$X = X(\omega), Y = Y(\omega), (X, Y)$ is a bivariate random variable
joint mass function
$P\{ X = x_i, Y = y_j\} = p_ij, i, j = 1, 2, …$
monotone increase
continuous on the right
$F\{ a<X≤b, c<Y≤d\} = F(b,d) - F(b,c) -F(a,d) +F(a,c)$
marginal mass function
$P\{X = x_i\} = P(\bigcup ^{+ \infty}{j =1} \{ X = x_i, Y =y_j\}) = \sum ^{+ \infty}{j= 1} p_{ij} \triangleq p_{i \cdot}, i =1, 2, …$
$\Rightarrow$ $F(x_i, + \infty)$ probabilities that $x =x_i$, all $y$
| $Y$ | $P\{X = x_i\}$ | ||||||
|---|---|---|---|---|---|---|---|
| $y_1$ | $y_2$ | … | $y_j$ | … | |||
| $x_1$ | $p_{11}$ | $p_{12}$ | … | $p_{1j}$ | … | $p_{1\cdot}$ | |
| $x_2$ | $p_{21}$ | $p_{22}$ | … | $p_{2j}$ | … | $p_{2\cdot}$ | |
| $X$ | … | … | … | … | … | ||
| $x_i$ | $p_{i1}$ | $p_{i2}$ | … | $p_{ij}$ | … | $p_{i\cdot}$ | |
| … | … | … | … | … | |||
| $P\{Y = y_j\}$ | $p_{\cdot 1}$ | $p_{\cdot 2}$ | … | $p_{\cdot j}$ | … | $1$ |
conditional mass function
given $Y = y_j$, the conditional mass function of $X$ is $P\{X =x_i|Y =y_j\} = \frac{p_{ij}}{p_{\cdot j}}, i, j= 1, 2, …$
likewise, given $X = x_i$, the conditional mass function of $Y$ is $P\{ Y = y_j | X =x_i \} = \frac{p_{ij}}{p_{i \cdot}}, i,j = 1,2,…$
<aside> ✔️ e.g.1 The joint mass function of $(x, y)$ is $F(x,y) = \left \{ \begin{array} {ll} 0 & , x<0 \text{ or } y<0 \\ 0.5y &, 0≤x<1, 0≤y<1 \\ y & ,x≥1, 0≤y<1 \\0.5 & ,0≤ x<1, y≥1 \\ 1 & , x≥1, y≥1 \end{array} \right.$, solve: (1) $F_X(x), F_Y(y)$; (2) $P(X≤0.5, Y>0.5)$. (1) from definition, when $x<0, F_X(x) = F(x, + \infty) =0$; and so we gets $F_X(x), F_Y(y)$. (2) $P(X≤ 0.5, Y>0.5) = P(X≤ 0.5) - P(X≤0.5, Y≤0.5) =F_X(0.5) - F(0.5, 0.5)$.
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<aside> ✔️ e.g.2 The probability of a shooter concentrates on the target is $p (0<p<1)$, and he shoots until he hit the target twice. Let $X$ be the times of shooting until he first hits the target and $Y$ be the total times of shooting, solve: (1) the joint mass function of $X$ and $Y$; (2) the marginal mass function of $X$ and $Y$; (3) the conditional mass function of $X$ and $Y$. (1) $P(s, t) = P(X=s)P(Y=t|X=s) = q^{s-1}\cdot p \cdot q^{t-s-1}\cdot p = p^2 q^{t-2}$ (2) $P_X(s) = pq^{s-1}$ (directly get or through $Y=t$) $P_Y(t) = C^{1}_{t-1}pq^{t-2}p = (t-1)p^2q^{t-2}$ (3) $P(X=s|Y=t) = \frac{P(s, t)}{P_Y(t)} = \frac{1}{t-1}, s= 1,2,…,t-1$. $P(Y=t|X=s) = \frac{P(s,t)}{P_X(s)} = pq^{t-s-1}$. We see in (1) that $P(s,t)$ is irrelevant with $s$, therefore for given $t$, any $s$ has a same $p$ $\Rightarrow$ in (3) we have that $P(X=s|Y=t) = \frac{P(s, t)}{P_Y(t)} = \frac{1}{t-1}$ has a value irrelevant with $s$.
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joint probability density function
$f(x,y)$ that satisfies $F(x, y) = \int^x_{- \infty} \int ^y_{- \infty}f(u,v)dudv$
(1) $\int^{\infty}{- \infty} \int^{\infty}{ -\infty} f(x, y)dxdy =1$
(2) $P\{(X, Y) \in D\} = \iint_{D} f(x,y)dxdy$
(3) at continuous point of $f(x, y)$, $\frac{\partial ^2 F(x, y)}{\partial x \partial y} = f(x,y)$
marginal probability density function