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--- b:head_first_statistics:using_discrete_probability_distributions [2025/09/23 13:22] – [Fat Dan changed his prices] hkimscil
+++ b:head_first_statistics:using_discrete_probability_distributions [2025/09/23 14:17] (current) – hkimscil
@@ Line 92: / Line 92: @@
 <code>
 k <- c(-1,4,9,14,19)
-prob <- c(.977,.008,.008,.006,.001)
+pk <- c(.977,.008,.008,.006,.001)
-k * prob
+k * pk
-sum(k * prob)
+sum(k * pk)
-mu <- sum(k * prob)
+mu <- sum(k * pk)
 k - mu
 (k - mu)^2
-((k - mu)^2) * prob
+((k - mu)^2) * pk
-sum(((k - mu)^2) * prob)
+sum(((k - mu)^2) * pk)
 </code>
 <code>
 > k <- c(-1,4,9,14,19)
-> prob <- c(.977,.008,.008,.006,.001)
+> pk <- c(.977,.008,.008,.006,.001)
-> k * prob
+> k * pk
 [1] -0.977  0.032  0.072  0.084  0.019
-> sum(k * prob)
+> sum(k * pk)
 [1] -0.77
-> mu <- sum(k * prob)
+> mu <- sum(k * pk)
 > k - mu
 [1] -0.23  4.77  9.77 14.77 19.77
 > (k - mu)^2
 [1]   0.0529  22.7529  95.4529 218.1529 390.8529
-> ((k - mu)^2) * prob
+> ((k - mu)^2) * pk
 [1] 0.0516833 0.1820232 0.7636232 1.3089174 0.3908529
-> sum(((k - mu)^2) * prob)
+> sum(((k - mu)^2) * pk)
 [1] 2.6971
 >
@@ Line 221: / Line 221: @@
 <code>
-k <- c(-2,   23,   48,   73,   98)
+n <- c(-2,   23,   48,   73,   98)
-p <- c(0.977, 0.008, 0.008, 0.006, 0.001)
+pn <- c(0.977, 0.008, 0.008, 0.006, 0.001)
-k * p
+n * pn
-sum(k * p)
+sum(n * pn)
-k.mu <- sum(k * p)
+n.mu <- sum(n * pn)
-k.mu
+n.mu
 </code>
 <code>
-> k <- c(-2,   23,   48,   73,   98)
+> n <- c(-2,   23,   48,   73,   98)
-> p <- c(0.977, 0.008, 0.008, 0.006, 0.001)
+> pn <- c(0.977, 0.008, 0.008, 0.006, 0.001)
-> k * p
+> n * pn
 [1] -1.954  0.184  0.384  0.438  0.098
-> sum(k * p)
+> sum(n * pn)
 [1] -0.85
-> k.mu <- sum(k * p)
+> n.mu <- sum(n * pn)
-> k.mu
+> n.mu
 [1] -0.85
->
 >
 </code>
 The expectation is slightly lower, so in the long term, we can expect to lose %%$%% 0.85 each game. The variance is
 much larger. This means that we stand to lose more money in the long term on this machine, but there’s less
 certainty.
 Do
 <code>
-kj.mu <- sum(k*j)
+n.mu <- sum(n * pn)
-kj.var <- sum((k-kj.mu)^2*j)
+n.var <- sum((n - n.mu)^2 * pn)
-kj.var
+n.var
 </code>
-<code>> kj.mu <- sum(k*j)
+<code>
-> kj.var <- sum((k-kj.mu)^2*j)
+> n.mu <- sum(n * pn)
-> kj.var
+> n.var <- sum((n - n.mu)^2 * pn)
+> n.var
 [1] 67.4275
-> </code>
+>
+>
+</code>
 <WRAP col2>
@@ Line 315: / Line 316: @@
 \end{eqnarray*}
+===== 확률의 연산 =====
 \begin{eqnarray*}
@@ Line 326: / Line 328: @@
 E(X1 + X2 + \ldots Xn) & = & nE(X) \\
 Var(X1 + X2 + \ldots Xn) & = & nVar(X) \\
+\\
 \end{eqnarray*}
@@ Line 340: / Line 343: @@
 \end{eqnarray*}
 ----
+===== e.g. =====
 Eg.
