0 \leq y \leq \frac{1}{2}) with 0x120 \leq x \leq \frac{1}{2} and 0y120 \leq y \leq \frac{1}{2} can be obtained as follows.

P(0x12,0y12)=012012(x+y)dxdy=012[12x2+yx]012dy=012(18+12y)dy=[18y+14y2]012=18 \begin{aligned} P(0 \leq x \leq \frac{1}{2}, 0 \leq y \leq \frac{1}{2}) &= \int^{\frac{1}{2}}_{0} \int^{\frac{1}{2}}_{0} (x+y)dxdy \\ &= \int^{\frac{1}{2}}_{0}[\frac{1}{2}x^2 + yx]_0^{\frac{1}{2}} dy \\ &= \int^{\frac{1}{2}}_{0}(\frac{1}{8} + \frac{1}{2}y)dy \\ &= [\frac{1}{8}y + \frac{1}{4}y^2]^{\frac{1}{2}}_0 \\ &= \frac{1}{8} \end{aligned}

Ryusei Kakujo

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