 {{discrete.prob.fortune.cookie.eg.jpg?500}}
@@ Line 348: / Line 351: @@
 p.fc <- c(0.8, 0.1, 0.07, 0.03)
-exp.fc <- sum(v.fc*p.fc)
+exp.fc <- sum(v.fc * p.fc)
-var.fc <- sum((v.fc-exp.fc)^2*p.fc)
+var.fc <- sum((v.fc - exp.fc)^2 * p.fc)
 exp.fc
 var.fc
@@ Line 358: / Line 361: @@
 p.fc <- c(0.8, 0.1, 0.07, 0.03)
-exp.fc2 <- sum(v.fc2*p.fc)
+exp.fc2 <- sum(v.fc2 * p.fc)
-var.fc2 <- sum((v.fc2-exp.fc2)^2*p.fc)
+var.fc2 <- sum((v.fc2 - exp.fc2)^2 * p.fc)
 exp.fc2
 var.fc2
@@ Line 400: / Line 403: @@
 \end{eqnarray*}
-----
+===== e.g.2 =====
 A restaurant offers two menus, one for weekdays and the other for weekends. Each menu offers four set prices, and the probability distributions for the amount someone pays is as follows:
@@ Line 512: / Line 516: @@
 </code>
-====== Theorems ======
+SEE [[:Expected value and variance properties]]
-| $E(X)$ | $\sum{X}\cdot P(X=x)$  |
-| $E(X^2)$ | $\sum{X^{2}}\cdot P(X=x)$  |
-| $E(aX + b)$ | $aE(X) + b$  |
-| $E(f(X))$ | $\sum{f(X)} \cdot P(X=x)$  |
-| $E(aX - bY)$ | $aE(X)-bE(Y)$  |
-| $E(X1 + X2 + X3)$ | $E(X) + E(X) + E(X) = 3E(X) \;\;\; $ ((X1,X2,X3는 동일한 statistics을 갖는 (X의 특성을 갖는, 즉, 집합 X의 동일한 mean, variance, sdev 값을 갖는) 집합))   |
-| $Var(X)$ | $E(X-\mu)^{2} = E(X^{2})-E(X)^{2} \;\;\; $   see $\ref{var.theorem.1} $ |
-| $Var(c)$  | $0 \;\;\; $ see $\ref{var.theorem.41}$   |
-| $Var(aX + b)$ | $a^{2}Var(X) \;\;\; $  see $\ref{var.theorem.2}$ and $\ref{var.theorem.3}$ |
-| $Var(aX - bY)$ | $a^{2}Var(X) + b^{2}Var(Y)$ see 1 |
-| $Var(X1 + X2 + X3)$ | $Var(X) + Var(X) + Var(X) = 3 Var(X) \;\;\; $ ((X1, x2, x3는 동일한 특성을 (statistic, 가령 Xbar = 0, sd=1) 갖는 독립적인 세 집합이다. 따라서 세집합의 분산은 모두 1인 상태이고, 이들의 분삽값은 모두 동일하므로 Var(3X)의 성질을 갖는다.))  |
-| $Var(X1 + X1 + X1)$  | $Var(3X) = 3^2 Var(X) = 9 Var(X) $  |
-\begin{eqnarray*}
-Var(aX - bY) & = & Var(aX + -bY) \\
-& = & Var(aX) + Var(-bY) \\
-& = & a^{2}Var(X) + b^{2}Var(Y)
-\end{eqnarray*}
-see also [[:why n-1]]
-====== Variance Theorem 1 ======
-\begin{align}
-Var[X] & = {E{(X-\mu)^2}}  \nonumber \\
-& = E[(X^2 - 2 X \mu + \mu^2)] \nonumber \\
-& = E[X^2] - 2 \mu E[X] + E[\mu^2] \nonumber \\
-& = E[X^2] - 2 \mu E[X] + E[\mu^2], \; \text{because E[X]=} \mu \text{, and E[} \mu^2 \text{] = } \mu^2, \nonumber \\
-& = E[X^2] - 2 \mu^2 + \mu^2   \nonumber \\
-& = E[X^2] - \mu^2 \nonumber \\
-& = E[X^2] - E[X]^2 \label{var.theorem.1} \tag{variance theorem 1} \\
-\end{align}
-====== Theorem 2: Why square ======
-$ \ref{var.theorem.1} $ 에 따르면
-$$ Var[X] = E[X^2] − E[X]^2 $$
-이므로
-\begin{align*}
-Var[aX] & = & E[a^2X^2] − (E[aX])^2 \\
- & = & a^2 E[X^2] - (a E[X])^2 \\
- & = & a^2 E[X^2] - (a^2 E[X]^2) \\
- & = & a^2 (E[X^2] - (E[X])^2) \\
- & = & a^2 (Var[X]) \label{var.theorem.2} \tag{variance theorem 2} \\
-\end{align*}
-====== Theorem 3: Why Var[X+c] = Var[X] ======
-\begin{align}
-Var[X + c] = Var[X] \nonumber
-\end{align}
-$ \ref{var.theorem.1} $ 에 따르면
-$$ Var[X] = E[X^2] − E[X]^2 $$
-이므로
-\begin{align}
-Var[X + c]
-= & E[(X+c)^2] - E[X+c]^2 \nonumber \\
-= & E[(X^2 + 2cX + c^2)] \label{tmp.1} \tag{temp 1} \\
-  & − E(X + c)E(X + c)  \label{tmp.2} \tag{temp 2} \\
-\end{align}
-$ \ref{tmp.1} $ 에서
-\begin{align}
-E (X^2 + 2cX + c^2) = E (X^2) + 2cE(X) + c^2  \\
-\end{align}
-그리고 $\ref{tmp.2}$ 에서 보면
-\begin{align}
-E(X + c)E(X + c) = & E(X)(E(X + c)) + E(c)(E(X + c)) \nonumber \\
-= & E(X)^2 + cE(X) + cE(X) + c^2 \nonumber \\
-= & E(X)^2 + 2cE(X) + c^2 \\
-\end{align}
-위의 둘을 모두 보면
-\begin{align}
-Var(X + c) = & E(X^2) + 2cE(X) + c^2 − E(X)^2 − 2cE(X) − c^2 \nonumber \\
-= & E(X^2) − E(X)^2 \nonumber \\
-= & Var(X) \label{var.theorem.3} \tag{variance theorem 3} \\
-\end{align}
-====== Theorem 4: Var(c) = 0 ======
-\begin{align}
-Var(X) = & 0; \;\;\;\; \text{if   X = c, a constant }  \label{var.theorem.41} \tag{variance theorem 4.1} \\
-\text{otherwise    } \nonumber \\
-Var(X) \ge & 0 \label{var.theorem.42} \tag{variance theorem 4.2} \\
-\end{align}
-Variance는 기본적으로 아래와 같다. 이 때 $X=c$ 라고 (c=상수) 하면
-\begin{align}
-Var(X) & = E[(X − E(X))^2] \text{    because  X = c, and E(X) = c}    \nonumber \\
-& = E[(c-c)^2] \nonumber  \\
-& = 0   \nonumber  \\
-\text{if X  } \ne \text{c, then} \nonumber  \\
-&  \text{because   }  (X − E(X))^2 \ge 0 \nonumber \\
-& Var(X) \ge 0 \nonumber
-\end{align}
-====== Theorem 5: Var(X+Y) ======
-\begin{align}
-Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y) \label{var.theorem.51} \tag{variance theorem 5-1} \\
-Var(X − Y) = Var(X) + Var(Y) − 2Cov(X, Y) \label{var.theorem.52} \tag{variance theorem 5-2} \\
-\end{align}
-$ \ref{var.theorem.1} $ 에서
-$$ Var[X] = E[X^2] - E[(X)]^2  $$
-이므로 X <- X+Y 를 대입해보면
-\begin{align}
-Var[X+Y] = & E[(X + Y)^2]  \label{tmp.03} \tag{temp 3} \\
-- & E[(X + Y)]^2  \label{tmp.04} \tag{temp 4}
-\end{align}
-$\ref{tmp.03}$과 $\ref{tmp.04}$ 는 아래처럼 정리된다
-\begin{align*}
-  &  E[(X + Y)^2] = E[X^2 + 2XY + Y^2] = E[X^2] + 2E[XY] + E[Y^2] \\
-- & [E(X + Y)]^2 = [E(X) + E(Y)]^2 = E(X)^2 + 2E(X)E(Y) + E(Y)^2 \\
-\end{align*}
-각 줄의 가장 오른쪽 정리식을 보면,
-\begin{align*}
-Var[(X+Y)] =
-  & E[X^2] & + & 2E[XY] & + & E[Y^2] \\
-- & E(X)^2 & - & 2E(X)E(Y) & - & E(Y)^2 \\
-  & Var[X] & + & 2 E[XY]-2E(X)E(Y) & + & Var[Y] \\
-\end{align*}
-가운데 부분은
-\begin{align}
-E(XY)- E(X)E(Y) = Cov[X,Y] \label{cov} \tag{covariance} \\
-\end{align}
-따라서
-\begin{align*}
-Var[(X+Y)] = Var[X] + 2 Cov[X,Y] + Var[Y] \\
-\end{align*}
-====== Questions ======
-Which one is correct?
-\begin{align}
-Var(X+X) & = Var(X) + Var(X) & = 2 * Var(X)  \label{tmp.05} \tag{1} \\
-Var(X+X) & = Var(2X) & = 2^2 * Var(X) \label{tmp.06} \tag{2}
-\end{align}
-$\ref{var.theorem.51}$ 을 다시 보면
-\begin{align*}
-Var(X+Y) = Var(X) + 2 Cov(X,Y) + Var(Y) \\
-\end{align*}
-X와 Y가 independent 한 event라고 (group) 하면
-$ Cov(X,Y) = 0 $ 이므로
-\begin{align*}
-Var[(X+Y)] = Var[X] + Var[Y] \\
-\end{align*}
-보통 X1, X2 집합은 같은 특성을 (statistic) 갖는 두 독립적인 집합을 의미하므로
-\begin{align*}
-Var(X1 + X2)  = Var(X1) + Var(X2)  \\
-\end{align*}
-X1, X2는 같은 분포를 갖는 서로 독립적인 집합이고 (가령 각 집합은 n=10000이고 mean=0, var=4의 특성을 갖는) 이 때의 두 집합을 합한 집합의 Variance는 각 Variance를 더한 값과 같다는 뜻. 반면에 아래는 동일한 집합을 선형적인 관계로 옮긴 것 (X to 2X).
-\begin{align}
-Var(X1 + X1) & = Var(2*X1) \nonumber \\
-& = 2^2 Var(X1) \nonumber \\
-& = 4 Var(X1)  \nonumber \\
-\end{align}
-따라서 수식 $(\ref{tmp.06})$ 가 참이다.
-이것은 아래처럼 생각해 볼 수도 있다.
-$\ref{var.theorem.51}$ 에서 $Y$ 대신에 $X$를 대입하면
-\begin{align*}
-Var(X + X) & = Var(X) + 2 Cov(X, X) + Var(X)  \\
-& \;\;\;\;\; \text{because } \\
-& \;\;\;\;\; \text{according to the below } \ref{cov.xx}, \\
-& \;\;\;\;\; Cov(X,X) = Var(X) \\
-& = Var(X) + 2 Var(X) + Var(X)   \\
-& = 4 Var(X)
-\end{align*}
-\begin{align}
-Cov[X,Y] & = E(XY) - E(X)E(Y) \nonumber \\
-Cov[X,X] & = E(XX) - E(X)E(X) \nonumber \\
-& = E(X^2) - E(X)^2 \nonumber \\
-& = V(X) \label{cov.xx} \tag{3}
-\end{align}
-====== e.gs in R  ======
-R에서 이를 살펴보면
-<code>
-# variance theorem 4-1, 4-2
-# http://commres.net/wiki/variance_theorem
-# need a function, rnorm2
-rnorm2 <- function(n,mean,sd) { mean+sd*scale(rnorm(n)) }
-m <- 0
-v <- 1
-n <- 10000
-set.seed(1)
-x1 <- rnorm2(n, m, sqrt(v))
-x2 <- rnorm2(n, m, sqrt(v))
-x3 <- rnorm2(n, m, sqrt(v))
-m.x1 <- mean(x1)
-m.x2 <- mean(x2)
-m.x3 <- mean(x3)
-m.x1
-m.x2
-m.x3
-v.x1 <- var(x1)
-v.x2 <- var(x2)
-v.x3 <- var(x3)
-v.x1
-v.x2
-v.x3
-v.12 <- var(x1 + x2)
-v.12
-######################################
-## v.12 should be near var(x1)+var(x2)
-######################################
-## 정확히 2*v가 아닌 이유는 x1, x2가
-## 아주 약간은 (random하게) dependent하기 때문
-##(혹은 상관관계가 있기 때문)
-## theorem 5-1 에서
-## var(x1+x2) = var(x1)+var(x2)+ (2*cov(x1,x2))
-cov.x1x2 <- cov(x1,x2)
-var(x1 + x2)
-var(x1) + var(x2) + (2*cov.x1x2)
-# theorem 5-2 도 확인
-var(x1 - x2)
-var(x1) + var(x2) - (2 * cov.x1x2)
-# only when x1, x2 are independent (orthogonal)
-# var(x1+x2) == var(x1) + var(x2)
-########################################
-## 그리고 동일한 (독립적이지 않은) 집합 X1에 대해서는
-v.11 <- var(x1 + x1)
-# var(2*x1) = 2^2 var(X1)
-v.11
-</code>